From 25af29fecb825465233a9bbe537e9a14a7d1f350 Mon Sep 17 00:00:00 2001 From: Mathilde Miroir Date: Thu, 12 Sep 2024 13:16:17 -0400 Subject: [PATCH 1/7] Create MMiroir - P1.md --- projects/project-1/assignment/MMiroir - P1.md | 151 ++++++++++++++++++ 1 file changed, 151 insertions(+) create mode 100644 projects/project-1/assignment/MMiroir - P1.md diff --git a/projects/project-1/assignment/MMiroir - P1.md b/projects/project-1/assignment/MMiroir - P1.md new file mode 100644 index 0000000..0b756f8 --- /dev/null +++ b/projects/project-1/assignment/MMiroir - P1.md @@ -0,0 +1,151 @@ +# Project 1: OWL Cheat Sheet + +## Note +My results for assignment 2 vary significantly from the example chart provided. I believe I set everything up correctly but only achieved a single successful combination. Still learning how to write axioms/arguments... + +## Assignment Part 1 +### Role Constraint Combinations** +| | Funct. | iFunct. | Trans. | Symm. | Asymm. | Ref. |Irref. | +|-------- |-------- |---------|--------|------- |-------- |----- |-------- | +| **Funct.** |- | OK |XNS| OK | OK | OK | OK | +| **iFunct.**|OK | - |XNS| OK | OK | OK | OK | +| **Trans.** |XNS| XNS| - | OK |XNS | OK |XNS | +| **Symm.** |OK | OK | OK | - |XUNSAT| OK | OK | +| **Asymm.** |OK | OK |XNS|XUNSAT| - |XUNSAT| OK | +| **Ref.** |OK | OK | OK | OK |XUNSAT| - |XUNSAT| +| **Irref.** | OK | OK |XNS| OK | OK |XUNSAT| - | + +**_provided by the assignment instructions_ + +For example, to explain why `R` cannot be both asymmetric and symmetric, your explanation may take the form: +- Suppose `R` is both symmetric and asymmetric. Then by symmetry for any x and y, if x`R`y it follows that y`R`x. However, by asymmetry it also follows that it is not the case that y`R`x. Hence, `R` cannot be both symmetric and asymmetric. +Similarly, to explain why `R` cannot be both transitive and inverse functional. + +#### Transitive + Functional -> NS +- Suppose `R` is both transitive and functional. By transitivity for any x and y, if x`R`y and y`R`z, it follows that x`R`z. By functionality, if x`R`y and x`R`z, then it follows that y=z. + +#### Transitive + Inverse Functional -> NS +- Suppose `R` is both transitive and inverse functional. By transitivity for any x and y, if x`R`y and y`R`z, it follows that x`R`z. By inverse functional, if xRy and zRy, it follows that x=z. + + +#### Assymetric + Transitive -> NS +- Suppose `R` is both assymetric and transitive. By assymetry, if xRy, then it is not the case that yRx. By transitivity, if xRy and yRz, then xRz. + +#### Assymetric + Symmetric -> UNSAT +- _Explained in the assignment instructions_ : Suppose `R` is both symmetric and asymmetric. Then by symmetry for any x and y, if x`R`y it follows that y`R`x. However, by asymmetry it also follows that it is not the case that y`R`x. Hence, `R` cannot be both symmetric and asymmetric. + + +#### Assymetric + Reflexive -> UNSAT +- Suppose `R` is both assymetric and reflexive. By assymetry, f xRy, then it is not the case that yRx. By reflexivity, xRx. + +#### Reflexive + Irreflexive -> UNSAT +- Suppose `R` is both reflexive and irreflexive. Then by reflexivity for any x, x`R`x. However by irreflexivity, it also follows that it is not the case that xRx. Hence, `R` cannot be both reflexive and irreflexive. Similary for the case of why `R` cannot be both symmetric and assymetric. +xRx +It is not the case that xRx + +#### Irreflexive + Transitive -> NS +- Suppose `R` is both irreflexive and transitive. By irreflexivity, it is not the case that xRx. By transitivity, for any x and y, if x`R`y and y`R`z, it follows that x`R`z. + +## Assignment Part 2 + +### Variable Constraints + +- `B` owl:subPropertyOf of `A` +- `Bi` owl:subPropertyOf of `Ai` +- `Ai` is the inverse of `A` +- `Bi` is the inverse of `B` + +### Trials for Overlapping Transitive and Irreflexive Properties +| | A Trans | B Trans | Ai Trans | Bi Trans | +|------------|---------|---------|----------|----------| +| **A Irr** | N1 | N2 | N3 | N4 | +| **B Irr** | N5 | N6 | N7 | N8 | +| **Ai Irr** | N9 | N10| N11 | N12 | +| **Bi Irr** | Y13| N14| Y15 | N16 | + +*_N denotes an error while Y denotes a successful run_ + +**_Subscripts denote corresponding explantion below_ + +### [Object Property Expression Axioms](https://www.w3.org/TR/owl2-direct-semantics/) + +SubObjectPropertyOf( OPE1 OPE2 ) +- (OPE1)OP ⊆ (OPE2)OP + +InverseObjectProperties( OPE1 OPE2 ) +- (OPE1)OP = { ( x , y ) | ( y , x ) ∈ (OPE2)OP } + +IrreflexiveObjectProperty( OPE ) +- ∀ x : x ∈ ΔI implies ( x , x ) ∉ (OPE)OP + +TransitiveObjectProperty( OPE ) +- ∀ x , y , z : ( x , y ) ∈ (OPE)OP and ( y , z ) ∈ (OPE)OP imply ( x , z ) ∈ (OPE)OP + + +### Explanations (IN PROGRESS) + +- (1) + +- (2) + +- (3) + +- (4) + +- (5) + +- (6) + +- (7) + +- (8) + +- (9) + +- (10) + +- (11) + +- (12) + +- (13) + +- (14) + +- (15) + +- (16) + +### Sentences +- (1) Suppose `A` is both irreflexive and transitive. Because `A` is irreflexive, it is not the case for any `x` that `` is in `A`. Because `A` is transitive, if `` and ``, then ``. + +- (2) Suppose `A` is irreflexive and `B` is transitive. Because `A` is irreflexive, it is not the case for any `x` that `` is in `A`. Because `B` is transitive, if `` and ``, then ``. + +- (3) Suppose `A` is irreflexive and its inverse , `Ai`, is transitive. Because `A` is irreflexive, it is not the case for any `x` that `` is in `A`. Because `Ai` is transitive, if `` and ``, then ``. + +- (4) Suppose `A` is irreflexive and `Bi` is transitive. Because `A` is irreflexive, it is not the case for any `x` that `` is in `A`. Because `Bi` is transitive, if `` and ``, then ``. + +- (5) Suppose `B` is irreflexive and `A` is transitive. Because `A` is irreflexive, it is not the case for any `x` that `` is in `A`. Because `A` is transitive, if `` and ``, then ``. + +- (6) Suppose `B` is both irreflexive and transitive. Because `B` is irreflexive, it is not the case for any `x` that `` is in `B`. Because `B` is transitive, if `` and ``, then ``. + +- (7) Suppose `B` is irreflexive and `Ai` is transitive. Because `B` is irreflexive, it is not the case for any `x` that `` is in `B`. Because `Ai` is transitive, if `` and ``, then ``. + +- (8) Suppose `B` is irreflexive and its inverse , `Bi`, is transitive. Because `B` is irreflexive, it is not the case for any `x` that `` is in `B`. Because `Bi` is transitive, if `` and ``, then ``. + +- (9) Suppose `Ai` is irreflexive and its inverse , `A`, is transitive. Because `Ai` is irreflexive, it is not the case for any `x` that `` is in `Ai`. Because `A` is transitive, if `` and ``, then ``. + +- (10) Suppose `Ai` is irreflexive and `B` is transitive. Because `Ai` is irreflexive, it is not the case for any `x` that `` is in `Ai`. Because `B` is transitive, if `` and ``, then ``. + +- (11) Suppose `Ai` is both irreflexive and transitive. Because `Ai` is irreflexive, it is not the case for any `x` that `` is in `Ai`. Because `Ai` is transitive, if `` and ``, then ``. + +- (12) Suppose `Ai` is irreflexive and `Bi` is transitive. Because `Ai` is irreflexive, it is not the case for any `x` that `` is in `Ai`. Because `Bi` is transitive, if `` and ``, then ``. + +- (13) Suppose `Bi` is irreflexive and `A` is transitive. Because `Bi` is irreflexive, it is not the case for any `x` that `` is in `Bi`. Because `A` is transitive, if `` and ``, then ``. + +- (14) Suppose `Bi` is irreflexive and its inverse , `B`, is transitive. Because `Bi` is irreflexive, it is not the case for any `x` that `` is in `Bi`. Because `B` is transitive, if `` and ``, then ``. + +- (15) Suppose `Bi` is irreflexive and `Ai` is transitive. Because `Bi` is irreflexive, it is not the case for any `x` that `` is in `Bi`. Because `Ai` is transitive, if `` and ``, then ``. + +- (16) Suppose `Bi` is both irreflexive and transitive. Because `Bi` is irreflexive, it is not the case for any `x` that `` is in `Bi`. Because `Bi` is transitive, if `` and ``, then ``. + From a574bcc49a863289e2683b0cecd704d70ccd3a0b Mon Sep 17 00:00:00 2001 From: Mathilde Miroir Date: Mon, 16 Sep 2024 18:43:43 -0400 Subject: [PATCH 2/7] Update MMiroir - P1.md --- projects/project-1/assignment/MMiroir - P1.md | 49 ++++++++++++++----- 1 file changed, 36 insertions(+), 13 deletions(-) diff --git a/projects/project-1/assignment/MMiroir - P1.md b/projects/project-1/assignment/MMiroir - P1.md index 0b756f8..c3ce49b 100644 --- a/projects/project-1/assignment/MMiroir - P1.md +++ b/projects/project-1/assignment/MMiroir - P1.md @@ -5,21 +5,44 @@ My results for assignment 2 vary significantly from the example chart provided. ## Assignment Part 1 ### Role Constraint Combinations** -| | Funct. | iFunct. | Trans. | Symm. | Asymm. | Ref. |Irref. | -|-------- |-------- |---------|--------|------- |-------- |----- |-------- | -| **Funct.** |- | OK |XNS| OK | OK | OK | OK | -| **iFunct.**|OK | - |XNS| OK | OK | OK | OK | -| **Trans.** |XNS| XNS| - | OK |XNS | OK |XNS | -| **Symm.** |OK | OK | OK | - |XUNSAT| OK | OK | -| **Asymm.** |OK | OK |XNS|XUNSAT| - |XUNSAT| OK | -| **Ref.** |OK | OK | OK | OK |XUNSAT| - |XUNSAT| -| **Irref.** | OK | OK |XNS| OK | OK |XUNSAT| - | +| | Funct. | iFunct.| Trans. | Symm. | Asymm. | Ref. |Irref. | +|-------- |--------|--------|--------|------- |-------- |----- |-------- | +| **Funct.** |- | OK |XNS| OK | OK | OK | OK | +| **iFunct.**|OK | - |XNS| OK | OK | OK | OK | +| **Trans.** |XNS|XNS| - | OK |XNS | OK |XNS | +| **Symm.** |OK | OK | OK | - |XUNSAT| OK | OK | +| **Asymm.** |OK | OK |XNS|XUNSAT| - |XUNSAT| OK | +| **Ref.** |OK | OK | OK | OK |XUNSAT| - |XUNSAT| +| **Irref.** | OK | OK |XNS| OK | OK |XUNSAT| - | **_provided by the assignment instructions_ -For example, to explain why `R` cannot be both asymmetric and symmetric, your explanation may take the form: -- Suppose `R` is both symmetric and asymmetric. Then by symmetry for any x and y, if x`R`y it follows that y`R`x. However, by asymmetry it also follows that it is not the case that y`R`x. Hence, `R` cannot be both symmetric and asymmetric. -Similarly, to explain why `R` cannot be both transitive and inverse functional. +### [Object Property Expression Axioms](https://www.w3.org/TR/owl2-direct-semantics/) + + FunctionalObjectProperty( OPE ) +- ∀ x , y1 , y2 : ( x , y1 ) ∈ (OPE)OP and ( x , y2 ) ∈ (OPE)OP imply y1 = y2 + +InverseFunctionalObjectProperty( OPE ) +- ∀ x1 , x2 , y : ( x1 , y ) ∈ (OPE)OP and ( x2 , y ) ∈ (OPE)OP imply x1 = x2 + +TransitiveObjectProperty( OPE ) = +- ∀ x , y , z : ( x , y ) ∈ (OPE)OP and ( y , z ) ∈ (OPE)OP imply ( x , z ) ∈ (OPE)OP + +SymmetricObjectProperty( OPE ) +- ∀ x , y : ( x , y ) ∈ (OPE)OP implies ( y , x ) ∈ (OPE)OP + + AsymmetricObjectProperty( OPE ) +- ∀ x , y : ( x , y ) ∈ (OPE)OP implies ( y , x ) ∉ (OPE)OP + + ReflexiveObjectProperty( OPE ) +- ∀ x : x ∈ ΔI implies ( x , x ) ∈ (OPE)OP + + IrreflexiveObjectProperty( OPE ) +- ∀ x : x ∈ ΔI implies ( x , x ) ∉ (OPE)OP + +### Explanations + +_EXAMPLE GIVEN: To explain why `R` cannot be both asymmetric and symmetric, your explanation may take the form: Suppose `R` is both symmetric and asymmetric. Then by symmetry for any x and y, if x`R`y it follows that y`R`x. However, by asymmetry it also follows that it is not the case that y`R`x. Hence, `R` cannot be both symmetric and asymmetric. Similarly, to explain why `R` cannot be both transitive and inverse functional._ #### Transitive + Functional -> NS - Suppose `R` is both transitive and functional. By transitivity for any x and y, if x`R`y and y`R`z, it follows that x`R`z. By functionality, if x`R`y and x`R`z, then it follows that y=z. @@ -59,7 +82,7 @@ It is not the case that xRx | | A Trans | B Trans | Ai Trans | Bi Trans | |------------|---------|---------|----------|----------| | **A Irr** | N1 | N2 | N3 | N4 | -| **B Irr** | N5 | N6 | N7 | N8 | +| **B Irr** | Y5 | N6 | Y7 | N8 | | **Ai Irr** | N9 | N10| N11 | N12 | | **Bi Irr** | Y13| N14| Y15 | N16 | From b4f123ef82dadc656e64dcc3578a89916b48df8a