Prove: $((P ⇒ R) ∧ (Q ⇒ R)) ⇔ ((P ∨ Q) ⇒ R)$
Notation: $P[x]$ and $f[x$] for “$P$ applied to $x$” and “$f$ applied to $x$”, instead of $P(x)$ and $f(x)$.
So far I tried the following:
Direction $=>$
Assume ( (P => R)^ (Q => R) ) (1) and Prove ( (P v Q)=> R )
From (1) we get (P => R)(2) and (Q => R)(3)
From (2) we get P (4) From (3) we get Q (5)
Direction $<=$
Assume ( (P v Q) => R ) (6) and Prove ( (P => R) ^ (Q => R) )
From (6) by material implication we get ~(P v Q) v R (7)
We can rewrite (7) as (~P ^ ~Q) v R (8) by negation rules
We can retwrite (8) as (~P v R) ^ (~Q v R) (9) by distributivity laws
We can rewrite (9) as (P => R ) ^ (Q => R) by material implication.
I got stuck solving first direction $=>$. What should I do next?
Regarding the other direction $<=$, is there any other way to do it without rewriting?
Thanks.
