Download - CS2013 Mathematics for Computing Science Adam Wyner University of Aberdeen Computing Science Slides adapted from Michael P. Frank ' s course based on the.

CS2013Mathematics for Computing Science

Adam WynerUniversity of Aberdeen

Computing Science

Slides adapted fromMichael P. Frank's course based on the textDiscrete Mathematics & Its Applications

(5th Edition)by Kenneth H. Rosen

ProofReplacement & Quantifiers

2013 Frank / van Deemter / Wyner 3

Topics

• Equivalences that can be used to replace formulae in 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


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 :ogic

• 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

ß


Longer Quantifier Proof