Philosophy - Proofs

Submission Instructions. Complete the assignment below using Canvas text entry.  When you are finished, upload the homework as a MSWord (.docx) or Adobe (.pdf) document to the appropriate assignment in Canvas. If you wish, you may print the homework and scan the completed pages. You may also take photos of the pages and upload them, but be sure your answers are legible. This module corresponds to material from Chapter 8, section 1 of the book. 

You are intended to start this module after reading Chapter 8, Section 1 of the book and watching VLecture #10: Annotating Proofs.

Negation: ∼ Conjunction: ⦁ Disjunction: ⌵ Conditional: → Biconditional: ↔

 

Module #11                   Philosophy 117 CH 8: Justifying Proofs                         

 

Directions: Each of the symbolic arguments below needs to annotated. This requires you to fill in the empty brackets by  citing the lines that were used, and also implicational rule employed. In the example below, there were three lines that needed to be annotated: 5, 6, and 7. Line 5 was a justified step in the proof, because P followed from lines 1 and 3, using the implicational rule MP -- Modus Ponens. Line 6 was a justified step in the proof, because it followed from lines 2 and 3, also via Modus Ponens. Lastly, line 7, which is identical to the conclusion and therefore the final line for the proof, followed from lines 5 and 6 using the new implicational rule Conjunction, or CONJ. Try and complete the annotations for the proofs below by filling in the lines used and rule employed. If you are completely confused (even after reading Chapter 8 section 1), wait for the VLecture and I’ll walk you through justifying proofs.

EX:  

1. O → P

2. O→ Q

3. O

4. ∴P ⦁ Q 

5. P {1,3 MP}

6. Q {2,3 MP}

7. P ⦁Q {5,6 CONJ}  

 

1. P → Q

2. Q → R

∴P → R {                     }

 

 

1. P ⌵ ∼∼Q

2. ~P

∴~~Q  {                           }

 

 

1. (P→Q) ⦁ (R→S)

2. P

∴Q

3. P→Q     {                        }

4. Q           {                        }

 

1. (X → Y) ⦁ ∼Y

∴∼X

2. X→Y {                           }

3. ~Y    {                           }

4. ∼X    {                           }

 

 

1. A → B

2. ∼A → (C v D)

3. ∼B

4. ∼C

∴D

5. ~A      {                           }

6. C v D  {                           }

7. D         {                           }

 

 

1. P ⌵ Q

2. Q → R

3. ∼R

∴P  

4. ~Q  {                                        }

5. P     {                                         }

 

1. W ⦁ (X ↔ Y)

2. (X ↔ Y) → ∼Z

∴~Z ⌵ T

3. X ↔ Y   {                                    }

4. ~Z        {                                    }

5. ~Z ⌵ T {                                    }

 

 

1. (A → B) ⦁ (B → C)

∴ A → C

2. A → B  {                                    }

3. B → C  {                                    }

4. A → C  {                                    }

 

1. (P → Q) ⦁  (R → S)

2. P ⌵ R

∴Q ⌵ S

3. P → Q  {                                    }

4. R → S  {                                    }

5. Q ⌵ S  {                                     }