Question #255913

Use a truth table to verify this De Morgan’s law:

¬(p ∧ q) ≡ ¬p ∨ ¬q


Expert's answer

Let us use a truth table to verify this De Morgan’s law:

¬(p∧q)≡¬p∨¬q.¬(p ∧ q) ≡ ¬p ∨ ¬q.


We have the following trush table:


pqp∧q¬p¬q¬(p∧q)¬p∨¬qFFFTTTTFTFTFTTTFFFTTTTTTFFFF\begin{array}{||c|c||c|c|c|c|c||} \hline\hline p & q & p\land q & \neg p & \neg q & \neg(p\land q) & \neg p\lor\neg q\\ \hline\hline F & F & F & T & T & T & T \\ \hline F & T & F & T & F & T & T \\ \hline T & F & F & F & T & T & T\\ \hline T & T & T & F & F & F & F \\ \hline\hline \end{array}


Since the last two columns are coinside, we conclude that the De Morgan’s law is indeed true.


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