- ∀x ∈ Z, P(x) → Q(x)
- ∃x ∈ Z such that Q(x) ∧ ∼ R(x)
- ∃x ∈ Z such that P(x) → S(x)
(1) The first one base on the rule "For all x, A(x)" "There exist x such that not A(x)" So what I did is "∃x ∈ Z such that P(x) → Q(x) is false."
(2) For the second, base on the rule "There exists x such that A(x)" "For every x, not A(x)" So what I did is "∀x ∈ Z, Q(x) ∧ ∼ R(x) is false" (no such that at here?)
(3) For the third, also base on the rule above, it should be "∀x ∈ Z, P(x) → S(x) is false"
I am not sure whether it is correct or not, could anyone help me to check it? Thank you!
In all three cases, you need to work the negation 'further in'.
For example, for the first one, you go from
$\neg \forall x \in \mathbb{Z} (P(x) \rightarrow Q(x))$
To
$\exists z \in \mathbb{Z} \neg (P(x) \rightarrow Q(x))$
But you can bring the negation further in, since in general $\neg (P \rightarrow Q) \Leftrightarrow P \land \neg Q$, and so you get:
$\exists z \in \mathbb{Z} (P(x) \land \neg Q(x))$
Can you do the other two?