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Make a truth table for the conditional statement
~(q + ~p)

1 Answer

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Answer:

see below

Explanation:

The statement is equivalent to ~q·p (by DeMorgan's theorem).


\begin{array}{ccc}p&q&\\eg(q+\\eg p)\\T&T&F\\T&F&T\\F&T&F\\F&F&F\end{array}

User Roman Byshko
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