Can I determine $(\forall x \in \mathbb{N}) [P(x) \implies (Q(x) \lor \lnot R(x))]$ without using a counter-example? Consider the following open propositions:

$P(x) : 2 < x  \le 10 $
$Q(x) : x$ is odd
$R(x) : x$ is prime

Determine the truth value of

$(\forall x \in \mathbb{N}) [P(x) \implies (Q(x) \lor \lnot R(x))]$

What I did was simplify it to

$(\forall x \in \mathbb{N}) [\lnot P(x) \lor Q(x) \lor \lnot R(x)]$

So now all I have to do is show a number in $\mathbb{N}$ that turns $F_0$ for all three propositions. In this case, $4$ works, since
$\lnot P(4) \equiv F_0$ because $P(4) \equiv V_0$ since $2 < 4 \le 10$
$Q(4) \equiv F_0$ because $4$ is even
$R(4) \equiv F_0$ because $4$ is divisible by $2$
However, is it possible to determine the truth value of this without using a counter-example? I know it is easier to just show the counter-example - but I'm not sure how would I do it in a different way.
 A: The statement is actually true: it says that if $n$ is an integer between $3$ and $10$ inclusive, then $n$ is odd or $n$ is composite. Since every even integer greater than $2$ is composite, the statement is true.
You went astray in testing $4$: while $R(4)$ is indeed false, it’s $\neg R(n)$ that you have to falsify, not $R(n)$.
A: Yes, you don't have to use a counter-example here.
For many conditional statements you can assume the antecedent and then show that the consequent follows.  Here you can assume that x belongs {3, 4, 5, 6, 7, 8, 9, 10}.  Then, does it follow that x will either qualify as odd or not prime?  Well, if x belongs to {3, 5, 7, 9}, then x qualifies as odd.  So, x qualifies as odd or not prime.  If x belongs to {4, 6, 8, 10}, then x is not prime as each of the members of that set has a divisor other than 1 or itself.  So, x qualifies as odd or not prime in these cases also.  Since we've covered all cases, the conditional "if x belongs to {3, 4, ...}, then ..." holds as true.
