hw6Solutions

# hw6Solutions - Homework 6 Solutions Problem 1 ∃ x.f(x ∨...

This preview shows pages 1–2. Sign up to view the full content.

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

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Homework 6 Solutions Problem 1 ( ∃ x.f(x)) ∨ ( ∃ x.g(x)) |– ∃ x.(f(x) ∨ g(x)) f(x) f(x) ------------- { ∨ IL} --------------{ ∨ IR} f ( x ) ∨ g(x) f(x) ∨ g(x) ------------------ { ∃ I} ------------------ { ∃ I} ∃ x.f(x) ∃ x.(f(x) ∨ g(x)) ∃ x.g(x) ∃ x(f(x) ∨ g(x)) ----------------------------- { ∃ E} ----------------------------- { ∃ E} ( ∃ x.f(x) ∨ ∃ x.g(x)) ∃ x.(f(x) ∨ g(x)) ∃ x(f(x) ∨ g(x)) ---------------------------------------------------------------------------------------------------- { ∨ E} ∃ x.(f(x) ∨ g(x)) Problem 2 ( ∀ x.(f(x) ∨ g(x))) |– (( ∀ x.f(x)) ∨ ( ∀ x.g(x))) — This is not a theorem. counterexample: Let the universe of discourse consist of the symbols 1 and 2. Let f(1) = True, f(2) = False. Let g(1) = False, g(2) = True T h e n B u t ( ∀ x.(f(x) ∨ g(x))) = ((T ∧ F) ∨ (F ∧ T)) (( ∀ x.f(x)) ∨ ( ∀ x.g(x))) = ((f(1) ∨ g(1)) ∧ ((f(2) ∨ g(2)) = ((f(1) ∧ f(2)) ∨ (g(1) ∧ g(2)) = (T ∨ F)...
View Full Document

{[ snackBarMessage ]}

### Page1 / 3

hw6Solutions - Homework 6 Solutions Problem 1 ∃ x.f(x ∨...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online