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October 28th, 2011, 10:00 AM   #1
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Truth Table problem

Hello, I'm currently working on a truth table, and I seem to have a problem with grasping some of the concepts.
I'm trying to figure this out: (p -> q) <-> (-q -> -p). If you're used to different symbols, this is the text version: (p if then q) if and only if (not q if then not p)

p q p -> q -q => -p (p -> q) <-> (-q -> -p)
0 0 1 1 1
0 1 1 1 0
1 0 0 0 0
1 1 1 1 1

Am I right, or way off?

-Akhanoth
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October 28th, 2011, 10:12 AM   #2
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Re: Truth Table problem

p q.........p -> q.......-q => -p.........(p -> q) <-> (-q -> -p)
0 0.........1.................1..................... ....1
0 1.........1.................1..................... ....0
1 0.........0.................0..................... ....0
1 1.........1.................1..................... ....1

had to do it like this, couldn't get the multiple spaces to work =p
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October 28th, 2011, 11:29 AM   #3
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Re: Truth Table problem

Your method is right, and your first two (or four if you count the variables) columns are right. But the last column is p <-> q when it should be (p -> q) <-> (-q -> -p). Just put a 1 when the two columns immediately to the left are the same and a 0 when they are different.
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October 28th, 2011, 11:44 AM   #4
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Re: Truth Table problem

Thanks a lot for the help!

Does that go for every situation? Or simply because it's a <-> in between?
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October 28th, 2011, 06:10 PM   #5
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Re: Truth Table problem

Hello, Akhanoth!

Quote:

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October 29th, 2011, 07:38 AM   #6
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Re: Truth Table problem

Quote:
Originally Posted by Akhanoth
Thanks a lot for the help!

Does that go for every situation? Or simply because it's a <-> in between?
In every situation you can reduce a big problem to smaller problems and fill out the truth table by combining two existing columns with some standard operation. In this case the operation was <-> "are the two columns the same". Was the operation something else you'd do something else.
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