Mon Sep 17 00:00:00 2001 From: Mathilde Miroir Date: Tue, 17 Sep 2024 20:45:46 -0400 Subject: [PATCH 3/7] Update MMiroir - P1.md --- projects/project-1/assignment/MMiroir - P1.md | 444 ++++++++++++++---- 1 file changed, 364 insertions(+), 80 deletions(-) diff --git a/projects/project-1/assignment/MMiroir - P1.md b/projects/project-1/assignment/MMiroir - P1.md index c3ce49b..34c8c4a 100644 --- a/projects/project-1/assignment/MMiroir - P1.md +++ b/projects/project-1/assignment/MMiroir - P1.md @@ -1,7 +1,5 @@ # Project 1: OWL Cheat Sheet - -## Note -My results for assignment 2 vary significantly from the example chart provided. I believe I set everything up correctly but only achieved a single successful combination. Still learning how to write axioms/arguments... +Collaborative attempt by Jay Bittner, Tim Coleman, and Mathilde Miroir ## Assignment Part 1 ### Role Constraint Combinations** @@ -17,60 +15,42 @@ My results for assignment 2 vary significantly from the example chart provided. **_provided by the assignment instructions_ -### [Object Property Expression Axioms](https://www.w3.org/TR/owl2-direct-semantics/) - - FunctionalObjectProperty( OPE ) -- ∀ x , y1 , y2 : ( x , y1 ) ∈ (OPE)OP and ( x , y2 ) ∈ (OPE)OP imply y1 = y2 - -InverseFunctionalObjectProperty( OPE ) -- ∀ x1 , x2 , y : ( x1 , y ) ∈ (OPE)OP and ( x2 , y ) ∈ (OPE)OP imply x1 = x2 - -TransitiveObjectProperty( OPE ) = -- ∀ x , y , z : ( x , y ) ∈ (OPE)OP and ( y , z ) ∈ (OPE)OP imply ( x , z ) ∈ (OPE)OP - -SymmetricObjectProperty( OPE ) -- ∀ x , y : ( x , y ) ∈ (OPE)OP implies ( y , x ) ∈ (OPE)OP - - AsymmetricObjectProperty( OPE ) -- ∀ x , y : ( x , y ) ∈ (OPE)OP implies ( y , x ) ∉ (OPE)OP - - ReflexiveObjectProperty( OPE ) -- ∀ x : x ∈ ΔI implies ( x , x ) ∈ (OPE)OP - - IrreflexiveObjectProperty( OPE ) -- ∀ x : x ∈ ΔI implies ( x , x ) ∉ (OPE)OP - ### Explanations -_EXAMPLE GIVEN: To explain why `R` cannot be both asymmetric and symmetric, your explanation may take the form: Suppose `R` is both symmetric and asymmetric. Then by symmetry for any x and y, if x`R`y it follows that y`R`x. However, by asymmetry it also follows that it is not the case that y`R`x. Hence, `R` cannot be both symmetric and asymmetric. Similarly, to explain why `R` cannot be both transitive and inverse functional._ +The following example, provided in the assignment, serves as the guide for which all following explanations adhere to: +- To explain why `R` cannot be both asymmetric and symmetric, your explanation may take the form: Suppose `R` is both symmetric and asymmetric. Then by symmetry for any x and y, if x`R`y it follows that y`R`x. However, by asymmetry it also follows that it is not the case that y`R`x. Hence, `R` cannot be both symmetric and asymmetric. Similarly, to explain why `R` cannot be both transitive and inverse functional. -#### Transitive + Functional -> NS +#### (1) Transitive + Functional (NS) - Suppose `R` is both transitive and functional. By transitivity for any x and y, if x`R`y and y`R`z, it follows that x`R`z. By functionality, if x`R`y and x`R`z, then it follows that y=z. -#### Transitive + Inverse Functional -> NS +#### (2) Transitive + Inverse Functional (NS) - Suppose `R` is both transitive and inverse functional. By transitivity for any x and y, if x`R`y and y`R`z, it follows that x`R`z. By inverse functional, if xRy and zRy, it follows that x=z. -#### Assymetric + Transitive -> NS +#### (3) Assymetric + Transitive (NS) - Suppose `R` is both assymetric and transitive. By assymetry, if xRy, then it is not the case that yRx. By transitivity, if xRy and yRz, then xRz. -#### Assymetric + Symmetric -> UNSAT +#### (4) Assymetric + Symmetric (UNSAT) - _Explained in the assignment instructions_ : Suppose `R` is both symmetric and asymmetric. Then by symmetry for any x and y, if x`R`y it follows that y`R`x. However, by asymmetry it also follows that it is not the case that y`R`x. Hence, `R` cannot be both symmetric and asymmetric. - -#### Assymetric + Reflexive -> UNSAT +#### (5) Assymetric + Reflexive (UNSAT) - Suppose `R` is both assymetric and reflexive. By assymetry, f xRy, then it is not the case that yRx. By reflexivity, xRx. -#### Reflexive + Irreflexive -> UNSAT +#### (6)Reflexive + Irreflexive (UNSAT) - Suppose `R` is both reflexive and irreflexive. Then by reflexivity for any x, x`R`x. However by irreflexivity, it also follows that it is not the case that xRx. Hence, `R` cannot be both reflexive and irreflexive. Similary for the case of why `R` cannot be both symmetric and assymetric. -xRx -It is not the case that xRx -#### Irreflexive + Transitive -> NS + +#### (7) Irreflexive + Transitive (NS) - Suppose `R` is both irreflexive and transitive. By irreflexivity, it is not the case that xRx. By transitivity, for any x and y, if x`R`y and y`R`z, it follows that x`R`z. ## Assignment Part 2 +### Note + +- reflexive vs irreflexive (low priority) + +- symm vs asymm (low priority) + ### Variable Constraints - `B` owl:subPropertyOf of `A` @@ -78,7 +58,7 @@ It is not the case that xRx - `Ai` is the inverse of `A` - `Bi` is the inverse of `B` -### Trials for Overlapping Transitive and Irreflexive Properties +### (A) Transitive and Irreflexive Trials | | A Trans | B Trans | Ai Trans | Bi Trans | |------------|---------|---------|----------|----------| | **A Irr** | N1 | N2 | N3 | N4 | @@ -90,85 +70,389 @@ It is not the case that xRx **_Subscripts denote corresponding explantion below_ -### [Object Property Expression Axioms](https://www.w3.org/TR/owl2-direct-semantics/) +### (B) Transitive and Functional Trials (j's previous contribution - replace A with x or y) + +| | A Trans | B Trans | Ai Trans | Bi Trans | +|-------------|---------|---------|----------|----------| +| **A Func** | A1 | A2 | A3 | A4 | +| **B Func** | A5 | A6 | A7 | A8 | +| **Ai Func** | A9 | A10| A11 | A12 | +| **Bi Func** | A13| A14| A15 | A16 | + +*_N denotes an error while Y denotes a successful run_ + +**_Subscripts denote corresponding explantion below_ + + +### (C) trans vs infunc??? same as previous? +| | A Trans | B Trans | Ai Trans | Bi Trans | +|--------------|---------|---------|----------|----------| +| **A iFunc** | A1 | A2 | A3 | A4 | +| **B iFunc** | A5 | A6 | A7 | A8 | +| **Ai iFunc** | A9 | A10| A11 | A12 | +| **Bi iFUnc** | A13| A14| A15 | A16 | + +*_N denotes an error while Y denotes a successful run_ + +**_Subscripts denote corresponding explantion below_ + +### (D) Transitive and Asymmetric Trials (Tim's HW - replace A with x or y) +| | A Trans | B Trans | Ai Trans | Bi Trans | +|--------------|---------|---------|----------|----------| +| **A Asymm** | A1 | A2 | A3 | A4 | +| **B Asymm** | A5 | A6 | A7 | A8 | +| **Ai Asymm** | A9 | A10| A11 | A12 | +| **Bi Asymm** | A13| A14| A15 | A16 | -SubObjectPropertyOf( OPE1 OPE2 ) -- (OPE1)OP ⊆ (OPE2)OP +*_N denotes an error while Y denotes a successful run_ + +**_Subscripts denote corresponding explantion below_ -InverseObjectProperties( OPE1 OPE2 ) -- (OPE1)OP = { ( x , y ) | ( y , x ) ∈ (OPE2)OP } -IrreflexiveObjectProperty( OPE ) -- ∀ x : x ∈ ΔI implies ( x , x ) ∉ (OPE)OP +### (E) Transitive and Asymmetric Trials (J's HW - replace A with x or y) +| |A Reflex |B Reflex |Ai Reflex |Bi Reflex | +|--------------|---------|---------|----------|----------| +| **A Asymm** | A1 | A2 | A3 | A4 | +| **B Asymm** | A5 | A6 | A7 | A8 | +| **Ai Asymm** | A9 | A10| A11 | A12 | +| **Bi Asymm** | A13| A14| A15 | A16 | + +*_N denotes an error while Y denotes a successful run_ + +**_Subscripts denote corresponding explantion below_ -TransitiveObjectProperty( OPE ) -- ∀ x , y , z : ( x , y ) ∈ (OPE)OP and ( y , z ) ∈ (OPE)OP imply ( x , z ) ∈ (OPE)OP ### Explanations (IN PROGRESS) -- (1) +#### (A) Transitive and Irreflexive Explanations +- (A-1) Suppose `A` is both irreflexive and transitive. Because `A` is irreflexive, it is not the case for any `x` that `` is in `A`. Because `A` is transitive, if `` and ``, then ``. + +- (A-2) Suppose `A` is irreflexive and `B` is transitive. Because `A` is irreflexive, it is not the case for any `x` that `` is in `A`. Because `B` is transitive, if `` and ``, then ``. + +- (A-3) Suppose `A` is irreflexive and its inverse , `Ai`, is transitive. Because `A` is irreflexive, it is not the case for any `x` that `` is in `A`. Because `Ai` is transitive, if `` and ``, then ``. + +- (A-4) Suppose `A` is irreflexive and `Bi` is transitive. Because `A` is irreflexive, it is not the case for any `x` that `` is in `A`. Because `Bi` is transitive, if `` and ``, then ``. + +- (A-5) Suppose `B` is irreflexive and `A` is transitive. Because `A` is irreflexive, it is not the case for any `x` that `` is in `A`. Because `A` is transitive, if `` and ``, then ``. + +- (A-6) Suppose `B` is both irreflexive and transitive. Because `B` is irreflexive, it is not the case for any `x` that `` is in `B`. Because `B` is transitive, if `` and ``, then ``. + +- (A-7) Suppose `B` is irreflexive and `Ai` is transitive. Because `B` is irreflexive, it is not the case for any `x` that `` is in `B`. Because `Ai` is transitive, if `` and ``, then ``. + +- (A-8) Suppose `B` is irreflexive and its inverse , `Bi`, is transitive. Because `B` is irreflexive, it is not the case for any `x` that `` is in `B`. Because `Bi` is transitive, if `` and ``, then ``. + +- (A-9) Suppose `Ai` is irreflexive and its inverse , `A`, is transitive. Because `Ai` is irreflexive, it is not the case for any `x` that `` is in `Ai`. Because `A` is transitive, if `` and ``, then ``. + +- (A-10) Suppose `Ai` is irreflexive and `B` is transitive. Because `Ai` is irreflexive, it is not the case for any `x` that `` is in `Ai`. Because `B` is transitive, if `` and ``, then ``. + +- (A-11) Suppose `Ai` is both irreflexive and transitive. Because `Ai` is irreflexive, it is not the case for any `x` that `` is in `Ai`. Because `Ai` is transitive, if `` and ``, then ``. + +- (A-12) Suppose `Ai` is irreflexive and `Bi` is transitive. Because `Ai` is irreflexive, it is not the case for any `x` that `` is in `Ai`. Because `Bi` is transitive, if `` and ``, then ``. + +- (A-13) Suppose `Bi` is irreflexive and `A` is transitive. Because `Bi` is irreflexive, it is not the case for any `x` that `` is in `Bi`. Because `A` is transitive, if `` and ``, then ``. + +- (A-14) Suppose `Bi` is irreflexive and its inverse , `B`, is transitive. Because `Bi` is irreflexive, it is not the case for any `x` that `` is in `Bi`. Because `B` is transitive, if `` and ``, then ``. + +- (A-15) Suppose `Bi` is irreflexive and `Ai` is transitive. Because `Bi` is irreflexive, it is not the case for any `x` that `` is in `Bi`. Because `Ai` is transitive, if `` and ``, then ``. + +- (A-16) Suppose `Bi` is both irreflexive and transitive. Because `Bi` is irreflexive, it is not the case for any `x` that `` is in `Bi`. Because `Bi` is transitive, if `` and ``, then ``. + + + +#### (B) Transitive and Functional + +- (B-1) Suppose `A` is both transitive and functional. + +Because `A` is transitive, it is + +Because `A` is functional, if + +- (B-2) Suppose `A` is transitive and `B` is functional. + +Because `A` is transitive, + +Because `B` is functional, if + +- (B-3) Suppose `A` is transitive and its inverse , `Ai`, is functional. + +Because `A` is transitive, it is + +Because `Ai` is functional, if + +- (B-4) Suppose `A` is transitive and `Bi` is functional. + +Because `A` is transitive, + +Because `Bi` is functional, + +- (B-5) Suppose `B` is transitive and `A` is functional. + +Because `A` is transitive, + +Because `A` is functional, + +- (B-6) Suppose `B` is both transitive and functional. + +Because `B` is transitive, + +Because `B` is functional, + +- (B-7) Suppose `B` is transitive and `Ai` is functional. + +Because `B` is transitive, + +Because `Ai` is functional, + +- (B-8) Suppose `B` is transitive and its inverse , `Bi`, is functional. + +Because `B` is transitive, + +Because `Bi` is functional, + +- (B-9) Suppose `Ai` is transitive and its inverse , `A`, is functional. + +Because `Ai` is transitive, + +Because `A` is functional, + +- (B-10) Suppose `Ai` is transitive and `B` is functional. + +Because `Ai` is transitive, + +Because `B` is functional, + +- (B-11) Suppose `Ai` is both transitive and functional. + +Because `Ai` is transitive, + +Because `Ai` is functional, + +- (B-12) Suppose `Ai` is transitive and `Bi` is functional. + +Because `Ai` is transitive, + +Because `Bi` is functional, + +- (B-13) Suppose `Bi` is transitive and `A` is functional. + +Because `Bi` is transitive, + +Because `A` is functional, + +- (B-14) Suppose `Bi` is transitive and its inverse , `B`, is functional. + +Because `Bi` is transitive, + +Because `B` is functional, + +- (B-15) Suppose `Bi` is transitive and `Ai` is functional. + +Because `Bi` is transitive, + +Because `Ai` is functional, + +- (B-16) Suppose `Bi` is both transitive and functional. + +Because `Bi` is transitive, + +Because `Bi` is functional, + +#### (C) Transitive and Inverse Functional + +#### (D) Transitive and Asymmetric + +- (D-1) Suppose `A` is both transitive and functional. + +Because `A` is transitive, it is + +Because `A` is functional, if + +- (D-2) Suppose `A` is transitive and `B` is functional. + +Because `A` is transitive, + +Because `B` is functional, if + +- (D-3) Suppose `A` is transitive and its inverse , `Ai`, is functional. + +Because `A` is transitive, it is + +Because `Ai` is functional, if + +- (D-4) Suppose `A` is transitive and `Bi` is functional. + +Because `A` is transitive, + +Because `Bi` is functional, + +- (D-5) Suppose `B` is transitive and `A` is functional. + +Because `A` is transitive, + +Because `A` is functional, + +- (D-6) Suppose `B` is both transitive and functional. + +Because `B` is transitive, + +Because `B` is functional, + +- (D-7) Suppose `B` is transitive and `Ai` is functional. + +Because `B` is transitive, + +Because `Ai` is functional, + +- (D-8) Suppose `B` is transitive and its inverse , `Bi`, is functional. + +Because `B` is transitive, + +Because `Bi` is functional, + +- (D-9) Suppose `Ai` is transitive and its inverse , `A`, is functional. + +Because `Ai` is transitive, + +Because `A` is functional, + +- (D-10) Suppose `Ai` is transitive and `B` is functional. + +Because `Ai` is transitive, + +Because `B` is functional, + +- (D-11) Suppose `Ai` is both transitive and functional. + +Because `Ai` is transitive, + +Because `Ai` is functional, + +- (D-12) Suppose `Ai` is transitive and `Bi` is functional. + +Because `Ai` is transitive, + +Because `Bi` is functional, + +- (D-13) Suppose `Bi` is transitive and `A` is functional. + +Because `Bi` is transitive, + +Because `A` is functional, + +- (D-14) Suppose `Bi` is transitive and its inverse , `B`, is functional. + +Because `Bi` is transitive, + +Because `B` is functional, + +- (D-15) Suppose `Bi` is transitive and `Ai` is functional. + +Because `Bi` is transitive, + +Because `Ai` is functional, + +- (D-16) Suppose `Bi` is both transitive and functional. + +Because `Bi` is transitive, + +Because `Bi` is functional, + + +#### (E) Transitive and Asymmetric + +- (E-1) Suppose `A` is both transitive and asymmetric. + +Because `A` is transitive, it is + +Because `A` is asymmetric, if + +- (E-2) Suppose `A` is transitive and `B` is functional. + +Because `A` is transitive, + +Because `B` is asymmetric, if + +- (E-3) Suppose `A` is transitive and its inverse , `Ai`, is functional. + +Because `A` is transitive, it is + +Because `Ai` is asymmetric, if + +- (E-4) Suppose `A` is transitive and `Bi` is functional. + +Because `A` is transitive, + +Because `Bi` is asymmetric, + +- (E-5) Suppose `B` is transitive and `A` is functional. + +Because `A` is transitive, + +Because `A` is asymmetric, + +- (E-6) Suppose `B` is both transitive and functional. -- (2) +Because `B` is transitive, -- (3) +Because `B` is asymmetric, -- (4) +- (E-7) Suppose `B` is transitive and `Ai` is functional. -- (5) +Because `B` is transitive, -- (6) +Because `Ai` is asymmetric, -- (7) +- (E-8) Suppose `B` is transitive and its inverse , `Bi`, is functional. -- (8) +Because `B` is transitive, -- (9) +Because `Bi` is asymmetric, -- (10) +- (E-9) Suppose `Ai` is transitive and its inverse , `A`, is functional. -- (11) +Because `Ai` is transitive, -- (12) +Because `A` is asymmetric, -- (13) +- (E-10) Suppose `Ai` is transitive and `B` is functional. -- (14) +Because `Ai` is transitive, -- (15) +Because `B` is asymmetric, -- (16) +- (E-11) Suppose `Ai` is both transitive and functional. -### Sentences -- (1) Suppose `A` is both irreflexive and transitive. Because `A` is irreflexive, it is not the case for any `x` that `` is in `A`. Because `A` is transitive, if `` and ``, then ``. +Because `Ai` is transitive, -- (2) Suppose `A` is irreflexive and `B` is transitive. Because `A` is irreflexive, it is not the case for any `x` that `` is in `A`. Because `B` is transitive, if `` and ``, then ``. +Because `Ai` is asymmetric, -- (3) Suppose `A` is irreflexive and its inverse , `Ai`, is transitive. Because `A` is irreflexive, it is not the case for any `x` that `` is in `A`. Because `Ai` is transitive, if `` and ``, then ``. +- (E-12) Suppose `Ai` is transitive and `Bi` is functional. -- (4) Suppose `A` is irreflexive and `Bi` is transitive. Because `A` is irreflexive, it is not the case for any `x` that `` is in `A`. Because `Bi` is transitive, if `` and ``, then ``. +Because `Ai` is transitive, -- (5) Suppose `B` is irreflexive and `A` is transitive. Because `A` is irreflexive, it is not the case for any `x` that `` is in `A`. Because `A` is transitive, if `` and ``, then ``. +Because `Bi` is asymmetric, -- (6) Suppose `B` is both irreflexive and transitive. Because `B` is irreflexive, it is not the case for any `x` that `` is in `B`. Because `B` is transitive, if `` and ``, then ``. +- (E-13) Suppose `Bi` is transitive and `A` is functional. -- (7) Suppose `B` is irreflexive and `Ai` is transitive. Because `B` is irreflexive, it is not the case for any `x` that `` is in `B`. Because `Ai` is transitive, if `` and ``, then ``. +Because `Bi` is transitive, -- (8) Suppose `B` is irreflexive and its inverse , `Bi`, is transitive. Because `B` is irreflexive, it is not the case for any `x` that `` is in `B`. Because `Bi` is transitive, if `` and ``, then ``. +Because `A` is asymmetric, -- (9) Suppose `Ai` is irreflexive and its inverse , `A`, is transitive. Because `Ai` is irreflexive, it is not the case for any `x` that `` is in `Ai`. Because `A` is transitive, if `` and ``, then ``. +- (E-14) Suppose `Bi` is transitive and its inverse , `B`, is functional. -- (10) Suppose `Ai` is irreflexive and `B` is transitive. Because `Ai` is irreflexive, it is not the case for any `x` that `` is in `Ai`. Because `B` is transitive, if `` and ``, then ``. +Because `Bi` is transitive, -- (11) Suppose `Ai` is both irreflexive and transitive. Because `Ai` is irreflexive, it is not the case for any `x` that `` is in `Ai`. Because `Ai` is transitive, if `` and ``, then ``. +Because `B` is asymmetric, -- (12) Suppose `Ai` is irreflexive and `Bi` is transitive. Because `Ai` is irreflexive, it is not the case for any `x` that `` is in `Ai`. Because `Bi` is transitive, if `` and ``, then ``. +- (E-15) Suppose `Bi` is transitive and `Ai` is functional. -- (13) Suppose `Bi` is irreflexive and `A` is transitive. Because `Bi` is irreflexive, it is not the case for any `x` that `` is in `Bi`. Because `A` is transitive, if `` and ``, then ``. +Because `Bi` is transitive, -- (14) Suppose `Bi` is irreflexive and its inverse , `B`, is transitive. Because `Bi` is irreflexive, it is not the case for any `x` that `` is in `Bi`. Because `B` is transitive, if `` and ``, then ``. +Because `Ai` is asymmetric, -- (15) Suppose `Bi` is irreflexive and `Ai` is transitive. Because `Bi` is irreflexive, it is not the case for any `x` that `` is in `Bi`. Because `Ai` is transitive, if `` and ``, then ``. +- (E-16) Suppose `Bi` is both transitive and functional. -- (16) Suppose `Bi` is both irreflexive and transitive. Because `Bi` is irreflexive, it is not the case for any `x` that `` is in `Bi`. Because `Bi` is transitive, if `` and ``, then ``. +Because `Bi` is transitive, +Because `Bi` is asymmetric, \ No newline at end of file From 326b16c7ebd8ad91b074cbf0a596814bcaf92ef9 Mon Sep 17 00:00:00 2001 From: Mathilde Miroir Date: Wed, 18 Sep 2024 12:05:29 -0400 Subject: [PATCH 4/7] Update MMiroir - P1.md --- projects/project-1/assignment/MMiroir - P1.md | 456 +++++++----------- 1 file changed, 181 insertions(+), 275 deletions(-) diff --git a/projects/project-1/assignment/MMiroir - P1.md b/projects/project-1/assignment/MMiroir - P1.md index 34c8c4a..9431ac2 100644 --- a/projects/project-1/assignment/MMiroir - P1.md +++ b/projects/project-1/assignment/MMiroir - P1.md @@ -2,7 +2,24 @@ Collaborative attempt by Jay Bittner, Tim Coleman, and Mathilde Miroir ## Assignment Part 1 -### Role Constraint Combinations** + +### Table I. OWL2 DL Direct Semantics Role Constraints + +[OWL2 DL Direct Semantics](https://www.w3.org/TR/owl2-direct-semantics/) allows for constraints on object properties of the following sort, where `R` is an arbitrary object property: + +| **Constraint** | **Definition** | **Example** | +|------------------------|------------------------------------------------------------|-------------| +| **Functional** | If xRy and xRz, then y=z | If x *has birthdate* y and x *has birthdate* z, then y=z. | +| **Inverse Functional** | If xRy and zRy, then x=z | If x *has social security number* y and z *has social security number* y, then x=z. | +| **Transitive** | If xRy and yRz, then xRz | If x *is contained in* y and y *is contained in* z, then x *is contained in* z. | +| **Symmetric** | If xRy, then yRx | If x *is a friend of* y, then y *is a friend of* x. | +| **Asymmetric** | If xRy, then it is not the case that yRx | If x *is the parent of* y, then it is not the case that y *is the parent of* x. | +| **Reflexive** | xRx | x *is as tall as* itself. | +| **Irreflexive** | It is not the case that xRx | No x *is taller than* itself. | + + + +### Table II. Role Constraint Combinations*** | | Funct. | iFunct.| Trans. | Symm. | Asymm. | Ref. |Irref. | |-------- |--------|--------|--------|------- |-------- |----- |-------- | | **Funct.** |- | OK |XNS| OK | OK | OK | OK | @@ -13,43 +30,74 @@ Collaborative attempt by Jay Bittner, Tim Coleman, and Mathilde Miroir | **Ref.** |OK | OK | OK | OK |XUNSAT| - |XUNSAT| | **Irref.