Logical Equivalence: Show that R OR P implies R OR Q is equivalent to NOT R implies (P implies Q)?

Question

I'm practicing logical equivalence and I've come across a question that I'm struggling to answer:

Show that (R or P -> R or Q) is equivalent to (not R -> (P -> Q)).

I've examined the truth tables of both implications but the question states that I should use equivalence laws to show that the implications are equivalent.

If anyone could help me out, I would appreciate it.

Thank you.

Answer 1

Intuitive

A formal proof (included below) which only allows one to follow the steps one by one is less useful than a proof that helps us understand why both expressions are equivalent. Consider the first expression:

(R or P) -> (R or Q)

and think about its meaning...

The expression is trivial when R=true, isn't it? Therefore the only information it encloses is that when R=false, P -> (R or Q). But when R=false, (R or Q) = Q. So, the precise meaning of the expression is that when R=false, P -> Q. In other words, not R -> (P -> Q).

Formal

(R or P) -> (R or Q) = not (R or P) or (R or Q)             ;X -> Y = not X or Y
                     = (not R and not P) or (R or Q)        ;not (X or Y) = not X or not Y
                     = ((not R and not P) or R) or Q        ;X or (Y or Z) = (X or Y) or Z
                     = ((not R or R) and (not P or R)) or Q ;(X and Y) or Z = (X or Z) and (Y or Z)
                     = (not P or R) or Q                    ;(not X or X) = true
                     = (R or not P) or Q
                     = R or (not P or Q)
                     = R or (P -> Q)
                     = not (not R) or (P -> Q)
                     = not R -> (P -> Q)                    ;not X or Y = X -> Y