Prove that (A => ~B) & (B => A) = ~B

[carl89]

(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.

[stapel_eliz]

(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.

[carl89]

PRECEDENCE RULES HIGHEST TO LOWEST: ~, &, V(OR), =>, <=>

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]

LAW OF ALGEBRA FROM MY BOOK

[stapel_eliz]

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]

So you've proved the statement to be not true? Is this what you were supposed to do?

Also, by what rule did you replace the "if-then" statements?

[carl89]

Im only trying to prove that Left hand side of the statement is logically equivalent to ~B. I tried the truth table method and it worked just fine.

[stapel_eliz]

I tried the truth table method and it worked just fine.

If you're allowed to use a truth table, then I'd stick with that!

[carl89]

truth table method is not allowed it was just for confirmation so i have to go to this long way and keep getting stuck at same spot every time.