show that $(p \to q) \vee (p \to r) \to (q \vee r)$ and $p\vee q\vee r$ are logically equivalent

Question

without using the truth table:

Show that

$(p \to q) \vee (p \to r) \to (q \vee r)$ and $p\vee q\vee r$ are logically equivalent.

Answer 1

Your answer... \begin{equation*} \begin{split} (p\to q)\vee (p \to r)\to (q\vee r) & \equiv \neg [(\neg p\vee q)\vee (\neg p \vee r)]\vee (q\vee r) \quad \text{by implication rule}\\ &\equiv \neg[(\neg p\vee( q\vee r)]\vee (q\vee r) \quad \text{by associative rule}\\ & \equiv [( p\wedge \neg( q\vee r)]\vee (q\vee r) \quad\text{by De Morgan's rule}\\ & \equiv( p\vee ( q\vee r))\wedge (\neg(q\vee r)\vee (q\vee r))\quad\text{by distributive rule}\\ & \equiv ( p\vee ( q\vee r))\wedge T\quad\text{by negation rule}\\ & \equiv ( p\vee ( q\vee r))\quad\text{by identity}\\ & \equiv p\vee q\vee r\quad\text{by associativity rule}\\ \end{split} \end{equation*}