Solution to Show that -p → (q + r) and q→ (p V r) are logically equivalent. - Sikademy
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Archangel Macsika

Show that -p → (q + r) and q→ (p V r) are logically equivalent.

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Let us find the truth table for  q\to (p \lor r):


\begin{array}{|c|c|c|c|c|c|c|c|} \hline p & q & r & p\lor r & q\to(p\lor r)\\ \hline 0 & 0 & 0 & 0 & 1 \\ \hline 0 & 0 & 1 & 1 & 1 \\ \hline 0 & 1 & 0 & 0 & 0 \\ \hline 0 & 1 & 1 & 1 & 1 \\ \hline 1 & 0 & 0 & 1 & 1 \\ \hline 1 & 0 & 1 & 1 & 1 \\ \hline 1 & 1 & 0 & 1 & 1 \\ \hline 1 & 1 & 1 & 1 & 1 \\ \hline \end{array}


It follows that the truth value of |q\to (p \lor r)|=0 if and only if |p|=|r|=0|q|=1. On the other hand, for |p|=|r|=0|q|=1 we have that |q+r|=1+0=1, and therefore by definition of implication, |-p → (q + r)|=1. Consequently, the formulas -p → (q + r) and q→ (p\lor r) are not logically equivalent.

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