Question:
Prove that the following clause is a true constant clause: $ (p \wedge q) \Rightarrow p $
Answer
Based on the formulas: $ \begin {array} {l}
p \Rightarrow q \equiv \bar p \vee q \\
p \wedge q \equiv \bar p \vee \bar q \\
\bar p \Rightarrow \bar q \equiv p \wedge bar q
\end {array} $
And De Morgan’s theorem to prove or use the truth table to verify.
We have
$ \begin {array} {l}
\left ({p \wedge q} \right) \Rightarrow p \equiv \overline {p \wedge q} \vee p
\equiv \bar p \vee \bar q \vee p
\equiv \bar p \vee p \vee \bar q = 1
\end {array} $
So: $ \left ({p \wedge q } \right) \Rightarrow p = 1 $