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

Sequences, counting (including probability), logic and truth tables, algorithms, number theory, set theory, etc.
carl89
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### Prove that (A => ~B) & (B => A) = ~B

(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
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(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
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Joined: Thu Sep 19, 2013 1:38 am
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### Re: Prove that (A => ~B) & (B => A) = ~B

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]```
[/size]

LAW OF ALGEBRA FROM MY BOOK

stapel_eliz
Posts: 1628
Joined: Mon Dec 08, 2008 4:22 pm
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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]```
[/size]
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
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### Re: Prove that (A => ~B) & (B => A) = ~B

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
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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
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Joined: Thu Sep 19, 2013 1:38 am
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### Re: Prove that (A => ~B) & (B => A) = ~B

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.

stapel_eliz
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For interested readers:

Assume B. Then (by "if B, then A") A. Then (by "if A, then not-B") not-B.
Assume not-B. Then (by definition) not-B.

Assume A. Then (by "if A, then not-B") not-B.
Assume not-A. Then (by contrapositive of "if B, then A") not-B.

No matter what you start with on the left-hand side, you end up with the right-hand side.

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