I came across this problem, it asks to use logical equivalences (see image), show that $(p → r) ∨ (q → r)$ logically equivalent to the statement $(p ∧ q) → r$ (aka definition of biconditional)
After apply $p→q ≡ ¬p∨q$, I can't go any further can anyone show me any steps further?
Thank you
On
Fill in the blanks $$\begin{align} (p\to r)\wedge (q\to r) \iff & (\neg p\vee r)\wedge (q\to r) & \textsf{Implication Equivalence} \\[1ex] \iff & (\neg p\vee r)\wedge (\neg q\vee r) &\textsf{Implication Equivalence} \\[1ex] \iff & ~ & \textsf{Distribution} \\[1ex] \iff & ~ & \textsf{Association & Commutation} \\[1ex] \iff & ~ & \textsf{De Morgan's Duality} \\[1ex] \iff & (p\vee q)\to r & \textsf{Implication Equivalence} \end{align}$$
$(p → r) ∨ ( q → r)=(¬p ∨ r) ∨ (¬q ∨ r)=¬p ∨ r ∨ ¬q ∨ r=(¬p ∨ ¬q) ∨ r=¬(p∧q)∨r=(p∧q) → r$