https://photos.app.goo.gl/u5CTGAINE7vrsLen2
$\because$ $x$ and $y$ are universally quantified variable, we can write the given arguments as follows :-
- $\forall x(Q(x)\rightarrow (P(x)\ V \sim R(a)))$
- $(R(a)\ \Lambda \sim Q(a))$
- $Q(a)$
- $\forall y (\sim P(y))$
Now using Universal instantiation, 1. becomes
- $ Q(a)\rightarrow (P(a)\ V \sim R(a))$ where $a$ is an arbitrary constant given in question.
Similarly 4. becomes
- $ \sim P(a)$
Using Modus Ponens 5. and 3.
$Q(a)\rightarrow (P(a)\ V \sim R(a))$
$\underline {Q(a)\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ }$
- $\therefore\ P(a)\ V \sim R(a)$
Using resolution 7. and 2.
$R(a)\ V \sim Q(a))$
$\underline { \sim R(a)\ V P(a))\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ }$
- $\therefore\ P(a)\ V \sim Q(a)$
Using 6. and 8.
$ P(a)\ V \sim Q(a)$
$\underline{\sim P(a) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ }$
$\therefore\ \sim Q(a)$
After applying appropriate rules of inference, at last we get $\sim Q(a)$, which is inconsistent
(as it will give True if $Q(a)$ is false and vice versa.
$\therefore$ always true/always false is not possible).