CS2013 Mathematics for Computing Science Adam Wyner University of Aberdeen Computing Science Slides...

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

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