** | OK | OK |XNS| OK | OK |XUNSAT| - | -**_provided by the assignment instructions_ +*_NS denotes a combination creates an object property chain that may lead to undecidability_ + +**_UNSAT denotes a combination that is unsatisfiable_ + +***_Provided by the assignment instructions_ + +For the purposes of this assignment, the diagonal of the matrix in the above table (II) was ignored as these are not combinations of properties. Given that the matrix is also reflective across the diagonal, only side of the diagonal was observed and tested. Please note that this assignment does not explore Transitive & Inverse Functional (**C**) of the **NS** results or any of the **UNSAT** results. + +According to the table above (II), there are four property combinations which yielded an undecidable (**NS**) result: + +- (**A**) Irreflexive **&** Transitive + +- (**B**) Transitive **&** Functional + +- (**C**) Transitive **&** Inverse Functional + +- (**D**) Transitive **&** Assymetric + +There are three property combinations which yield an unsatifiable (**UNSAT**) result: + +- (**E**) Assymetric **&** Symmetric + +- (**F**)) Assymetric **&** Reflexive + +- (**G**) Reflexive **&** Irreflexive + ### Explanations The following example, provided in the assignment, serves as the guide for which all following explanations adhere to: - To explain why `R` cannot be both asymmetric and symmetric, your explanation may take the form: Suppose `R` is both symmetric and asymmetric. Then by symmetry for any x and y, if x`R`y it follows that y`R`x. However, by asymmetry it also follows that it is not the case that y`R`x. Hence, `R` cannot be both symmetric and asymmetric. Similarly, to explain why `R` cannot be both transitive and inverse functional. -#### (1) Transitive + Functional (NS) +#### (A) Transitive + Irreflexive (NS) +- Suppose `R` is both irreflexive and transitive. By irreflexivity, it is not the case that xRx. By transitivity, for any x and y, if x`R`y and y`R`z, it follows that x`R`z. + +ADD EXPLANATION HERE + +#### (B) Transitive + Functional (NS) - Suppose `R` is both transitive and functional. By transitivity for any x and y, if x`R`y and y`R`z, it follows that x`R`z. By functionality, if x`R`y and x`R`z, then it follows that y=z. -#### (2) Transitive + Inverse Functional (NS) +ADD EXPLANATION HERE + +#### (C) Transitive + Inverse Functional (NS) - Suppose `R` is both transitive and inverse functional. By transitivity for any x and y, if x`R`y and y`R`z, it follows that x`R`z. By inverse functional, if xRy and zRy, it follows that x=z. +ADD EXPLANATION HERE -#### (3) Assymetric + Transitive (NS) +#### (D) Transitive + Assymetric (NS) - Suppose `R` is both assymetric and transitive. By assymetry, if xRy, then it is not the case that yRx. By transitivity, if xRy and yRz, then xRz. -#### (4) Assymetric + Symmetric (UNSAT) -- _Explained in the assignment instructions_ : Suppose `R` is both symmetric and asymmetric. Then by symmetry for any x and y, if x`R`y it follows that y`R`x. However, by asymmetry it also follows that it is not the case that y`R`x. Hence, `R` cannot be both symmetric and asymmetric. +ADD EXPLANATION HERE -#### (5) Assymetric + Reflexive (UNSAT) -- Suppose `R` is both assymetric and reflexive. By assymetry, f xRy, then it is not the case that yRx. By reflexivity, xRx. - -#### (6)Reflexive + Irreflexive (UNSAT) -- Suppose `R` is both reflexive and irreflexive. Then by reflexivity for any x, x`R`x. However by irreflexivity, it also follows that it is not the case that xRx. Hence, `R` cannot be both reflexive and irreflexive. Similary for the case of why `R` cannot be both symmetric and assymetric. +#### (E) Assymetric + Symmetric (UNSAT) - _Explained in the assignment instructions_ +- Suppose `R` is both symmetric and asymmetric. Then by symmetry for any x and y, if x`R`y it follows that y`R`x. However, by asymmetry it also follows that it is not the case that y`R`x. Hence, `R` cannot be both symmetric and asymmetric. +ADD EXPLANATION HERE -#### (7) Irreflexive + Transitive (NS) -- Suppose `R` is both irreflexive and transitive. By irreflexivity, it is not the case that xRx. By transitivity, for any x and y, if x`R`y and y`R`z, it follows that x`R`z. +#### (F) Assymetric + Reflexive (UNSAT) +- Suppose `R` is both assymetric and reflexive. By assymetry, f xRy, then it is not the case that yRx. By reflexivity, xRx. -## Assignment Part 2 +ADD EXPLANATION HERE -### Note +#### (G) Reflexive + Irreflexive (UNSAT) +- Suppose `R` is both reflexive and irreflexive. Then by reflexivity for any x, x`R`x. However by irreflexivity, it also follows that it is not the case that xRx. Hence, `R` cannot be both reflexive and irreflexive. Similary for the case of why `R` cannot be both symmetric and assymetric. -- reflexive vs irreflexive (low priority) +ADD EXPLANATION HERE -- symm vs asymm (low priority) +## Assignment Part 2 ### Variable Constraints @@ -58,7 +106,7 @@ The following example, provided in the assignment, serves as the guide for which - `Ai` is the inverse of `A` - `Bi` is the inverse of `B` -### (A) Transitive and Irreflexive Trials +### Table III. (A) Transitive and Irreflexive Trials | | A Trans | B Trans | Ai Trans | Bi Trans | |------------|---------|---------|----------|----------| | **A Irr** | N1 | N2 | N3 | N4 | @@ -70,52 +118,41 @@ The following example, provided in the assignment, serves as the guide for which **_Subscripts denote corresponding explantion below_ -### (B) Transitive and Functional Trials (j's previous contribution - replace A with x or y) +### Table IV. (B) Transitive and Functional Trials (j's previous contribution) | | A Trans | B Trans | Ai Trans | Bi Trans | |-------------|---------|---------|----------|----------| -| **A Func** | A1 | A2 | A3 | A4 | -| **B Func** | A5 | A6 | A7 | A8 | -| **Ai Func** | A9 | A10| A11 | A12 | -| **Bi Func** | A13| A14| A15 | A16 | +| **A Func** | N1 | N2 | N3 | N4 | +| **B Func** | N5 | N6 | N7 | N8 | +| **Ai Func** | Y9 | Y10| N11 | N12 | +| **Bi Func** | Y13| Y14| N15 | N16 | *_N denotes an error while Y denotes a successful run_ **_Subscripts denote corresponding explantion below_ -### (C) trans vs infunc??? same as previous? -| | A Trans | B Trans | Ai Trans | Bi Trans | -|--------------|---------|---------|----------|----------| -| **A iFunc** | A1 | A2 | A3 | A4 | -| **B iFunc** | A5 | A6 | A7 | A8 | -| **Ai iFunc** | A9 | A10| A11 | A12 | -| **Bi iFUnc** | A13| A14| A15 | A16 | - -*_N denotes an error while Y denotes a successful run_ +### Table V. (C) trans vs infunc??? same as previous? -**_Subscripts denote corresponding explantion below_ - -### (D) Transitive and Asymmetric Trials (Tim's HW - replace A with x or y) | | A Trans | B Trans | Ai Trans | Bi Trans | |--------------|---------|---------|----------|----------| -| **A Asymm** | A1 | A2 | A3 | A4 | -| **B Asymm** | A5 | A6 | A7 | A8 | -| **Ai Asymm** | A9 | A10| A11 | A12 | -| **Bi Asymm** | A13| A14| A15 | A16 | +| **A iFunc** | -1 | -2 | -3 | -4 | +| **B iFunc** | -5 | -6 | -7 | -8 | +| **Ai iFunc** | -9 | -10| -11 | -12 | +| **Bi iFUnc** | -13| -14| -15 | -16 | *_N denotes an error while Y denotes a successful run_ **_Subscripts denote corresponding explantion below_ +### Table VI. (D) Transitive and Asymmetric Trials (Tim's HW) -### (E) Transitive and Asymmetric Trials (J's HW - replace A with x or y) -| |A Reflex |B Reflex |Ai Reflex |Bi Reflex | -|--------------|---------|---------|----------|----------| -| **A Asymm** | A1 | A2 | A3 | A4 | -| **B Asymm** | A5 | A6 | A7 | A8 | -| **Ai Asymm** | A9 | A10| A11 | A12 | -| **Bi Asymm** | A13| A14| A15 | A16 | +| | A Trans | B Trans | Ai Trans | Bi Trans | +|------------|---------|---------|----------|----------| +| **A Asym** | N1 | N2 | N3 | N4 | +| **B Asym** | N5 | N6 | N7 | N8 | +| **Ai Asym**| Y9 | Y10| N11 | N12 | +| **Bi Asym**| Y13| Y14| N15 | N16 | *_N denotes an error while Y denotes a successful run_ @@ -123,336 +160,205 @@ The following example, provided in the assignment, serves as the guide for which + ### Explanations (IN PROGRESS) #### (A) Transitive and Irreflexive Explanations - (A-1) Suppose `A` is both irreflexive and transitive. Because `A` is irreflexive, it is not the case for any `x` that `` is in `A`. Because `A` is transitive, if `` and ``, then ``. +ADD EXPLANATION HERE + - (A-2) Suppose `A` is irreflexive and `B` is transitive. Because `A` is irreflexive, it is not the case for any `x` that `` is in `A`. Because `B` is transitive, if `` and ``, then ``. +ADD EXPLANATION HERE + - (A-3) Suppose `A` is irreflexive and its inverse , `Ai`, is transitive. Because `A` is irreflexive, it is not the case for any `x` that `` is in `A`. Because `Ai` is transitive, if `` and ``, then ``. +ADD EXPLANATION HERE + - (A-4) Suppose `A` is irreflexive and `Bi` is transitive. Because `A` is irreflexive, it is not the case for any `x` that `` is in `A`. Because `Bi` is transitive, if `` and ``, then ``. +ADD EXPLANATION HERE + - (A-5) Suppose `B` is irreflexive and `A` is transitive. Because `A` is irreflexive, it is not the case for any `x` that `` is in `A`. Because `A` is transitive, if `` and ``, then ``. +ADD EXPLANATION HERE + - (A-6) Suppose `B` is both irreflexive and transitive. Because `B` is irreflexive, it is not the case for any `x` that `` is in `B`. Because `B` is transitive, if `` and ``, then ``. +ADD EXPLANATION HERE + - (A-7) Suppose `B` is irreflexive and `Ai` is transitive. Because `B` is irreflexive, it is not the case for any `x` that `` is in `B`. Because `Ai` is transitive, if `` and ``, then ``. +ADD EXPLANATION HERE + - (A-8) Suppose `B` is irreflexive and its inverse , `Bi`, is transitive. Because `B` is irreflexive, it is not the case for any `x` that `` is in `B`. Because `Bi` is transitive, if `` and ``, then ``. +ADD EXPLANATION HERE + - (A-9) Suppose `Ai` is irreflexive and its inverse , `A`, is transitive. Because `Ai` is irreflexive, it is not the case for any `x` that `` is in `Ai`. Because `A` is transitive, if `` and ``, then ``. +ADD EXPLANATION HERE + - (A-10) Suppose `Ai` is irreflexive and `B` is transitive. Because `Ai` is irreflexive, it is not the case for any `x` that `` is in `Ai`. Because `B` is transitive, if `` and ``, then ``. +ADD EXPLANATION HERE + - (A-11) Suppose `Ai` is both irreflexive and transitive. Because `Ai` is irreflexive, it is not the case for any `x` that `` is in `Ai`. Because `Ai` is transitive, if `` and ``, then ``. +ADD EXPLANATION HERE + - (A-12) Suppose `Ai` is irreflexive and `Bi` is transitive. Because `Ai` is irreflexive, it is not the case for any `x` that `` is in `Ai`. Because `Bi` is transitive, if `` and ``, then ``. +ADD EXPLANATION HERE + - (A-13) Suppose `Bi` is irreflexive and `A` is transitive. Because `Bi` is irreflexive, it is not the case for any `x` that `` is in `Bi`. Because `A` is transitive, if `` and ``, then ``. +ADD EXPLANATION HERE + - (A-14) Suppose `Bi` is irreflexive and its inverse , `B`, is transitive. Because `Bi` is irreflexive, it is not the case for any `x` that `` is in `Bi`. Because `B` is transitive, if `` and ``, then ``. +ADD EXPLANATION HERE + - (A-15) Suppose `Bi` is irreflexive and `Ai` is transitive. Because `Bi` is irreflexive, it is not the case for any `x` that `` is in `Bi`. Because `Ai` is transitive, if `` and ``, then ``. -- (A-16) Suppose `Bi` is both irreflexive and transitive. Because `Bi` is irreflexive, it is not the case for any `x` that `` is in `Bi`. Because `Bi` is transitive, if `` and ``, then ``. +ADD EXPLANATION HERE +- (A-16) Suppose `Bi` is both irreflexive and transitive. Because `Bi` is irreflexive, it is not the case for any `x` that `` is in `Bi`. Because `Bi` is transitive, if `` and ``, then ``. +ADD EXPLANATION HERE #### (B) Transitive and Functional -- (B-1) Suppose `A` is both transitive and functional. - -Because `A` is transitive, it is - -Because `A` is functional, if - -- (B-2) Suppose `A` is transitive and `B` is functional. - -Because `A` is transitive, - -Because `B` is functional, if - -- (B-3) Suppose `A` is transitive and its inverse , `Ai`, is functional. - -Because `A` is transitive, it is - -Because `Ai` is functional, if - -- (B-4) Suppose `A` is transitive and `Bi` is functional. - -Because `A` is transitive, - -Because `Bi` is functional, - -- (B-5) Suppose `B` is transitive and `A` is functional. +- (B-1) Suppose `A` is both transitive and functional. Because `A` is transitive, if x`R`y and y`R`z, then x`R`z. Because `A` is functional, if x`R`y and x`R`z, then y=z. -Because `A` is transitive, +ADD EXPLANATION HERE -Because `A` is functional, +- (B-2) Suppose `A` is transitive and `B` is functional. Because `A` is transitive, if x`R`y and y`R`z, then x`R`z. Because `B` is functional, if x`R`y and x`R`z, then y=z. -- (B-6) Suppose `B` is both transitive and functional. +ADD EXPLANATION HERE -Because `B` is transitive, +- (B-3) Suppose `A` is transitive and its inverse , `Ai`, is functional. Because `A` is transitive, if x`R`y and y`R`z, then x`R`z. Because `Ai` is functional, if x`R`y and x`R`z, then y=z. -Because `B` is functional, +ADD EXPLANATION HERE -- (B-7) Suppose `B` is transitive and `Ai` is functional. +- (B-4) Suppose `A` is transitive and `Bi` is functional. Because `A` is transitive, if x`R`y and y`R`z, then x`R`z. Because `Bi` is