1. CS2013 Maths for Computing Science Proof (part b) Adam Wyner University of Aberdeen Computing Science

2. October 2014CS2013 Maths for Computing2 Topics Equivalences that can be used to replace formulae in proofs Tableau proofs Further examples of Propositional Logic proofs. Proof rules with quantifiers.

3. October 2014CS2013 Maths for Computing3 Equivalences Equivalence expressions can be substituted since they do not change truth.

4. October 2014CS2013 Maths for Computing4 Equivalences

5. October 2014CS2013 Maths for Computing5 A Direct Proof 1. ((A ∨ ¬ B) ∨ C)  (D  (E  F)) 2. (A ∨ ¬ B)  ((F  G)  H) 3. A  ((E  F)  (F  G)) 4. A 5. Show: D  H 6. A ∨ ¬ B 7. (A ∨ ¬ B) ∨ C 8. (D  (E  F)) 9. (E  F)  (F  G) 10. D  (F  G) 11. (F  G)  H 12. D  H

6. October 2014CS2013 Maths for Computing6 A Direct Proof 1. ((A ∨ ¬ B) ∨ C)  (D  (E  F)) 2. (A ∨ ¬ B)  ((F  G)  H) 3. A  ((E  F)  (F  G)) 4. A 5. Show: D  HDD 12 6. A ∨ ¬ BDI 4 7. (A ∨ ¬ B) ∨ CDI 6 8. (D  (E  F))IE 1,7 9. (E  F)  (F  G)IE 3,4 10. D  (F  G)HS 8,9 11. (F  G)  HIE 2,6 12. D  HHS 10,11

7. October 2014CS2013 Maths for Computing7 A Conditional Proof 1. (A ∨ B)  (C ∧ D) 2. (D ∨ E)  F 3. Show: A  F 4. A 5. Show: F 6. A ∨ B 7. C ∧ D 8. D 9. (D ∨ E) 10. F

8. October 2014CS2013 Maths for Computing8 A Conditional Proof 1. (A ∨ B)  (C ∧ D) 2. (D ∨ E)  F 3. Show: A  FCD 4, 5 4. AAssumption 5. Show: FDD 10 6. A ∨ BDI 4 7. C ∧ DIE 1,6 8. DCE 7 9. (D ∨ E)DI 8 10. FIE 2,9

9. October 2014CS2013 Maths for Computing9 An Indirect Proof 1. A  (B ∧ C) 2. (B ∨ D)  E 3. (D ∨ A) 4. Show: EID 13 5. Assumption 6. IE 2,5 7. Second De Morgan 6 8. CE 7 9. DE 4,8 10. IE 1,9 11. CE 10 12. CE 7 13. ContraI 11,12

10. October 2014CS2013 Maths for Computing10 An Indirect Proof 1. A  (B ∧ C) 2. (B ∨ D)  E 3. (D ∨ A) 3. Show: E 4. ¬ E 5. ¬ (B ∨ D) 6. ¬ B ∧ ¬ D 7. ¬ D 8. A 9. B ∧ C 10. B 11. ¬ B 12. B ∧ ¬ B

11. October 2014CS2013 Maths for Computing11 An Indirect Proof 1. A  (B ∧ C) 2. (B ∨ D)  E 3. (D ∨ A) 4. Show: EID 13 5. ¬ EAssumption 6. ¬ (B ∨ D)IE 2,5 7. ¬ B ∧ ¬ DSecond De Morgan 6 8. ¬ DCE 7 9. ADE 4,8 10. B ∧ CIE 1,9 11. BCE 10 12. ¬ BCE 7 13. B ∧ ¬ BContraI 11,12

12. October 2014CS2013 Maths for Computing12 A Logical Equivalence Prove: ¬ (p ∨ (¬ p ∧ q)) is equivalent to ¬ p ∧ ¬ q 1. ¬ (p ∨ (¬ p ∧ q)) ....

