I came across resolution inference rule stating:
$((p\lor q)\land (\lnot p\lor r))\rightarrow(q\lor r)$
I googled a lot but what I get is either the proof using truth table or using this to prove others.
Then I tried something like this:
$LHS \equiv (p\lor q)\land (\lnot p\lor r)$
$\equiv (p\land \lnot p)\lor(p\land r)\lor (q\land \lnot p)\lor (q\land r)$
$\equiv (p\land r)\lor (q\land \lnot p)\lor (q\land r)$
But I cannot move further.
I also tried to prove the whole statement to true:
$(p\lor q)\land (\lnot p\lor r) \leftrightarrow (q\lor r)$
$\equiv \lnot((p\lor q)\land (\lnot p\lor r)) \lor (q\lor r)$
$\equiv \lnot(p\lor q)\lor \lnot(\lnot p\lor r) \lor (q\lor r)$
$\equiv (\lnot p\land \lnot q)\lor (p\land \lnot r) \lor (q\lor r)$
But I am unable to solve this further to equate this to TRUE. Where I am going wrong?
Best Answer
Resolution is not an equivalence.
The easiest way to derive it is to use the tautological equivalence between : $A \to B$ and $\lnot A \lor B$.
Thus :
is equivalent to :
Then, applying Hypothetical syllogism, we get :
i.e. : $q \lor r$.