1 / 73

Predicate Logic

Predicate Logic. Truth-table reminders. The problem people had the most trouble with was 1e: construct a truth-table for: (P & (~Q & R))

denise
Télécharger la présentation

Predicate Logic

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Predicate Logic

  2. Truth-table reminders

  3. The problem people had the most trouble with was 1e: construct a truth-table for: (P & (~Q & R)) Many of you only had four rows. The very first principle of truth-tables is to write out ALL your variables to the left, and ALL the possible combinations of truth values under them.

  4. This is OK

  5. This is also OK

  6. This is also OK

  7. The important point is that each possible combination of truth-values for P, Q, and R shows up in the rows. Notice that if I want to find, for example, P = T, Q = F, and R = T, I can.

  8. For a WFF containing only one variable, like “~(P&~~~P),” you only need two rows on the truth-table, because there are only two possibilities: P is true or P is false.

  9. Only Two Rows

  10. For a WFF with 2 variables, you’re going to need 4 rows, to represent the four possibilities: • P = T, Q = T • P = T, Q = F • P = F, Q = T • P = F, Q = F

  11. For a WFF with 3 variables (for example, P, Q, and R) you need 8 rows, because there are two possibilities for P (T or F), 2 possibilities for Q (T or F) and two possibilities for R (T or F), for a total of 8 possibilities: 2 x 2 x 2 = 8

  12. But not just any 8 lines will do. You have to have 8 lines that represent all 8 unique possibilities.

  13. Bad Truth-Table Construction

  14. Only Represents 4 Different Possibilities

  15. Standard Form This is why logicians always use a standard way of writing truth-tables. Suppose I had to write a table for 4 variables (16 rows). Here’s what I’d do: P rows: TTTTTTTTFFFFFFFF Q rows: TTTTFFFFTTTTFFFF R rows: TTFFTTFFTTFFTTFF S rows: TFTFTFTFTFTFTFTF

  16. Homework #2

  17. (P → Q) ├ ((P & R) → Q) 1 1. (P → Q) A 2 2. (P & R) A 2 3. P 2 &E 1,2 4. Q 1,3 →E 1 5. ((P & R) → Q) 2,4 →I

  18. Proving Conditionals Remember our strategy: if you want to prove a conditional: • Assume the antecedent. • Prove the consequent. • Use →I

  19. ~~P ├ P 1 1. ~~P A 2 2. ~P A (for ~E) 1,2 3. (~P & ~~P) 1,2 &I 1 4. P 2,3 ~E

  20. Hard Problems If you run into a problem that looks difficult, often your best option is to use ~E or ~I. • Figure out what you want to prove. • Assume the opposite. • Prove a contradiction. • Use ~E or ~I.

  21. (P → (Q → R)) ├ ((P & Q) → R) 1 1. (P → (Q → R)) A 2 2. (P & Q) A (for →I) 2 3. P 2 &E 2 4. Q 2 &E 1,2 5. (Q → R) 1,3 →E 1,2 6. R 4,5 →E 1 7. ((P & Q) → R) 2,6 →I

  22. Assume the Antecedent for →I It’s very important here to continue to follow our strategy for proving conditionals. In line #2 I assumed the antecedent. This is because the rule for →I requires that I assume the antecedent. It says that if I assumeφ and derive ψ, then I can write down (φ→ψ) on any future line (depending on whatever ψ depended on except φ).

  23. Bad Strategy 1 1. (P → (Q → R)) A 2 2. P A 3 3. Q A 1,2 4. (Q → R) 1,2 →E 1,2,3 5. R 3,4 →E ? 6. ??? ?

  24. Correct Step, Bad Result 1 1. (P → (Q → R)) A 2 2. P A 3 3. Q A 1,2 4. (Q → R) 1,2 →E 1,2,3 5. R 3,4 →E 1,3 6. (P → R) 2,5 →I

  25. Correct Step… 1 1. (P → (Q → R)) A 2 2. P A 3 3. Q A 1,2 4. (Q → R) 1,2 →E 1,2,3 5. R 3,4 →E 2,3 6. (P & Q) 2,3 &I

  26. Incorrect Step! 1 1. (P → (Q → R)) A 2 2. P A 3 3. Q A 1,2 4. (Q → R) 1,2 →E 1,2,3 5. R 3,4 →E 2,3 6. (P & Q) 2,3 &I 1 7. ((P & Q) → R) 6,5 →I

  27. ~(P v Q) ├ ~P 1 1. ~(P v Q) A 2 2. P A (for ~I) 2 3. (P v Q) 2 vI 1,2 4. ((P v Q) & ~(P v Q)) 1,3 &I 1 5. ~P 2,4 ~I