13. October 2014CS2013 Maths for Computing13 A Logical Equivalence Prove: ¬ (p ∨ (¬ p ∧ q)) is equivalent to ¬ p ∧ ¬ q 1. ¬ (p ∨ (¬ p ∧ q))  second De Morgan 2.  first De Morgan 3.  double negation 4.  second distributive 5.  negation 6.  commutativity 7.  identity law for F

14. October 2014CS2013 Maths for Computing14 A Logical Equivalence Prove: ¬ (p ∨ (¬ p ∧ q)) is equivalent to ¬ p ∧ ¬ q 1. ¬ (p ∨ (¬ p ∧ q))  ¬ p ∧ ¬ (¬ p ∧ q) 2.  ¬ p ∧ (¬ (¬ p) ∨ ¬ q) 3.  ¬ p ∧ (p ∨ ¬ q) 4.  (¬ p ∧ p) ∨ ( ¬ p ∧ ¬ q) 5.  F ∨ ( ¬ p ∧ ¬ q) 6.  ( ¬ p ∧ ¬ q) ∨ F 7.  ( ¬ p ∧ ¬ q)

15. October 2014CS2013 Maths for Computing15 A Logical Equivalence Prove: ¬ (p ∨ (¬ p ∧ q)) is equivalent to ¬ p ∧ ¬ q 1. ¬ (p ∨ (¬ p ∧ q))  ¬ p ∧ ¬ (¬ p ∧ q)second De Morgan 2.  ¬ p ∧ (¬ (¬ p) ∨ ¬ q)first De Morgan 3.  ¬ p ∧ (p ∨ ¬ q)double negation 4.  (¬ p ∧ p) ∨ ( ¬ p ∧ ¬ q)second distributive 5.  F ∨ ( ¬ p ∧ ¬ q)negation 6.  ( ¬ p ∧ ¬ q) ∨ Fcommutativity 7.  ( ¬ p ∧ ¬ q)identity law for F

16. Tableau Proofs Another formal method for logical proofs. Thanks to Ulle Endriss for some slides. October 2014CS2013 Maths for Computing16

17. Tableau Method – How? A method for proving that a given set of formulas is not satisfiable : Given: set of premises ∆ and conclusion φ Task: prove ∆ |= φ How? show ∆ ∪ {¬φ} is not satisfiable (which is equivalent), i.e. add the complement of the conclusion to the premises and derive a contradiction (“refutation procedure”). If a proof fails, we can sometimes conclude that the set in question is satisfiable, or that the conclusion in question does not follow (ok for propositional logic; for first-order logic only in special cases). October 2014CS2013 Maths for Computing17

18. Construction Data structure: a proof is represented as a tableau—a binary tree, the nodes of which are labelled with formulas. Start: we start by putting the premises and the negated conclusion into the root of an otherwise empty tableau. Expansion: we apply expansion rules to the formulas on the tree, thereby adding new formulas and splitting branches. Use all premises and copy analysed formulas into all relevant branches. Closure: we close branches that are contradictory. Success: a proof is successful iff we can close all branches. October 2014CS2013 Maths for Computing18

19. Construction We close a branch if it contains both a formula and the same formula with a "¬" in front. Each such branch represents a way in which the original sentences might perhaps be true. Closing a branch signifies that this branch is not a way in which the original sentences could all be true. Where all the branches close, there was no way in which the original sentences could all be true. There is at most one rule that can be applied to a given formula. October 2014CS2013 Maths for Computing19

20. Propositional Tableau Rules October 2014CS2013 Maths for Computing20

21. Example (Truth Tables say?) Prove: ((P → Q) → P) → P ¬(((P → Q) → P) → P) ((P → Q) → P) ¬P ¬(P → Q)P x P ¬Q x October 2014CS2013 Maths for Computing21

22. Prove: P → (Q ∧ R), ¬Q ∨ ¬R |= ¬P P → (Q ∧ R), ¬Q ∨ ¬R, P ¬P(Q ∧ R) x Q R ¬Q¬Rx October 2014CS2013 Maths for Computing22

