$P(x) ⇒ Q(x)$
$\neg R(x) \lor \neg S(x)$
$\neg S(x) ⇒ \neg Q(x)$
$∴ P(x) ⇒ \neg R(x) $
I'm wondering how I can $P(x)$ instead of $Q(x)$. I tried using hypothetical reasoning such that we need $Q(x)$ so $S(x)$ but the question only gives $\neg S(x)$.
Logic Argument proof
0
$\begingroup$
discrete-mathematics
logic
predicate-logic
1 Answers
1
Hypothetical reasoning is a good approach!
Ok, so let's assume $P(x)$
then by the first statement it follows that $Q(x)$.
We can do Contraposition on the third statement, giving us $Q(x) => S(x)$, so with $Q(x)$ We get $S(x)$.
By the second statement, we then have to get $\neg R(x)$, since we have $S(x)$, and thus you can't have that $\neg S(x)$.
So, assuming that $P(x)$ leads to $\neg R(x)$.
So, by conditional proof: $P(x) => \neg R(x)$.