  28. (P ↔ ~P) ├ Q 1 1. (P ↔ ~P) A 2 2. P A (for ~I) 1 3. ((P → ~P) & (~P → P)) 1 ↔E 1 4. (P → ~P) 3 &E 1,2 5. ~P 2,4 →E 1,2 6. (P & ~P) 2,5 &I 1 7. ~P 2,6 ~I

  29. (P ↔ ~P) ├ Q 1 7. ~P A 1 8. (~P → P) 3 &E 1 9. P 7,8 →E 1 10. (P v Q) 9 vI 1 11. Q 7,10 vE

  30. Dependencies Even though I had P and ~P at lines 2 and 5 in my proof and I could have proved Q (my goal), I did not. Why? Because both P and ~P depended on line 2.

  31. (P ↔ ~P) ├ Q 1 1. (P ↔ ~P) A 2 2. P A (for ~I) 1 3. ((P → ~P) & (~P → P)) 1 ↔E 1 4. (P → ~P) 3 &E 1,2 5. ~P 2,4 →E

  32. (P ↔ ~P) ├ Q 1 1. (P ↔ ~P) A 2 2. P A (for ~I) 1 3. ((P → ~P) & (~P → P)) 1 ↔E 1 4. (P → ~P) 3 &E 1,2 5. ~P 2,4 →E 2 6. (P v Q) 2 vI 1,2 7. Q 2,6 ~I

  33. Dependencies 1,2 7. Q 2,6 ~I What does this line mean? Another way of writing it is: (P ↔ ~P), P ├ Q This is not what we want to show!

  34. Instead, what I do is prove ~P again– this time depending only on 1. Then I prove P again, this time depending only on 1. Then I’m free to prove Q, depending only on 1, and my proof is complete.

  35. Prove ~P Again 1 1. (P ↔ ~P) A 2 2. P A (for ~I) 1 3. ((P → ~P) & (~P → P)) 1 ↔E 1 4. (P → ~P) 3 &E 1,2 5. ~P 2,4 →E 1,2 6. (P & ~P) 2,5 &I 1 7. ~P 2,6 ~I

  36. Prove P Again 1 7. ~P A 1 8. (~P → P) 3 &E 1 9. P 7,8 →E

  37. Then Prove Q 1 7. ~P A 1 8. (~P → P) 3 &E 1 9. P 7,8 →E 1 10. (P v Q) 9 vI 1 11. Q 7,10 vE

  38. Very Hard Bonus Question! The very hard bonus question asked you to prove: ├ (((P → Q) → P) → P) Here’s some strategy. We want to prove a conditional, so we should assume its antecedent, ((P → Q) → P). Now we want to derive its consequent, P. One way to do this is if we had (P → Q). Since that’s a conditional, we assume its antecedent, P and try to prove Q.

  39. ├ (((P → Q) → P) → P) 1 1. ((P → Q) → P) A (for →I) 2 2. P A (for →I) 3 3. ~P A (for ~E) 2 4. (P v Q) 2 vI 2,3 5. Q 3,4 vE 3 6. (P → Q) 2,5 →I 1,3 7. P1,6 →E

  40. ├ (((P → Q) → P) → P) 1,3 8. (P & ~P) 3,7 &I 1 9. P 3,8 ~E 10. (((P → Q) → P) → P) 1,9 →I

  41. Here P shows up in the proof 3 times. Only on the third time am I done with the proof, because only then can I conclude (((P → Q) → P) → P) depending on no assumptions at all.

  42. ├ (((P → Q) → P) → P) 1 1. ((P → Q) → P) A (for →I) 2 2. P A (for →I) 2 3. (((P → Q) → P) → P) 1,2 →I

  43. ├ (((P → Q) → P) → P) 1 1. ((P → Q) → P) A (for →I) 2 2. P A (for →I) 3 3. ~P A (for ~E) 2 4. (P v Q) 2 vI 2,3 5. Q 3,4 vE 3 6. (P → Q) 2,5 →I 1,3 7. P1,6 →E 3 8. (((P → Q) → P) → P) 1,7 →I

  44. ├ (((P → Q) → P) → P) 1,3 8. (P & ~P) 3,7 &I 1 9. P 3,8 ~E ----- 10. (((P → Q) → P) → P) 1,9 →I

  45. Predicate Logic

  46. Remember that the goal of logic is to develop formal tests for validity. This is done by finding deductively valid argument forms. In SL, we have two tests to determine valid argument forms: the truth-table test, and derivations. Any argument that has a form that is valid according to the truth-table test is valid, and any argument that has a form that can be proven is valid.

  47. Write Down ALL Possibilities

More Related