23. More Examples {(P → Q) → R), S → ¬P, T, (¬S → T) → Q, ¬R} {P → (Q ∨ R), ¬R → (S ∧ T), ¬T, T → U} {P ∧ R, (¬P ∨ Q) ∧ (¬Q ∨ ¬R)} {(P ∨ Q) ∧ R, ¬Q ∨ ¬R, ¬P} {¬P ∨ Q, ¬P ∨ ¬Q, P ∨ R, ¬R} {P ∧ R, P ∨ Q, ¬P ∨ ¬Q} October 2014CS2013 Maths for Computing23

24. Termination Assuming we analyse each formula at most once, we have: Theorem 1 (Termination) For any propositional tableau, after a finite number of steps no more expansion rules will be applicable. This must be so, because each rule results in ever shorter formulas. Note: Importantly, termination will not hold in the first- order case. October 2014CS2013 Maths for Computing24

25. Quantifiers How to prove with universal and existential quantifiers. October 2014CS2013 Maths for Computing25

26. October 2014CS2013 Maths for Computing26 Universal Instantiation  x P(x)  P(o)(substitute any constant o) The same for any other variable than x.

27. October 2014CS2013 Maths for Computing27 Existential Generalization P(o)  x P(x) The same for any other variable than x.

28. October 2014CS2013 Maths for Computing28 Universal Generalisation P(g)  x P(x) This is not a valid inference of course. But suppose you can prove P(g) without using any information about g...... then the inference to  x P(x) is valid! In other words...

29. October 2014CS2013 Maths for Computing29 Universal Generalisation P(g) (for g an arbitrary or general constant)  x P(x) Concretely, your strategy should be to choose a new constant g (i.e., that did not occur in your proof so far) and to prove P(g).

30. October 2014CS2013 Maths for Computing30 Existential Instantiation  x P(x)  P(c) (substitute a new constant c) Once again, the inference is not generally valid, but we can regard it as valid if c is a new constant.

31. October 2014CS2013 Maths for Computing31 Simple Formal Proof in Predicate Logic Argument: –“ All TAs compose quizzes. Ramesh is a TA. Therefore, Ramesh composes quizzes. ” First, separate the premises from conclusions: –Premise #1: All TAs compose quizzes. –Premise #2: Ramesh is a TA. –Conclusion: Ramesh composes quizzes.

32. October 2014CS2013 Maths for Computing32 Rendering in Logic Render the example in logic notation. Premise #1: All TAs compose easy quizzes. –Let U.D. = all people –Let T(x) :≡ “ x is a TA ” –Let E(x) :≡ “ x composes quizzes ” –Then Premise #1 says:  x(T(x)→E(x))

33. October 2014CS2013 Maths for Computing33 Rendering cont… Premise #2: Ramesh is a TA. –Let r :≡ Ramesh –Then Premise #2 says: T(r) –And the Conclusion says: E(r) The argument is correct, because it can be reduced to a sequence of applications of valid inference rules, as follows:

34. October 2014CS2013 Maths for Computing34 Formal Proof Using Natural Deduction StatementHow obtained 1.  x(T(x) → E(x))(Premise #1) 2.T(r) → E(r)(Universal instantiation) 3.T(r)(Premise #2) 4.  E(r)(MP 2, 3)

35. October 2014CS2013 Maths for Computing35 A very similar proof Can you prove:  x(T(x) → E(x)) and  E(r)   T(r).

36. October 2014CS2013 Maths for Computing36  in Natural Deduction A very simple example: Theorem: From  xF(x) it follows that  yF(y) 1.  xF(x)(Premiss) 2.F(a)(Arbitrary a, Exist. Inst.) 3.   yF(y)(Exist. Generalisation)

37. October 2014CS2013 Maths for Computing37 Quantifier Rules

38. October 2014CS2013 Maths for Computing38 Longer Quantifier Proof

39. References and Credits Slides adapted from Michael P. Frank ' s course Textbook: Rosen. Discrete Mathematics & Its Applications. McGraw-Hill. Fifth edition. CS 2013 Maths for Computing39October 2014