$P(x): x ≤ 0,$
$Q(x): $x^2$ = 1,$
$R(x): x $ is odd,
$S(x): x = x + 1.$
Statement:
$∀x ∈ Z, S(x) → R(x) \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;(1)$
$∃x ∈ Z$ such that $Q(x) ∧ ∼ R(x) \;\;(2)$
$∃x ∈ Z$ such that $P(x) → S(x) \;\;\;\;(3)$
I try to let $x$ in $S(x)$ is even and odd to prove $S(x) → R(x)$, $x = (2n+1) + 1 =2n +2$ this is an even, so this statement is false?
The true set of $Q(x)$ is $\{-1, 1\}$, the set is not satisfactory for $~R(x)$, because $~R(x)$ is an even. Thus this is false.
I have no idea about this one, I find $P(x)\leq 0$, but there no result is satisfactory for $S(x)$, so this is false..?
The first statement, $$∀x ∈ Z, S(x) → R(x)$$ is true because $S(x)$ is always false.
You are correct on the second one.
The third statement is also true. You may pick $x=2$ for example.