{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

exercise10sol

# exercise10sol - CSE541 EXERCISE 10 SOLUTIONS Covers 12 Read...

This preview shows pages 1–5. Sign up to view the full content.

CSE541 EXERCISE 10 SOLUTIONS Covers Chapters 10, 11, 12 Read and learn all examples and exercises in the chapters as well! QUESTION 1 Let GL be the Gentzen style proof system for classical logic defined in chapter 11. Prove, by constructing a proper decomposition tree that (1) GL (( ¬ a b ) ( ¬ b a )). Solution: By definition we have that GL (( ¬ a b ) ( ¬ b a )) if and only if GL -→ (( ¬ a b ) ( ¬ b a )) . We construct the decomposition tree for -→ A as follows. T A -→ (( ¬ a b ) ( ¬ b a )) | ( →⇒ ) ( ¬ a b ) -→ ( ¬ b a ) | ( →⇒ ) ¬ b, ( ¬ a b ) -→ a | ( → ¬ ) ( ¬ a b ) -→ b, a ^ ( ⇒-→ ) -→ ¬ a, b, a | ( → ¬ ) a -→ b, a axiom b -→ b, a axiom All leaves of the tree T A are axioms, hence we have found the proof of A in GL . 1

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
(2) Let GL be the Gentzen style proof system defined in chapter 11. Prove, by constructing a proper decom- position tree that 6 ‘ GL (( a b ) ( ¬ b a )) . Solution: Observe that for any formula A , its decomposition tree T A in GL is not unique. Hence when constructing decomposition trees we have to cover all possible cases. We construct the decomposition tree for -→ A as follows. T 1 A -→ (( a b ) ( ¬ b a )) | ( →⇒ ) ( one choice ) ( a b ) -→ ( ¬ b a ) | ( →⇒ ) ( first of two choices ) ¬ b, ( a b ) -→ a | ( ¬ → ) ( one choice ) ( a b ) -→ b, a ^ ( ⇒-→ ) ( one choice ) -→ a, b, a non - axiom b -→ b, a axiom The tree contains a non- axiom leaf -→ a, b, a, hence it is not a proof of (( a b ) ( ¬ b a )) in GL . We have only one more tree to construct. Here it is. 2
T 2 A -→ (( a b ) ( ¬ b a )) | ( →⇒ ) ( one choice ) ( a b ) -→ ( ¬ b a ) ^ ( ⇒-→ ) ( second of two choices ) -→ ( ¬ b a ) , a ( -→⇒ ) ( one choice ) ¬ b -→ a, a | ( ¬ → ) ( one choice ) -→ b, a, a non - axiom b -→ ( ¬ b a ) | ( →⇒ ) ( one choice ) b, ¬ b -→ a | ( ¬ → ) ( one choice ) b -→ b, a axiom All possible trees end with an non-axiom leave whet proves that 6 ‘ GL (( a b ) ( ¬ b a )) . QUESTION 2 Does the tree below constitute a proof in GL ? Justify your answer. T A -→ ¬¬ (( ¬ a b ) ( ¬ b a )) | ( → ¬ ) ¬ (( ¬ a b ) ( ¬ b a )) -→ 3

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
| ( ¬ → ) -→ (( ¬ a b ) ( ¬ b a )) | ( →⇒ ) ( ¬ a b ) -→ ( ¬ b a ) | ( →⇒ ) ( ¬ a b ) , ¬ b -→ a | ( ¬ → ) ( ¬ a b ) -→ b, a ^ ( ⇒-→ ) -→ ¬ a, b, a | ( → ¬ ) a -→ b, a axiom b -→ b, a axiom Solution: The tree is a not a proof in GL because a rule corresponding to the decomposition step below does not exists in it. ( ¬ a b ) , ¬ b -→ a | ( ¬ → ) ( ¬ a b ) -→ b, a It is a proof is some system GL1 that has all the rules of GL except its ( ¬ → ). This rule has to be replaced by the rule: ( ¬ → ) 1 Γ , Γ 0 -→ Δ , A, Δ 0 Γ , ¬ A, Γ 0 -→ Δ , Δ 0 . Also the step above this one, i.e. ( ¬ a b ) -→ ( ¬ b a ) | ( →⇒ ) ( ¬ a b ) , ¬ b -→ a is incorrect for similar reason.
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### Page1 / 13

exercise10sol - CSE541 EXERCISE 10 SOLUTIONS Covers 12 Read...

This preview shows document pages 1 - 5. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online