I have been trying to prove that these two logical statements are equivalent using only standard logic equivalence transformations.
(p ∧ q) → r and (p → r) ∨ (q → r)
The following is my working however I'm certain I have made an error somewhere.
(p ∧ q) → r and (p → r) ∨ (q → r)
(p ∧ q) → r and (¬(p → ¬r)) ∨ (¬(q ∧ ¬r)
(p ∧ q) → r and (¬(p ∨ q)) ∧ ¬r)
¬(p ∧ q) ∨ r and (¬(p ∨ q)) ∧ ¬r)
¬p ∨ ¬q) ∨ r and (¬(p ∨ q)) ∧ ¬r)
¬(p ∨ q) → ¬r and ¬(p ∧ q) ∨ r
$a\to b$ is equivalent to $(\neg a)\vee b$. So $(p\to r)\vee(q\to r)$ is equivalent to $((\neg p)\vee r)\vee((\neg q)\vee r)$, which is equivalent to $(\neg p)\vee(\neg q)\vee r$. You've already shown that the other expression is equivalent to this.