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Consider the example

$\exists a, b \in Z$ such that $P(a, b)\rightarrow Q(a, b)$.

I follow the rules for negating universally quantified statements: I reverse the quantifier and I negate the inside:

$\neg \big(\exists a, b \in Z$ such that $P(a, b)\rightarrow Q(a, b)\big)=\forall a, b \in Z, \neg \big(P(a, b)\rightarrow Q(x, y)\big)$.

Following the rules for negating conditionals, I get:

$\neg \big(\exists a, b \in Z$ such that $P(a, b)\rightarrow Q(a, b)\big)=\forall a, b \in Z, \big(P(a, b)\wedge\neg Q(x, y)\big)$.

But let's look at a particular example. Take $P$ and $Q$ to represent the following predicates:

$P(a, b)=a$ is even or $b$ is even.

$Q(a, b)=a\cdot b$ is odd.

Then my original statement, $\exists a, b \in Z$ such that $P(a, b)\rightarrow Q(a, b)$, translates to

“There exists a pair of numbers $a$ and $b$ such that if $a$ or $b$ is even, then their product is odd.”

This statement is false. No such pair exists. So its negation ought to be true. But the negation I worked out above, $\forall a, b \in Z, \big(P(a, b)\wedge \neg Q(x, y)\big)$, translates to

“For all pairs of numbers $a$ and $b$, $a$ or $b$ is even and their product is even.”

But this is also false! What is going wrong here?

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But let's look at a particular example. Take $P$ and $Q$ to represent the following predicates:

$P(a, b)=a$ is even or $b$ is even.

$Q(a, b)=a\cdot b$ is odd.

Then my original statement, $\exists a, b \in Z$ such that $P(a, b)\rightarrow Q(a, b)$, translates to

“There exists a pair of numbers $a$ and $b$ such that if $a$ or $b$ is even, then their product is odd.”

This statement is false. No such pair exists.

Actually, the statement is true! Here is why: you can easily pick a pair of numbers $a$ and $b$ that are not both even (e.g. $a=b=1$). That means that $P(a,b)$ is false, and hence $P(a,b) \rightarrow Q(a,b)$ is true. In other words, there is a pair of numbers such that if $a$ or $b$ is even, then their product is odd.

So note what happens here is that I make the conditional trivially true by making sure its antecedent is false. This is in fact why you rarely see an existentially quantified conditional, since it's often easy to make the conditional trivially true by picking something for the existentially quantified variable, and hence the statement ends up saying nothing of interest.

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  • $\begingroup$ Thank you! We were both looking at this thinking, we're following all the rules, what's going wrong? But the only thing going wrong was that we weren't giving enough thought to how dumb the original statement was. $\endgroup$ – thecommexokid Feb 10 at 22:08

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