# How to prove logical equivalence: ((R OR P) -> (R OR Q)) <-> (NOT R -> (P -> Q))?

I've tried making the antecedent and the conclusion implications on the LHS expression, and I've tried making the implication into an OR but I'm stumped as to how to derive logical equivalence from the LHS expression.

Original Expression: $$((R \vee P) \Rightarrow (R \vee Q)) \Leftrightarrow (R \Rightarrow (P \Rightarrow q))$$

Attempt One: $$(R \vee P) \Rightarrow (R \vee Q)$$ $$(\neg R \Rightarrow P) \Rightarrow (\neg R \Rightarrow Q)$$ Then I'm stuck.

I've also tried: $$(R \vee P) \Rightarrow (R \vee Q)$$ $$\neg(R \vee P) \vee (R \vee Q)$$ $$(\neg R \wedge \neg P) \vee (Q \vee R )$$ $$(\neg R \Rightarrow P) \vee (\neg R \Rightarrow Q )$$

And stuck again

• Can you please add what you did to your post? – Bram28 May 22 '18 at 2:26

$(R \lor P) \to (R \lor Q)\\ \lnot (R \lor P) \lor (R \lor Q)\\ (\lnot R \land\lnot P) \lor (R \lor Q)\\ ((\lnot R \land \lnot P) \lor R) \lor Q\\ ((\lnot R \lor R) \land (\lnot P \lor R)) \lor Q\\ (\top \land (\lnot P \lor R)) \lor Q\\ (\lnot P \lor R) \lor Q\\ R \lor (\lnot P \lor Q)\\ R \lor (P \to Q)\\ \lnot \lnot R \lor (P \to Q)\\ \lnot R \to (P \to Q)$

How I thought through this problem was essentially I saw that there were two R's in one side of the statement and only one in another, so they must cancel out at some point. So I worked towards only getting one R and then did some cleanup

• Some additional parentheses wouldn't hurt (in appropriate places) and might well help. See the introduction to posting mathematical notation, which also allows one to use logical operators of a more conventional kind. – hardmath May 22 '18 at 3:09

LHS:

$(R \lor P) \rightarrow (R \lor Q) = \neg (R \lor P) \lor (R \lor Q)= (\neg R \land \neg P) \lor R \lor Q = ((\neg R \lor R) \land (\neg P \lor R) \lor Q= (\top \land (\neg P \lor R)) \lor Q = (\neg P \lor R) \lor Q= \neg P \lor R \lor Q$

RHS:

$\neg R \rightarrow (P \rightarrow Q) = \neg \neg R \lor (\neg P \lor Q) = R \lor \neg P \lor Q= \neg P \lor R \lor Q$