(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.
Please reply with your efforts so far, your book's version of the "Precedence" rule, and your book's definition of "statement algebra" and what its laws are. Thank you.(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.
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(~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]
So you've proved the statement to be not true? Is this what you were supposed to do?THIS IS WHAT I HAVE SO FAR:
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(~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]