(A => ~B) & (B => A) = ~B
Where, => is Implies, ~ is Not, & is AND, = is Logically Equivalent
Prove this using Precedence rule and Law of statement algebra.
carl89 wrote:(A => ~B) & (B => A) = ~B
Where, => is Implies, ~ is Not, & is AND, = is Logically Equivalent
Prove this using Precedence rule and Law of statement algebra.
(~A V (~B)) & (~B V A) [ELIMINATE =>]
[(~A V ~B) & ~B] V [(~A V ~B) & A] [DISTRIBUTION]
[~B & (~A V ~B)] V [A & (~A V ~B)] [COMMUNATIVE]
[(~B & ~A) V (~B & ~B)] V [(A & ~A) V (A & ~B)] [DISTRIBUTIVE]
[(~B & ~A) V (~B & ~B)] V [F V (A & ~B)] [CONTRADICTION]
carl89 wrote:THIS IS WHAT I HAVE SO FAR:
- Code: Select all
(~A V (~B)) & (~B V A) [ELIMINATE =>]
[(~A V ~B) & ~B] V [(~A V ~B) & A] [DISTRIBUTION]
[~B & (~A V ~B)] V [A & (~A V ~B)] [COMMUNATIVE]
[(~B & ~A) V (~B & ~B)] V [(A & ~A) V (A & ~B)] [DISTRIBUTIVE]
[(~B & ~A) V (~B & ~B)] V [F V (A & ~B)] [CONTRADICTION]
carl89 wrote:I tried the truth table method and it worked just fine.