functional, if x`R`y and x`R`z, then y=z. -Because `B` is transitive, +ADD EXPLANATION HERE -Because `Ai` is functional, +- (B-5) Suppose `B` is transitive and `A` is functional. Because `A` is transitive, if x`R`y and y`R`z, then x`R`z. Because `A` is functional, if x`R`y and x`R`z, then y=z. -- (B-8) Suppose `B` is transitive and its inverse , `Bi`, is functional. +ADD EXPLANATION HERE -Because `B` is transitive, +- (B-6) Suppose `B` is both transitive and functional. Because `B` is transitive, if x`R`y and y`R`z, then x`R`z. Because `B` is functional, if x`R`y and x`R`z, then y=z. -Because `Bi` is functional, +ADD EXPLANATION HERE -- (B-9) Suppose `Ai` is transitive and its inverse , `A`, is functional. +- (B-7) Suppose `B` is transitive and `Ai` is functional. Because `B` is transitive, if x`R`y and y`R`z, then x`R`z. Because `Ai` is functional, if x`R`y and x`R`z, then y=z. -Because `Ai` is transitive, +ADD EXPLANATION HERE -Because `A` is functional, +- (B-8) Suppose `B` is transitive and its inverse , `Bi`, is functional. Because `B` is transitive, if x`R`y and y`R`z, then x`R`z. Because `Bi` is functional, if x`R`y and x`R`z, then y=z. -- (B-10) Suppose `Ai` is transitive and `B` is functional. +ADD EXPLANATION HERE -Because `Ai` is transitive, +- (B-9) Suppose `Ai` is transitive and its inverse , `A`, is functional. Because `Ai` is transitive, if x`R`y and y`R`z, then x`R`z. Because `A` is functional, if x`R`y and x`R`z, then y=z. -Because `B` is functional, +ADD EXPLANATION HERE -- (B-11) Suppose `Ai` is both transitive and functional. +- (B-10) Suppose `Ai` is transitive and `B` is functional. Because `Ai` is transitive, if x`R`y and y`R`z, then x`R`z. Because `B` is functional, if x`R`y and x`R`z, then y=z. -Because `Ai` is transitive, +ADD EXPLANATION HERE -Because `Ai` is functional, +- (B-11) Suppose `Ai` is both transitive and functional. Because `Ai` is transitive, if x`R`y and y`R`z, then x`R`z. Because `Ai` is functional, if x`R`y and x`R`z, then y=z. -- (B-12) Suppose `Ai` is transitive and `Bi` is functional. +ADD EXPLANATION HERE -Because `Ai` is transitive, +- (B-12) Suppose `Ai` is transitive and `Bi` is functional. Because `Ai` is transitive, if x`R`y and y`R`z, then x`R`z. Because `Bi` is functional, if x`R`y and x`R`z, then y=z. -Because `Bi` is functional, +ADD EXPLANATION HERE -- (B-13) Suppose `Bi` is transitive and `A` is functional. +- (B-13) Suppose `Bi` is transitive and `A` is functional. Because `Bi` is transitive, if x`R`y and y`R`z, then x`R`z. Because `A` is functional, if x`R`y and x`R`z, then y=z. -Because `Bi` is transitive, +ADD EXPLANATION HERE -Because `A` is functional, +- (B-14) Suppose `Bi` is transitive and its inverse , `B`, is functional. Because `Bi` is transitive, if x`R`y and y`R`z, then x`R`z. Because `B` is functional, if x`R`y and x`R`z, then y=z. -- (B-14) Suppose `Bi` is transitive and its inverse , `B`, is functional. +ADD EXPLANATION HERE -Because `Bi` is transitive, +- (B-15) Suppose `Bi` is transitive and `Ai` is functional. Because `Bi` is transitive, if x`R`y and y`R`z, then x`R`z. Because `Ai` is functional, if x`R`y and x`R`z, then y=z. -Because `B` is functional, +ADD EXPLANATION HERE -- (B-15) Suppose `Bi` is transitive and `Ai` is functional. +- (B-16) Suppose `Bi` is both transitive and functional. Because `Bi` is transitive, if x`R`y and y`R`z, then x`R`z. Because `Bi` is functional, if x`R`y and x`R`z, then y=z. -Because `Bi` is transitive, +ADD EXPLANATION HERE -Because `Ai` is functional, -- (B-16) Suppose `Bi` is both transitive and functional. - -Because `Bi` is transitive, - -Because `Bi` is functional, - -#### (C) Transitive and Inverse Functional +#### (C) Transitive and Inverse Functional - Not to be developed in this assignment. #### (D) Transitive and Asymmetric -- (D-1) Suppose `A` is both transitive and functional. - -Because `A` is transitive, it is - -Because `A` is functional, if - -- (D-2) Suppose `A` is transitive and `B` is functional. - -Because `A` is transitive, - -Because `B` is functional, if - -- (D-3) Suppose `A` is transitive and its inverse , `Ai`, is functional. - -Because `A` is transitive, it is - -Because `Ai` is functional, if - -- (D-4) Suppose `A` is transitive and `Bi` is functional. - -Because `A` is transitive, - -Because `Bi` is functional, - -- (D-5) Suppose `B` is transitive and `A` is functional. - -Because `A` is transitive, - -Because `A` is functional, - -- (D-6) Suppose `B` is both transitive and functional. - -Because `B` is transitive, - -Because `B` is functional, - -- (D-7) Suppose `B` is transitive and `Ai` is functional. - -Because `B` is transitive, - -Because `Ai` is functional, - -- (D-8) Suppose `B` is transitive and its inverse , `Bi`, is functional. - -Because `B` is transitive, - -Because `Bi` is functional, - -- (D-9) Suppose `Ai` is transitive and its inverse , `A`, is functional. - -Because `Ai` is transitive, - -Because `A` is functional, - -- (D-10) Suppose `Ai` is transitive and `B` is functional. - -Because `Ai` is transitive, - -Because `B` is functional, - -- (D-11) Suppose `Ai` is both transitive and functional. - -Because `Ai` is transitive, - -Because `Ai` is functional, - -- (D-12) Suppose `Ai` is transitive and `Bi` is functional. - -Because `Ai` is transitive, - -Because `Bi` is functional, - -- (D-13) Suppose `Bi` is transitive and `A` is functional. - -Because `Bi` is transitive, - -Because `A` is functional, - -- (D-14) Suppose `Bi` is transitive and its inverse , `B`, is functional. - -Because `Bi` is transitive, - -Because `B` is functional, - -- (D-15) Suppose `Bi` is transitive and `Ai` is functional. - -Because `Bi` is transitive, - -Because `Ai` is functional, - -- (D-16) Suppose `Bi` is both transitive and functional. - -Because `Bi` is transitive, - -Because `Bi` is functional, - - -#### (E) Transitive and Asymmetric - -- (E-1) Suppose `A` is both transitive and asymmetric. - -Because `A` is transitive, it is - -Because `A` is asymmetric, if - -- (E-2) Suppose `A` is transitive and `B` is functional. - -Because `A` is transitive, - -Because `B` is asymmetric, if - -- (E-3) Suppose `A` is transitive and its inverse , `Ai`, is functional. - -Because `A` is transitive, it is - -Because `Ai` is asymmetric, if - -- (E-4) Suppose `A` is transitive and `Bi` is functional. - -Because `A` is transitive, - -Because `Bi` is asymmetric, - -- (E-5) Suppose `B` is transitive and `A` is functional. - -Because `A` is transitive, - -Because `A` is asymmetric, - -- (E-6) Suppose `B` is both transitive and functional. - -Because `B` is transitive, +- (D-1) Suppose `A` is both transitive and asymmetric. Because `A` is transitive, if x`R`y and y`R`z, then x`R`z. Because `A` is asymmetric, if xRy, then it is not the case that yRx. -Because `B` is asymmetric, +ADD EXPLANATION HERE -- (E-7) Suppose `B` is transitive and `Ai` is functional. +- (D-2) Suppose `A` is transitive and `B` is functional. Because `A` is transitive, if x`R`y and y`R`z, then x`R`z. Because `B` is asymmetric, if xRy, then it is not the case that yRx. -Because `B` is transitive, +ADD EXPLANATION HERE -Because `Ai` is asymmetric, +- (D-3) Suppose `A` is transitive and its inverse , `Ai`, is functional. Because `A` is transitive, if x`R`y and y`R`z, then x`R`z. Because `Ai` is asymmetric, if xRy, then it is not the case that yRx. -- (E-8) Suppose `B` is transitive and its inverse , `Bi`, is functional. +ADD EXPLANATION HERE -Because `B` is transitive, +- (D-4) Suppose `A` is transitive and `Bi` is functional. Because `A` is transitive, if x`R`y and y`R`z, then x`R`z. Because `Bi` is asymmetric, if xRy, then it is not the case that yRx. -Because `Bi` is asymmetric, +ADD EXPLANATION HERE -- (E-9) Suppose `Ai` is transitive and its inverse , `A`, is functional. +- (D-5) Suppose `B` is transitive and `A` is functional. Because `A` is transitive, if x`R`y and y`R`z, then x`R`z. Because `A` is asymmetric, if xRy, then it is not the case that yRx. -Because `Ai` is transitive, +ADD EXPLANATION HERE -Because `A` is asymmetric, +- (D-6) Suppose `B` is both transitive and functional. Because `B` is transitive, if x`R`y and y`R`z, then x`R`z. Because `B` is asymmetric, if xRy, then it is not the case that yRx. -- (E-10) Suppose `Ai` is transitive and `B` is functional. +ADD EXPLANATION HERE -Because `Ai` is transitive, +- (D-7) Suppose `B` is transitive and `Ai` is functional. Because `B` is transitive, if x`R`y and y`R`z, then x`R`z. Because `Ai` is asymmetric, if xRy, then it is not the case that yRx. -Because `B` is asymmetric, +ADD EXPLANATION HERE -- (E-11) Suppose `Ai` is both transitive and functional. +- (D-8) Suppose `B` is transitive and its inverse , `Bi`, is functional. Because `B` is transitive, if x`R`y and y`R`z, then x`R`z. Because `Bi` is asymmetric, if xRy, then it is not the case that yRx. -Because `Ai` is transitive, +ADD EXPLANATION HERE -Because `Ai` is asymmetric, +- (D-9) Suppose `Ai` is transitive and its inverse , `A`, is functional. Because `Ai` is transitive, if x`R`y and y`R`z, then x`R`z. Because `A` is asymmetric, if xRy, then it is not the case that yRx. -- (E-12) Suppose `Ai` is transitive and `Bi` is functional. +ADD EXPLANATION HERE -Because `Ai` is transitive, +- (D-10) Suppose `Ai` is transitive and `B` is functional. Because `Ai` is transitive, if x`R`y and y`R`z, then x`R`z. Because `B` is asymmetric, if xRy, then it is not the case that yRx. -Because `Bi` is asymmetric, +ADD EXPLANATION HERE -- (E-13) Suppose `Bi` is transitive and `A` is functional. +- (D-11) Suppose `Ai` is both transitive and functional. Because `Ai` is transitive, if x`R`y and y`R`z, then x`R`z. Because `Ai` is asymmetric, if xRy, then it is not the case that yRx. -Because `Bi` is transitive, +ADD EXPLANATION HERE -Because `A` is asymmetric, +- (D-12) Suppose `Ai` is transitive and `Bi` is functional. Because `Ai` is transitive, if x`R`y and y`R`z, then x`R`z. Because `Bi` is asymmetric, if xRy, then it is not the case that yRx. -- (E-14) Suppose `Bi` is transitive and its inverse , `B`, is functional. +ADD EXPLANATION HERE -Because `Bi` is transitive, +- (D-13) Suppose `Bi` is transitive and `A` is functional. Because `Bi` is transitive, if x`R`y and y`R`z, then x`R`z. Because `A` is asymmetric, if xRy, then it is not the case that yRx. -Because `B` is asymmetric, +ADD EXPLANATION HERE -- (E-15) Suppose `Bi` is transitive and `Ai` is functional. +- (D-14) Suppose `Bi` is transitive and its inverse , `B`, is functional. Because `Bi` is transitive, if x`R`y and y`R`z, then x`R`z. Because `B` is asymmetric, if xRy, then it is not the case that yRx. -Because `Bi` is transitive, +ADD EXPLANATION HERE -Because `Ai` is asymmetric, +- (D-15) Suppose `Bi` is transitive and `Ai` is functional. Because `Bi` is transitive, if x`R`y and y`R`z, then x`R`z. Because `Ai` is asymmetric, if xRy, then it is not the case that yRx. -- (E-16) Suppose `Bi` is both transitive and functional. +ADD EXPLANATION HERE -Because `Bi` is transitive, +- (D-16) Suppose `Bi` is both transitive and functional. Because `Bi` is transitive, if x`R`y and y`R`z, then x`R`z. Because `Bi` is asymmetric, if xRy, then it is not the case that yRx. -Because `Bi` is asymmetric, \ No newline at end of file +ADD EXPLANATION HERE From e3afd11206323dcdeead15621be0b703472fdfe7 Mon Sep 17 00:00:00 2001 From: Mathilde Miroir Date: Thu, 19 Sep 2024 13:02:28 -0400 Subject: [PATCH 5/7] Update MMiroir - P1.md --- projects/project-1/assignment/MMiroir - P1.md | 144 ++++++++---------- 1 file changed, 61 insertions(+), 83 deletions(-) diff --git a/projects/project-1/assignment/MMiroir - P1.md b/projects/project-1/assignment/MMiroir - P1.md index 9431ac2..fcc8abf 100644 --- a/projects/project-1/assignment/MMiroir - P1.md +++ b/projects/project-1/assignment/MMiroir - P1.md @@ -65,39 +65,37 @@ The following example, provided in the assignment, serves as the guide for which #### (A) Transitive + Irreflexive (NS) - Suppose `R` is both irreflexive and transitive. By irreflexivity, it is not the case that xRx. By transitivity, for any x and y, if x`R`y and y`R`z, it follows that x`R`z. -ADD EXPLANATION HERE +When `R` is irreflexive it can never relate back to itself but when `R` is transitive, it can relate to multiple variables via the same relationship including but not necessarily limited to itself. #### (B) Transitive + Functional (NS) - Suppose `R` is both transitive and functional. By transitivity for any x and y, if x`R`y and y`R`z, it follows that x`R`z. By functionality, if x`R`y and x`R`z, then it follows that y=z. -ADD EXPLANATION HERE +When `R` is functional, it relates to single variable where as when it is transitive it communicates a single type of relationship that can apply to multiple variables. These are in conflict. #### (C) Transitive + Inverse Functional (NS) - Suppose `R` is both transitive and inverse functional. By transitivity for any x and y, if x`R`y and y`R`z, it follows that x`R`z. By inverse functional, if xRy and zRy, it follows that x=z. -ADD EXPLANATION HERE +The argumentative limitations for the transitive and functional combination of properties applies here aswell. #### (D) Transitive + Assymetric (NS) - Suppose `R` is both assymetric and transitive. By assymetry, if xRy, then it is not the case that yRx. By transitivity, if xRy and yRz, then xRz. -ADD EXPLANATION HERE +When `R` is assymetric, the `R` relationship is unidirectional and when `R` is transitive, the `R` relationship can be applied circularly or bidirectionally which is in conflict with a limited unidirectional relationship. #### (E) Assymetric + Symmetric (UNSAT) - _Explained in the assignment instructions_ - Suppose `R` is both symmetric and asymmetric. Then by symmetry for any x and y, if x`R`y it follows that y`R`x. However, by asymmetry it also follows that it is not the case that y`R`x. Hence, `R` cannot be both symmetric and asymmetric. -ADD EXPLANATION HERE - #### (F) Assymetric + Reflexive (UNSAT) - Suppose `R` is both assymetric and reflexive. By assymetry, f xRy, then it is not the case that yRx. By reflexivity, xRx. -ADD EXPLANATION HERE +When `R` is assymetric, the `R` relationship is unidirectional, but by reflexivity, the `R` 'flows' back to itself and therefore works bidirectionaly. #### (G) Reflexive + Irreflexive (UNSAT) - Suppose `R` is both reflexive and irreflexive. Then by reflexivity for any x, x`R`x. However by irreflexivity, it also follows that it is not the case that xRx. Hence, `R` cannot be both reflexive and irreflexive. Similary for the case of why `R` cannot be both symmetric and assymetric. -ADD EXPLANATION HERE +The `R` relationship cannot be defined such that it relates something to itself and also cannot relate something to itself. -## Assignment Part 2 +## Assignment Part 2 **PLEASE NOTE THIS IS INCOMPLETE ### Variable Constraints @@ -106,7 +104,9 @@ ADD EXPLANATION HERE - `Ai` is the inverse of `A` - `Bi` is the inverse of `B` -### Table III. (A) Transitive and Irreflexive Trials +### (A) Transitive and Irreflexive + +#### Table III. (A) Transitive and Irreflexive Trials | | A Trans | B Trans | Ai Trans | Bi Trans | |------------|---------|---------|----------|----------| | **A Irr** | N1 | N2 | N3 | N4 | @@ -118,77 +118,27 @@ ADD EXPLANATION HERE **_Subscripts denote corresponding explantion below_ -### Table IV. (B) Transitive and Functional Trials (j's previous contribution) - -| | A Trans | B Trans | Ai Trans | Bi Trans | -|-------------|---------|---------|----------|----------| -| **A Func** | N1 | N2 | N3 | N4 | -| **B Func** | N5 | N6 | N7 | N8 | -| **Ai Func** | Y9 | Y10| N11 | N12 | -| **Bi Func** | Y13| Y14| N15 | N16 | - -*_N denotes an error while Y denotes a successful run_ - -**_Subscripts denote corresponding explantion below_ - - -### Table V. (C) trans vs infunc??? same as previous? - -| | A Trans | B Trans | Ai Trans | Bi Trans | -|--------------|---------|---------|----------|----------| -| **A iFunc** | -1 | -2 | -3 | -4 | -| **B iFunc** | -5 | -6 | -7 | -8 | -| **Ai iFunc** | -9 | -10| -11 | -12 | -| **Bi iFUnc** | -13| -14| -15 | -16 | - -*_N denotes an error while Y denotes a successful run_ - -**_Subscripts denote corresponding explantion below_ - -### Table VI. (D) Transitive and Asymmetric Trials (Tim's HW) - -| | A Trans | B Trans | Ai Trans | Bi Trans | -|------------|---------|---------|----------|----------| -| **A Asym** | N1 | N2 | N3 | N4 | -| **B Asym** | N5 | N6 | N7 | N8 | -| **Ai Asym**| Y9 | Y10| N11 | N12 | -| **Bi Asym**| Y13| Y14| N15 | N16 | - -*_N denotes an error while Y denotes a successful run_ - -**_Subscripts denote corresponding explantion below_ - +- The trials performed, for the combination of transitive and irreflexive roles, indicated that when a property (A or Ai) is transitive and its subproperty (B or Bi) or the inverse of its subproperty (Bi or B) is irreflexive there are no conflicts in OWL. - -### Explanations (IN PROGRESS) +- Conversley, any instance in which a property (A or Ai) was irreflexive and its inverse (Ai or A) or its subproperty (B or Bi) or the inverse of its subproperty (Bi or B) was transitive, there is a conflict in OWL. #### (A) Transitive and Irreflexive Explanations -- (A-1) Suppose `A` is both irreflexive and transitive. Because `A` is irreflexive, it is not the case for any `x` that `` is in `A`. Because `A` is transitive, if `` and ``, then ``. -ADD EXPLANATION HERE +- (A-1) Suppose `A` is both irreflexive and transitive. Because `A` is irreflexive, it is not the case for any `x` that `` is in `A`. Because `A` is transitive, if `` and ``, then ``. A single property cannot be both irreflexive and transitive as explained in the above part of this assignment. _This applies to instances in which A is both irreflexive and transitive (**A-1**), B is both irreflexive and transitive (**A-6**), Ai is both irreflexive and transitive (**A-11**), Bi is both irreflexive and transitive (**A-16**)._ - (A-2) Suppose `A` is irreflexive and `B` is transitive. Because `A` is irreflexive, it is not the case for any `x` that `` is in `A`. Because `B` is transitive, if `` and ``, then ``. -ADD EXPLANATION HERE - -- (A-3) Suppose `A` is irreflexive and its inverse , `Ai`, is transitive. Because `A` is irreflexive, it is not the case for any `x` that `` is in `A`. Because `Ai` is transitive, if `` and ``, then ``. - -ADD EXPLANATION HERE - -- (A-4) Suppose `A` is irreflexive and `Bi` is transitive. Because `A` is irreflexive, it is not the case for any `x` that `` is in `A`. Because `Bi` is transitive, if `` and ``, then ``. +ADD EXPLANATION HERE - -ADD EXPLANATION HERE +Defining `A` as irreflexive insinuates that `B`, which is a subproperty of `A` also bears the irreflexive role. Since `B` has been defined to bear the transitive role, this is in direct conflict since a property cannot bear both an irreflexive and transitive role, as has been explained above in part 1 of this assignment. _This explanation applies to where (), (), (), and () similarly as the combinations of subproperties and their roles result in a single subproperty bearing both the transitive and irreflexive role._ -- (A-5) Suppose `B` is irreflexive and `A` is transitive. Because `A` is irreflexive, it is not the case for any `x` that `` is in `A`. Because `A` is transitive, if `` and ``, then ``. -ADD EXPLANATION HERE +- (A-3) Suppose `A` is irreflexive and its inverse , `Ai`, is transitive. Because `A` is irreflexive, it is not the case for any `x` that `` is in `A`. Because `Ai` is transitive, if `` and ``, then ``. -- (A-6) Suppose `B` is both irreflexive and transitive. Because `B` is irreflexive, it is not the case for any `x` that `` is in `B`. Because `B` is transitive, if `` and ``, then ``. +ADD EXPLANATION HERE - -ADD EXPLANATION HERE - -- (A-7) Suppose `B` is irreflexive and `Ai` is transitive. Because `B` is irreflexive, it is not the case for any `x` that `` is in `B`. Because `Ai` is transitive, if `` and ``, then ``. +- (A-4) Suppose `A` is irreflexive and `Bi` is transitive. Because `A` is irreflexive, it is not the case for any `x` that `` is in `A`. Because `Bi` is transitive, if `` and ``, then ``. ADD EXPLANATION HERE @@ -204,31 +154,31 @@ ADD EXPLANATION HERE ADD EXPLANATION HERE -- (A-11) Suppose `Ai` is both irreflexive and transitive. Because `Ai` is irreflexive, it is not the case for any `x` that `` is in `Ai`. Because `Ai` is transitive, if `` and ``, then ``. - -ADD EXPLANATION HERE - (A-12) Suppose `Ai` is irreflexive and `Bi` is transitive. Because `Ai` is irreflexive, it is not the case for any `x` that `` is in `Ai`. Because `Bi` is transitive, if `` and ``, then ``. ADD EXPLANATION HERE -- (A-13) Suppose `Bi` is irreflexive and `A` is transitive. Because `Bi` is irreflexive, it is not the case for any `x` that `` is in `Bi`. Because `A` is transitive, if `` and ``, then ``. - -ADD EXPLANATION HERE - - (A-14) Suppose `Bi` is irreflexive and its inverse , `B`, is transitive. Because `Bi` is irreflexive, it is not the case for any `x` that `` is in `Bi`. Because `B` is transitive, if `` and ``, then ``. -ADD EXPLANATION HERE +ADD EXPLANATION HERE - SUCCESSFUL -- (A-15) Suppose `Bi` is irreflexive and `Ai` is transitive. Because `Bi` is irreflexive, it is not the case for any `x` that `` is in `Bi`. Because `Ai` is transitive, if `` and ``, then ``. +## EXPLANATIONS PAST THIS POINT TO BE EXPLORED (I just had surgery so could not explore this fully) -ADD EXPLANATION HERE +### (B) Transitive and Functional -- (A-16) Suppose `Bi` is both irreflexive and transitive. Because `Bi` is irreflexive, it is not the case for any `x` that `` is in `Bi`. Because `Bi` is transitive, if `` and ``, then ``. +#### Table IV. (B) Transitive and Functional Trials -ADD EXPLANATION HERE +| | A Trans | B Trans | Ai Trans | Bi Trans | +|-------------|---------|---------|----------|----------| +| **A Func** | N1 | N2 | N3 | N4 | +| **B Func** | N5 | N6 | N7 | N8 | +| **Ai Func** | Y9 | Y10| N11 | N12 | +| **Bi Func** | Y13| Y14| N15 | N16 | -#### (B) Transitive and Functional +*_N denotes an error while Y denotes a successful run_ + +**_Subscripts denote corresponding explantion below_ - (B-1) Suppose `A` is both transitive and functional. Because `A` is transitive, if x`R`y and y`R`z, then x`R`z. Because `A` is functional, if x`R`y and x`R`z, then y=z. @@ -295,9 +245,37 @@ ADD EXPLANATION HERE ADD EXPLANATION HERE -#### (C) Transitive and Inverse Functional - Not to be developed in this assignment. +### (C) Transitive and Inverse Functional - Not to be developed in this assignment. + + +#### Table V. (C) Transitive and Inverse Functional + +| | A Trans | B Trans | Ai Trans | Bi Trans | +|--------------|---------|---------|----------|----------| +| **A iFunc** | -1 | -2 | -3 | -4 | +| **B iFunc** | -5 | -6 | -7 | -8 | +| **Ai iFunc** | -9 | -10| -11 | -12 | +| **Bi iFUnc** | -13| -14| -15 | -16 | + +*_N denotes an error while Y denotes a successful run_ + +**_Subscripts denote corresponding explantion below_ + +### (D) Transitive and Asymmetric + +#### Table VI. (D) Transitive and Asymmetric Trials + +| | A Trans | B Trans | Ai Trans | Bi Trans | +|------------|---------|---------|----------|----------| +| **A Asym** | N1 | N2 | N3 | N4 | +| **B Asym** | N5 | N6 | N7 | N8 | +| **Ai Asym**| Y9 | Y10| N11 | N12 | +| **Bi Asym**| Y13| Y14| N15 | N16 | + +*_N denotes an error while Y denotes a successful run_ + +**_Subscripts denote corresponding explantion below_ -#### (D) Transitive and Asymmetric - (D-1) Suppose `A` is both transitive and asymmetric. Because `A` is transitive, if x`R`y and y`R`z, then x`R`z. Because `A` is asymmetric, if xRy, then it is not the case that yRx. From 27bb4fd6e92b55b7eab5f6470fc076721771ad80 Mon Sep 17 00:00:00 2001 From: Mathilde Miroir Date: Thu, 26 Sep 2024 18:27:49 -0400 Subject: [PATCH 6/7] Broke down all of the pitfalls