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

CS2013Maths for Computing Science

Proof (part b)

Adam WynerUniversity of Aberdeen

Computing Science

October 2014 CS2013 Maths for Computing 2

Topics

• Equivalences that can be used to replace formulae in proofs

• Tableau proofs• Further examples of Propositional Logic proofs.• Proof rules with quantifiers.


Equivalences

Equivalence expressions can be substituted since they do not change truth.


Equivalences


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 H6. A ∨ ¬ B

7. (A ∨ ¬ B) ∨ C

8. (D (E F))

9. (E F) (F G)

10. D (F G)

11. (F G) H12. D H


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 DD 12

6. A ∨ ¬ B DI 4

7. (A ∨ ¬ B) ∨ C DI 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) H IE 2,6

12. D H HS 10,11


A Conditional Proof

1. (A B)∨ (C ∧ D)

2. (D E)∨ F3. Show: A F4. A

5. Show: F

6. A ∨ B

7. C ∧ D

8. D

9. (D E)∨10. F


A Conditional Proof

1. (A B)∨ (C ∧ D)

2. (D E)∨ F3. Show: A F CD 4, 5

4. A Assumption

5. Show: F DD 10

6. A ∨ B DI 4

7. C ∧ D IE 1,6

8. D CE 7

9. (D E)∨ DI 8

10. F IE 2,9


An Indirect Proof

1. A (B ∧ C)

2. (B D)∨ E3. (D A)∨4. Show: E ID 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


An Indirect Proof

1. A (B ∧ C)

2. (B D)∨ E3. (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


An Indirect Proof

1. A (B ∧ C)

2. (B D)∨ E3. (D A)∨4. Show: E ID 13

5. ¬ E Assumption

6. ¬ (B D)∨ IE 2,5

7. ¬ B ∧ ¬ D Second De Morgan 6

8. ¬ D CE 7

9. A DE 4,8

10. B ∧ C IE 1,9

11. B CE 10

12. ¬ B CE 7

13. B ∧ ¬ B ContraI 11,12


A Logical Equivalence

Prove: ¬ (p (∨ ¬ p ∧ q)) is equivalent to ¬ p ∧ ¬ q

1. ¬ (p (∨ ¬ 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




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)




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) ∨ F commutativity

7. ( ¬ p ∧ ¬ q) identity law for F

Tableau Proofs

• Another formal method for logical proofs.• Thanks to Ulle Endriss for some slides.


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 ∆ |= φHow? 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).


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.


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.


Propositional Tableau Rules


Example (Truth Tables say?)Prove: ((P → Q) → P) → P

¬(((P → Q) → P) → P)

((P → Q) → P)

¬P

¬(P → Q) P

x

P

¬Q

x


Prove: P → (Q R), ¬Q ¬R |= ¬P∧ ∨

P → (Q R), ¬Q ¬R, P∧ ∨

¬P (Q R)∧x

Q

R

¬Q ¬R

x x


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}∧ ∨ ∨


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.


Quantifiers

• How to prove with universal and existential quantifiers.



Universal Instantiation

• x P(x)P(o) (substitute any constant o)

The same for any other variable than x.


Existential Generalization

• P(o) x P(x)

The same for any other variable than x.


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 ...


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).


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.


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.


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))


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:


Formal Proof UsingNatural Deduction

Statement How 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)


A very similar proof

Can you prove:• x(T(x) → E(x)) and E(r)• T(r).


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)


Quantifier Rules


Longer Quantifier Proof

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 Computing 39October 2014

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