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Math Help - Logically equivalent

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    Logically equivalent

    Use a truth table to show that p => (q^~q) is logically equivalent to ~p?

    Well I've done the truth tables and I don't see how they are logically equivalent for p => (q^~q) final values are F, F, T, T and ~p values are F, T. So they're not equivalent? Note that the => equals the conditional arrow and ~ is negation and ^ is 'and'.
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    Quote Originally Posted by brumby_3 View Post
    Use a truth table to show that p => (q^~q) is logically equivalent to ~p?

    Well I've done the truth tables and I don't see how they are logically equivalent for p => (q^~q) final values are F, F, T, T and ~p values are F, T. So they're not equivalent? Note that the => equals the conditional arrow and ~ is negation and ^ is 'and'.
    q^~q is always F as it is a contradiction. So we look at two rows in the truth table for p -> q. The rows are: T F F and F F T. (p is true, q is false gives (p -> q) is false. p is false, q is false gives (p -> q) is true.) So p -> (q^~q) is true if and only if p is false.
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