identified and tackled by the OOPS tool. Translated information into list & diagram form in order to pick and choose which ones to develop sparql queries for. --- projects/project-1/assignment/MMiroir - P1.md | 2 +- .../Miroir SPARQL Queries/N1.sparql | 13 ++ .../Miroir SPARQL Queries/ReadME.md | 182 ++++++++++++++++++ 3 files changed, 196 insertions(+), 1 deletion(-) create mode 100644 projects/project-2/assignment/Miroir SPARQL Queries/N1.sparql create mode 100644 projects/project-2/assignment/Miroir SPARQL Queries/ReadME.md diff --git a/projects/project-1/assignment/MMiroir - P1.md b/projects/project-1/assignment/MMiroir - P1.md index fcc8abf..ee7c308 100644 --- a/projects/project-1/assignment/MMiroir - P1.md +++ b/projects/project-1/assignment/MMiroir - P1.md @@ -148,7 +148,7 @@ ADD EXPLANATION HERE - (A-9) Suppose `Ai` is irreflexive and its inverse , `A`, is transitive. Because `Ai` is irreflexive, it is not the case for any `x` that `` is in `Ai`. Because `A` is transitive, if `` and ``, then ``. -ADD EXPLANATION HERE +ADD EXPLANATION HERE - - (A-10) Suppose `Ai` is irreflexive and `B` is transitive. Because `Ai` is irreflexive, it is not the case for any `x` that `` is in `Ai`. Because `B` is transitive, if `` and ``, then ``. diff --git a/projects/project-2/assignment/Miroir SPARQL Queries/N1.sparql b/projects/project-2/assignment/Miroir SPARQL Queries/N1.sparql new file mode 100644 index 0000000..9df49a2 --- /dev/null +++ b/projects/project-2/assignment/Miroir SPARQL Queries/N1.sparql @@ -0,0 +1,13 @@ +# Title: +# (descriptive title of the query) +# Constraint Description: +# (description of the query functionality) +# Severity: +# (select "Warning" or "Error") + + + + + + +# BIND (concat("WARNING: The following ontology elements have the same rdfs:label ", str(?element), " and ", str(?element2)) AS ?error) \ No newline at end of file diff --git a/projects/project-2/assignment/Miroir SPARQL Queries/ReadME.md b/projects/project-2/assignment/Miroir SPARQL Queries/ReadME.md new file mode 100644 index 0000000..5269cf4 --- /dev/null +++ b/projects/project-2/assignment/Miroir SPARQL Queries/ReadME.md @@ -0,0 +1,182 @@ +### NOTES FOR ASSIGNMENT + +PLEASE NOTE: this is a complilation of 42 problems/pitfalls identified by a preexisting software (OOPS) whose tools are behind a paywall. The paper attached, includes diagrams displaying the code patterns used to identify some of these pitfalls. PLEASE confirm that pitfall solution explored has not been defined by another tool/reasoner listed in the paper (e.g. XD-Tools, Moki, Onto-check, reasoner and/or specific test cases - TABLE III) + +OOPS () for sale +https://www.igi-global.com/article/oops-ontology-pitfall-scanner/116450 + + +https://oa.upm.es/35873/1/INVE_MEM_2014_192872.pdf (site for all pitfalls defined by OOPS) + +P1, P9, P14, P15, P15, P16, P17, P18, P23 have not been addressed by the OOPS tool. + + + +::: mermaid +classDiagram + Pitfalls -- Critical + Pitfalls -- Important + Pitfalls -- Minor + Pitfalls: identified by OOPS + + class Critical{ + P01. Creating Polysemous Elements + P03. Creating the Relationship “is” Instead of Using “rdfs:subClassOf”, “rdf:type” or “owl:sameAs” + P05. Defining Wrong Inverse Relationships + P06. Including Cycles in the Hierarchy + P14. Misusing “owl:allValuesFrom” + P15. Misusing “not some” and “some not” + P16. Misusing Primitive and Defined Classes + P19. Swapping Intersection and Union + P27. Defining Wrong Equivalent Relationships + P28. Defining Wrong Symmetric Relationships + P29. Defining Wrong Transitive Relationships + P31. Defining Wrong Equivalent Classes + P37. Ontology Not Available + P39. Ambiguous Namespace + P40. Namespace Hijacking + } + class Important{ + P10. Missing Disjointness + P11. Missing Domain or Range in Properties + P12. Missing Equivalent Properties + P17. Specializing a Hierarchy Exceedingly + P18. Specifying the Domain or the Range Exceedingly + P23. Using Incorrectly Ontology Elements + P24. Using Recursive Definition + P25. Defining a Relationship Inverse to Itself + P26. Defining Inverse Relationships for a Symmetric One + P30. Missing Equivalent Classes + P34. Untyped Class + P35. Untyped Property + P38. No OWLOntology Declaration + } + class Minor{ + P02. Creating Synonyms as Classes + P04. Creating Unconnected Ontology Elements + P07. Merging Different Concepts in the Same Class + P08. Missing Annotations + P09. Missing Basic Information + P13. Missing Inverse Relationships + P20. Misusing Ontology Annotations + P21. Using a Miscellaneous Class + P22. Using Different Naming Criteria in the Ontology + P32. Several Classes with the Same Label + P33. Creating a Property Chain with Just One Property + P36. URI Contains File Extension + } +::: + + + +P01. Creating Polysemous Elements: An ontology element whose name has different meanings is included in the ontology to represent more than one conceptual idea. For example, the class “Theatre” is used to represent both the artistic discipline and the place in which a play is performed. + +- P02. Creating Synonyms as Classes: Several classes whose identifiers are synonyms are created and defined as equivalent. For example, the classes “Waterfall” and “Cascade” are defined as equivalents. This pitfall is related to the guidelines presented in Noy, and McGuinness (2001), which explain that synonyms for the same concept do not represent different classes. + +- P03. Creating the Relationship “is” Instead of Using “rdfs:subClassOf”, “rdf:type” or “owl:sameAs”: The “is” relationship is created in the ontology instead of using OWL primitives for representing the subclass relationship (“subclassOf”), the membership to a class (“instanceOf”), or the equality between instances (“sameAs”). An example of this pitfall is to define the class “Actor” in the following way ‘Actor ≡ Person ⨅ ∃inter- prets.Actuation ⨅ ∃is.Man’. This pitfall is related to the guidelines for understanding the “is-a” relation provided in Noy, and McGuinness (2001). + +- P04. Creating Unconnected Ontology Elements: Ontology elements (classes, relationships or attributes) are created with no relation to the rest of the ontology. An example of this type of pitfall is to create the relationship “memberOfTeam” and to miss the class representing teams; thus, the relationship created is isolated in the ontology. + +- P05. Defining Wrong Inverse Relationships: Two relationships are defined as inverse relations when they are not necessarily inverse. An example of this type of pitfall is to define “isSoldIn” and “isBoughtIn” as inverse relationships. + +- P06. Including Cycles in the Hierarchy (Gómez-Pérez, 2004; Noy, & McGuin- ness, 2001): A cycle between two classes in the hierarchy is included in the ontology even though the ontology is not intended to have such classes as equivalent. That is, some class A has a subclass B, and at the same time B is a superclass of A. An example of this type of pitfall is represented by the class “Professor” as subclass of “Person”, and the class “Person” as subclass of “Professor”. + +- P07. Merging Different Concepts in the Same Class: A class whose identifier refers to two or more different concepts is cre- ated. An example of this type of pitfall is the creation of the class “StyleAndPeriod”. + +- P08. Missing Annotations: Ontology terms lack annotations properties such as rdfs:label or rdfs:comment. An example of this type of pitfall is to create a class and to fail to provide human readable annotations attached to such a class. + +- P09. Missing Basic Information: Some of the information needed is not included in the ontology. This pitfall may be related to the requirements in the ontology requirements specification document (ORSD) not covered by the ontology, or to knowledge that can be added to the ontology to make it more complete. An example of this type of pitfall is to create the relationship “startsIn” in order to represent that the routes have a starting point in a particular location and to miss the relationship “endsIn” in order to represent that a route has an end point. + +- P10. Missing Disjointness (Gómez- Pérez, 2004; Noy, & McGuinness, 2001; Rector et al., 2004): The ontology lacks disjoint axioms between classes or between properties that should be defined as disjoint. For example, we can create the classes “Odd” and “Even” (or the classes “Prime” and “Composite”) without being disjoint; such representation is incomplete with regard to the definition of these types of numbers. + +- P11. Missing Domain or Range in Properties: Relationships and/or attributes without domain or range (or none of them) are included in the ontology. An example of this type of pitfall is to create the relationship “hasWritten”, with no domain nor range specification, in an ontology about art in which the relationship domain should be “Writer” and the relationship range should be “LiteraryWork”. This pitfall is related to the common error that appears when defining the ranges and domains described in Rector et al. (2004). + +- P12. Missing Equivalent Properties: When an ontology is imported into another, classes duplicated in both ontologies are normally defined as equivalent classes. However, the ontology developer misses the definition of equivalent properties in the cases of duplicated relationships and attri- butes. An example of this type of pitfalls is to fail to define the relations “hasMember” and “has-Member” as equivalent. + +- P13. Missing Inverse Relationships: This pitfall appears when any relationship (ex- cept for the symmetric ones) does not have an inverse relationship defined within the ontology. For example, the case in which the ontology developer omits the inverse definition between the relations “hasLanguageCode” and “isCodeOf”. + +- P14. Misusing “owl:allValuesFrom” (Rector et al., 2004): This pitfall can appear in two different ways. Firstly, when the universal restriction (“allValuesFrom”) is used as the default qualifier instead of the existential restriction (“someValuesFrom”). Secondly, when “allValuesFrom” is included to close off the possibility of further additions for a given property. An example of this type of pitfall is to define the class “Book” in the following way ‘Book ≡ ∃producedBy.Writer ⨅ ∀uses.Paper’ thus closing the possibility of adding “Ink” as an element used in the writing. + +- P15. Misusing “not some” and “some not” (Rector et al., 2004): The pitfall here is to confuse the representation of “some not” with “not some”. An example of this type of pitfall is to define a vegetarian pizza as any pizza which has both some topping which is not meat and some topping which is not fish. This example is explained in more detail in Rector et al. (2004). + +- P16. Misusing Primitive and Defined Classes (Rector et al., 2004): This pitfall implies failing to make the definition ‘complete’ rather than ‘partial’ (or ‘necessary and sufficient’ rather than just ‘necessary). It is critical to understand that, in general, nothing will be inferred to be subsumed under a primitive class by the classifier (Rector et al., 2004). This pitfall implies that the developer does not understand the open world assumption. An example of this pitfall is to create the primitive class ‘CheesyPizza ⊏ Pizza ⨅ ∃hasTopping. Cheese’ instead of creating it as a defined class in the following way: ‘CheesyPizza ≡ Pizza ⨅ ∃hasTopping.Cheese’. This example is explained in more detail in Rector et al. (2004). + +- P17. Specializing a Hierarchy Exceedingly: The hierarchy in the ontology is specialized in such a way that the final leaves cannot have instances since they are actually instances and should have been created as such instead of as classes. Authors in Noy, and McGuinness (2001) provide guidelines for distinguishing between a class and an instance when modeling hierarchies. An example of this type of pitfall is to create the classes “Madrid”, “Barcelona” and “Sevilla”, among others, as subclasses of “Place”. + +- P18. Specifying the Domain or the Range Exceedingly (Noy, & McGuinness, 2001; Rector et al., 2004): This pitfall means failing to find a domain or a range general enough. An example of this type of pitfall is to restrict the domain of the relationship “isOfficialLanguage” to the class “City”, instead of allowing the class “Country” or a more general concept such as “GeopoliticalObject” to have an official language. + +- P19. Swapping Intersection and Union: The ranges and/or domains of the properties (relationships and attributes) are defined by intersecting several classes in cases in which the ranges and/or domains should be the union of those classes. This pitfall is related both to the common error that appears when defining ranges and domains described in Rector et al. (2004) and to the guidelines for defining these elements provided in Noy, and McGuinness (2001). An example of this type of pitfall is to create the relationship “takesPlaceIn” with one range declaration for the class “City” and other range declaration for the class “Nation”, as this implementation represents the intersection of both ranges instead of the union. + +- P20. Misusing Ontology Annotations: The contents of some annotation properties are swapped or misused. An example of this type of pitfall is to include in the rdfs:label annotation of the class “Crossroads” the following sentence ’the place of intersection of two or more roads’; and to include in the rdfs:comment annotation the word ‘Crossroads’. + +- P21. Using a Miscellaneous Class: This means creating in a hierarchy a class containing the instances that do not belong to the sibling classes instead of classifying such instances as instances of the class in the upper level of the hierarchy. An example of this type of pitfall is to create the class “HydrographicalResource”, and the subclasses “Stream” and “Waterfall”, among others, and also the subclass “OtherRiverElement”. + +- P22. Using Different Naming Criteria in the Ontology: Ontology elements are not named following the same convention within the whole ontology. Some notions about naming conventions are provided in Noy, and McGuinness (2001). For example, this pitfall appears when a class identifier starts with upper case, e.g. “Ingredient”, whereas its subclass identifiers start with lower case, e.g. “flour” and “milk”. + +- P23. Using Incorrectly Ontology Elements: An ontology element (class, relationship or attribute) is used to model a part of the ontology that should be modeled with a different element. A particular case of this pitfall regarding the misuse of classes and property values is addressed in Noy, and McGuinness (2001). An example of this type of pitfall is to create the relationship “isEcological” between an instance of “Car” and the instances “Yes” or “No”, instead of creating an attribute “isEcological” whose range is Boolean. + +- P24. Using Recursive Definition: An ontology element is used in its own definition. An example of this type of pitfall is to create the relationship “hasFork” and to establish as its range the following: The set of restaurants that have at least one value for the relationship “hasFork”. + +- P25. Defining a Relationship Inverse to Itself: A relationship is defined as inverse of itself. In this case, this property could have been defined as “owl:SymmetricProperty” instead. An example of this type of pitfall is to create the relationship “hasBorderWith” and to state that “hasBorderWith” is its inverse relationship. + +- P26. Defining Inverse Relationships for a Symmetric One: A relationship is defined as “owl:SymmetricProperty”, and such a relationship is defined as inverse of another relationship. For example, to create for the symmetric relationship “farFrom” an inverse relationship, e.g. itself, “farFrom”. + +- P27. Defining Wrong Equivalent Relationships: Two relationships are defined as equivalent relations when they are not necessarily equivalent. An example of this type of pitfalls is to mix up common relationships that could hold between several types of entities, as “hasPart” defined in one ontology between human body parts and the relation “hasPart” defined in another ontology between research plans and research projects. + +- P28. Defining Wrong Symmetric Relationships: A relationship is defined as symmetric when the relationship is not necessarily symmetric. This situation can appear because the domain and range are too specific; for example, if we define the symmetric relationship “hasSpouse” between the concepts “Man” and “Woman” instead of using the concept “Person” both as domain and range of such a relationship. + +- P29. Defining Wrong Transitive Relationships: A relationship is defined as transitive when the relationship is not necessarily transitive. An example of this type of pitfall is to create the relationship “participatesIn”, whose domain is the union of the concepts “Team” and “Individual” and whose range is the concept “Event”, and defining the relationship as transitive. + +- P30. Missing Equivalent Classes: When an ontology is imported into another, classes with the same conceptual meaning that are duplicated in both ontologies should be defined as equivalent classes in order to benefit the interoperability between both ontologies. However, the ontology developer may miss the definition of equivalent classes in the cases of duplicated concepts. An example of this pitfall is to fail to define the classes ‘Trainer’ (class in an imported ontology) and ‘Coach’ (class in the ontology about sports being developed) as equivalent classes. + +- P31. Defining Wrong Equivalent Classes: Two classes are defined as equivalent when they are not necessarily equivalent. For example, defining “Car” as equivalent to “Vehicle”. + +- P32. Several Classes with the Same Label: Two or more classes have the same content in the rdfs:label annotation. For example, to link the label “Theatre” both with the building and the literary discipline, adding no more labels to them. + +- P33. Creating a Property Chain with Just One Property: There is a property chain that includes only one property in the antecedent part. For example, to create the following property chain: isInChargeOf -> supervises. + +- P34. Untyped Class (Hogan et al., 2010): A resource is used as a class without having been declared as a Class. An example of this type of pitfall is to create individu- als of the class “Person” and to omit that “Person” is a class. + +- P35. Untyped Property (Hogan et al., 2010): A resource is used as a property with- out having been declared as a rdf:Property or as some subclass of it. An example of this type of pitfall is to link individual by the relation “hasPart” and to omit that “hasPart” is an object property. + +- P36. URI Contains File Extension (Archer, Goedertier, & Loutas, 2012): This involves including file extensions as “.owl”, “.rdf”, “.ttl”, “.n3” and “.rdfxml” in an ontology URI. An example of this pitfall is to define an ontology uri as “http://www. biopax.org/release/biopax-level3.owl” containing the extension “.owl” related to the technology used. + +- P37. Ontology Not Available: This involves omitting to provide online description or documentation of the ontology when looking up its URI. An example of this pitfall could be the following case: “Ontology Security (ontosec)” (URI: http://www.semanticweb.org/ontologies/2008/11/OntologySecurity.owl) which is not available online as RDF nor as HTML (at the moment of carrying out this work). + +- P38. No OWLOntology Declaration: This means failing to declare the owl:Ontology tag where the ontology metadata should be provided. An example of this pitfall could be found at the “Creative Commons Rights Expression Language (cc)” ontology (URI: http://creativecommons.org/ ns) that does not have any owl:Ontology declaration in its RDF file even though it has other OWL elements used as, for example, owl:equivalentProperty (at the moment of carrying out this work). + +- P39. Ambiguous Namespace: This means failing to define both the ontology URI and the xml:base namespace. An example of this pitfall could be found at “Basic Access Control ontology (acl)” (URI: http://www.w3.org/ns/auth/acl) that has no owl:Ontology tag nor xml:base definition. + +- P40. Namespace Hijacking (Heath, & Bizer, 2011): This means reusing or referring to terms from other namespaces not actually defined in such namespace. This pitfall is related to the Linked Data publishing guidelines provided in Heath, and Bizer (2011): “Only define new terms in a namespace that you control.” An example of this pitfall is to use “http://www.w3.org/2000/01/rdf-schema#Property” that is not defined in the rdf namespace (http://www.w3.org/2000/01/rdf-schema#) instead of using “http://www.w3.org/1999/02/22-rdf-syntax-ns#Property”, that is actually defined in the rdfs namespace (http://www.w3.org/1999/02/22-rdf-syntax-ns#). + + +# Process workflow for this assignment + +::: mermaid +flowchart TD +%% Nodes +A("Project Idea"):::green +B("Initial Planning"):::orange +C("Detailed Design
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Requirements"):::blue +D{"Decision: Continue or Stop?"}:::yellow +E("Development Phase"):::pink +F("Testing Phase"):::purple +G("Deployment"):::green +H("Feedback and Improvement"):::orange + +%% Edges +A --> B --> C --> D +D -- Continue --> E --> F --> G +D -- Stop --> H +G --> H +H --> B + +%% Styling +classDef green fill:#B2DFDB,stroke:#00897B,stroke-width:2px; +classDef orange fill:#FFE0B2,stroke:#FB8C00,stroke-width:2px; +classDef blue fill:#BBDEFB,stroke:#1976D2,stroke-width:2px; +classDef yellow fill:#FFF9C4,stroke:#FBC02D,stroke-width:2px; +classDef pink fill:#F8BBD0,stroke:#C2185B,stroke-width:2px; +classDef purple fill:#E1BEE7,stroke:#8E24AA,stroke-width:2px; + +::: \ No newline at end of file From ee6b4592f6d42fac7106e763f031a19e2abcb618 Mon Sep 17 00:00:00 2001 From: Mathilde Miroir Date: Thu, 26 Sep 2024 18:32:18 -0400 Subject: [PATCH 7/7] Update ReadME.md Minor formatting change made in order to test the difference between leaving comments on the commit via GitHub Desktop vs VSCode vs Web based GitHub. --- .../Miroir SPARQL Queries/ReadME.md | 30 +------------------ 1 file changed, 1 insertion(+), 29 deletions(-) diff --git a/projects/project-2/assignment/Miroir SPARQL Queries/ReadME.md b/projects/project-2/assignment/Miroir SPARQL Queries/ReadME.md index 5269cf4..fb68740 100644 --- a/projects/project-2/assignment/Miroir SPARQL Queries/ReadME.md +++ b/projects/project-2/assignment/Miroir SPARQL Queries/ReadME.md @@ -11,7 +11,6 @@ https://oa.upm.es/35873/1/INVE_MEM_2014_192872.pdf (site for all pitfalls define P1, P9, P14, P15, P15, P16, P17, P18, P23 have not been addressed by the OOPS tool. - ::: mermaid classDiagram Pitfalls -- Critical @@ -152,31 +151,4 @@ P01. Creating Polysemous Elements: An ontology element whose name has different # Process workflow for this assignment -::: mermaid -flowchart TD -%% Nodes -A("Project Idea"):::green -B("Initial Planning"):::orange -C("Detailed Design
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Requirements"):::blue -D{"Decision: Continue or Stop?"}:::yellow -E("Development Phase"):::pink -F("Testing Phase"):::purple -G("Deployment"):::green -H("Feedback and Improvement"):::orange - -%% Edges -A --> B --> C --> D -D -- Continue --> E --> F --> G -D -- Stop --> H -G --> H -H --> B - -%% Styling -classDef green fill:#B2DFDB,stroke:#00897B,stroke-width:2px; -classDef orange fill:#FFE0B2,stroke:#FB8C00,stroke-width:2px; -classDef blue fill:#BBDEFB,stroke:#1976D2,stroke-width:2px; -classDef yellow fill:#FFF9C4,stroke:#FBC02D,stroke-width:2px; -classDef pink fill:#F8BBD0,stroke:#C2185B,stroke-width:2px; -classDef purple fill:#E1BEE7,stroke:#8E24AA,stroke-width:2px; - -::: \ No newline at end of file +#### insert flowchart here when complete