Equivalence of (P -> R) V (Q -> R) and (P ∧ Q) -> R

[Dysnex]

I can see how the equivalence can formulated with

(P -> R) V (Q -> R)
= (¬P V R) V (¬Q V R)
= (¬P ∧ ¬Q) V R
= ¬(P ∧ Q) V R
= (P ∧ Q) -> R
(Sorry, I would've written this in LaTeX if I were more competent.)

although I still it counter-intuitive and, at a glance, first thought it was (P V Q) -> R. I asked someone else and they also arrived at (P V Q) -> R, which seems to contradict(?) the above formulation and the book's answer key. Am I missing something?

Any help would be appreciated, thanks!

 

[tauon]

You have a typo on the third line: it's supposed to be "(¬P V ¬Q) V R" and then by DeMorgan's rule you get the 4th line ¬(P ∧ Q) V R. Maybe that was bothering you?

As for the intuitiveness of it. Think about when any of (P -> R) V (Q -> R) and (P ∧ Q) -> R are false: only when both P and Q are true but R is false; in both expressions. In all the other cases both expressions are true simultaneously. Maybe if you write truth tables for them it would be easier to see it.

 

[gnarly]

If p or q are false the statement (P^Q)-> R will always be true. The same can be said for the statement (P -> R) V (Q -> R). If the first statement in P->Q is false then P ->Q will always be true.

 

[JeremyHorne]

tauon is right about the typo. Adding a step or so will reveal why:

Original -

(¬P V R) V (¬Q V R)
= (¬P ∧ ¬Q) V R

New -

(¬P V R) V (¬Q V R)
(¬P V (R V ¬Q) V R) Association
(¬P V (~Q V R) V R) Commutation
(¬P V ~Q) V (R V R) Association
(¬P V ~Q) V (R) Idempotence

The conjunct doesn't even enter into this up to here. Now, go on to DeMorgan etc., as tauon says.