Discrete Structures II: Proofs of Tautologies using CP and IP Rules, Study notes of Discrete Structures and Graph Theory

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Discrete Structures II

CS 251

Lecture 3

( Examples, Sections 6.4, 7.1.1)

Review example: CP rule

Use CP rule to prove that the wff is a tautology. ( A   B)  (B  C)  ( C  D)  (A  D).

1. ( A   B) P
2. (B  C) P
3. ( C  D) P
4. A P start subproof
5.  B 1, 4, DS
6. C 2, 5, DS
7. D 3, 6, DS
8. A  D 4, 7, CP QED 1, 2, 3, 8, CP.

Proof Notes

Let W denote the wff (A  B)  B. Note this wff is not a tautology. (eg. A=B= false). Now consider the proof.

1. (A  B) P
2. A P
3. B 1,2, MP QED 1,3 CP.

• What is wrong with the proof of W?
• Write down the statement that the “proof” proves. The conjunction of all the premises that you used to prove something is precisely the antecedent of the conditional that you proved.

Proof Notes

Suppose we have a wff that is not a tautology

A  ((A  B)  C)  B.

1. A P
2. (A  B)  C P
3. B 1,2 MP

QED 1,2,3, CP.

• What is wrong with the proof?

Don’t apply inference rules to subexpressions of wffs.

6.4.1 An example axiom system

• In the last class we studied that a formal reasoning system consists of: A set of wffs, a set of axioms, a set of inference rules.
• Is there a formal system for which the propositional calculus is both sound and complete?
• YES!

Frege-Lukasiewicz Axioms

• Example: Frege-Lukasiewicz Axioms, where A, B, C represent any wff generated by propositional variables and the two connectives ¬ and →

1. A → (B → A).
2. (A → (B → C)) → ((A → B) →(A → C)).
3. (¬ A → ¬B) → ( B → A).

Proof of CP rule

If A is a premise in a proof of B , then there is a

proof of A  B that does not use A as a premise.

Proof:

Assumption: A is a premise in the proof of B.

Suppose B 1 ,…., Bn is a proof of B that uses the premise A. We’ll show by induction for each k in the interval 1≤ k ≤ n that there is a proof of A  Bk that does not use A as the premise.

Proof of CP rule

• Proof Contd.: If k =1, then either B 1 = A or B 1 =axiom or premise other than A. if B 1 = A then A  B 1 = A  A which by lemma 1 has a proof that does not use A as a premise. If B 1 =axiom or premise other than A then 1. B 1 An axiom or premise other than A. 2. B 1  (A  B 1 ) Axiom 1 3. A  B 1 1, 2, MP.

Hypothetical Syllogism

From the premises A  B and B  C , there is a proof of A  C.

Proof: 1. A  B P 2_. B_  C P 3_. A P_ 4_. B_ 1, 3 MP 5_. C_ 2, 4 MP 6_. A_  C 3, 5 CP_._ QED Note: We can use CP rule to HS to say that from the premise A  B there is a proof of ( B  C)  (A  C)

Six Tautologies

1.  A  ( A  B).

2.   A  A.

3. A    A.

4. ( A  B)  ( B   A).

5. A ( B   ( A  B) ).

6. ( A  B) (( A  B)  B).

Proofs using only axioms, MP, HS, and previously proven results.

Proof 2

•   A  A.

Proof: 1.   A P

2.   A  ( A     A) Proof 1
3.  A     A 1, 2, MP
4. ( A     A)  (  A  A) Ax. 3
5.   A  A 3, 4 MP
6. A 1, 5, MP QED 1, 6, CP.

Proof 3

• A    A.

Proof: 1.    A   A Proof 2

2. (   A   A )  (A    A) Ax.
3. A    A 1, 2, MP QED

Proof 5

• A ( B   ( A  B)).

Proof: 1. A P

2. A B P
3. B 1, 2 , MP
4. (A B) B 2, 3 , CP
5. ((A B) B)  ( B   ( A  B)). Proof 4
6.  B   ( A  B) 4, 5, MP QED 1, 6, CP

Proof 6

• ( A  B) (( A  B)  B). Proof:

1. ( A  B) P
2. ( A  B) ( B   A) Proof 4
3.  B   A 1, 2, MP
4.  A  B P
5.  B   B P
6.  B (B  (A  B)) Proof 1
7. ( B (B  (A  B)))  (( B B )  ( B  (A  B))) Ax. 2
8. ( B B )  ( B  (A  B)) 6, 7 MP 9  B   (A  B) 5, 8 MP 10 ( B   (A  B))  ((A  B)  B)) Axiom 3 11 (A  B)  B 9, 10, MP 12 B 1, 11, MP 13 ( A  B)  B 4, 12, CP QED 1, 13, CP