Can you distribute implication statements?

Question

Logically, the following statement makes sense in my head, but I can't find a law or proof that gives this example or explains this.

$$ ((p \lor q) \to r) \;\;\text{is equivalent to}\;\; ((p \to r) \lor (q \to r)) $$

I would think this is sort of like the distributive property, but I cant find a law that allows this (or I keep skipping over it in my notes).

It is equivalent to $(p\implies r$ AND $q\implies r)$, while distributing OR gets changed to AND and AND gets changed to OR. You can demonstrate this equivalence by simply using the equivalence $A\implies B\Leftrightarrow A'B$ and De Morgan's laws. — Shubham Johri, Sep 06 '21 at 17:29
Please include some information about what you have tried in order to prove this. For example, have you tried proving it via truth tables? — Greg Nisbet, Sep 06 '21 at 17:32
Also, this question might be a duplicate of this question. The linked question is mainly about conjunction, but it covers disjunction as well. — Greg Nisbet, Sep 06 '21 at 17:34
Hint: it is a tautology that if $p_1 \to p_3 ; \to ; (p_2 \to p_3) \to (p_1 \lor p_2 \to p_3)$. Show that if the LHS of the equivalence is false then the tautology will be and it will also imply that the RHS is false, and also that if the RHS is true of the equivalence is false then the tautology is also, and it will imply that the LHS is also false. Proving that it is a logical equivalence. — , Sep 07 '21 at 07:15

score 1 · Answer 1 · edited Sep 06 '21 at 23:40

(Elaborating on the first two comments under the question.)

$$(p \lor q) \to r$$ means “the truth of either of $p$ or $q$ guarantees the truth of $r$”, whereas $$(p \to r) \lor (q \to r) $$ means “both $p$ and $q$ must be true to guarantee $r$'s truth” (having only one of $p$ and $q$ true is insufficient to guarantee $r$'s truth because its corresponding argument of the disjunction might be false).
$${\quad(p \lor q) \to r\\\equiv\lnot(p \lor q)\lor r\\\equiv(\lnot p\land\lnot q)\lor r\\\equiv (\lnot p\lor r) \land (\lnot q\lor r)\\\equiv (p \to r) \land (q \to r)\\~\\\quad(p\land q)\to r \\\equiv \lnot(p\land q)\lor r\\\equiv (\lnot p\lor\lnot q)\lor r\\\equiv (\lnot p\lor r)\lor(\lnot q\lor r)\\\equiv (p\to r)\lor(q\to r)}$$

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