Posted by **Jen** on Wednesday, July 13, 2011 at 6:08pm.

Construct a truth table for (p ^ q) ↔ ~p

- Math-geometry -
**MathMate**, Wednesday, July 13, 2011 at 6:25pm
the truth table for bi-conditional <-> is

it is true when both p & q are true, or both are false.

p q (p^q) <-> ~p

T T T T T **F** F

T F T F F **F** T

F T F F T **F** F

F F F F F **F** T

Since in all cases, the proposition is false, it is a contradiction.

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