Show that (p∧q=⇒ r and (p=⇒r)∧(q=⇒ r) are not logically equivalent.

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$\begin{matrix}p & q & r & p\land q & p \to r & q \to r & (p\land q ) \to r& (p \to r)\land (q \to r)\\ T & T & T & T & T & T & T & T\\ T & T & F & T & F & F & F & F\\ T & F &T &F & T & T & T & T\\ T & F & F &F &F &T & T & F\\ F&T&T& F&T&T&T&T\\ F&T&F&F&T&F&T&F\\ F&F&T&F&T&T&T&T\\ F&F&F&F&T&T&T&T \end{matrix}$

$(p∧q)⇒ r \; \textrm{and }(p⇒r)∧(q⇒ r)$ are not logically equivalent, because the last two columns of the truth table do not contain the same truth value in each raw.

Show that (p∧q=⇒ r and (p=⇒r)∧(q=⇒ r) are not logically equivalent.

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Question ID: mtid-5-stid-8-sqid-3745-qpid-2444