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

Postby carl89 » Thu Sep 19, 2013 1:46 am

(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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stapel_eliz
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Postby stapel_eliz » Thu Sep 19, 2013 12:17 pm

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.

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

Postby carl89 » Thu Sep 19, 2013 2:05 pm

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
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stapel_eliz
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Postby stapel_eliz » Thu Sep 19, 2013 7:14 pm

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]

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

Postby carl89 » Fri Sep 20, 2013 1:15 am

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.

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stapel_eliz
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Postby stapel_eliz » Fri Sep 20, 2013 10:33 am

carl89 wrote: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! :wink:

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

Postby carl89 » Fri Sep 20, 2013 2:05 pm

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. :oops:

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Postby stapel_eliz » Wed Mar 19, 2014 3:30 pm

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. :wink:


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