Disproving a predicate-logic claim

Question

The following question is taken from UCLA's Logic quiz:

Let $P(n,m)$ be a property about two integers $n$ and $m$. If we want to disprove the claim "There exists an integer $n$ such that $P(n,m)$ is true for all integers $m$", then we need to prove that ___?

I understand that the correct answer is

For every integer $n$, there exists an integer $m$ such that $P(n,m)$ is false.

I am not sure why this option is incorrect:

There exists an integer m such that $P(n,m)$ is false for all integers $n.$

After all, if I show the existence of an $m^*$ such that $P(n, m^*)$ is false for all $n,$ then there cannot exist an $n^*$ such that $P(n^*,m)$ is true for all $m$. What am I missing?

Answer 1

As an example, suppose $P(n,m)$ is "$n+m$ is even."

Then the statement to refute is

There exists an integer $n$ such that (($n+m$ is even) for all integers $m$).

This is indeed false. I'm adding parentheses to avoid ambiguities in English language meaning.

The correct negation is

For every integer $n$, (there exists an integer $m$ such that ($n+m$ is not even)).

This is true.

Your other statement would be

There exists an integer $m$ such that (($n+m$ is not even) for all integers $n$).

This is false! No single $m$ will make $n+m$ always odd for all possible $n$.

The order of quantifiers matters. In the correct negation, we need one $m$ for each $n$ which makes $P(n,m)$ false, but $m$ can depend on $n$. Your other statement claims there's one $m$ which will make $P(n,m)$ false no matter what $n$ is. That's a stricter condition, not in general equivalent.

Answer 2

If we want to disprove the claim "There exists an integer $n$ such that $P(n,m)$ is true for all integers $m$", then we need to prove that ___?

I understand that the correct answer is

For every integer $n$, there exists an integer $m$ such that $P(n,m)$ is false.

I am not sure why this option is incorrect:

There exists an integer m such that $P(n,m)$ is false for all integers $n.$

Since $$∃m∀n¬P(n,m) \models ∀n∃m¬P(n,m)\\∀n∃m¬P(n,m)\not\models∃m∀n¬P(n,m),$$

• proving that 'incorrect' option is indeed sufficient to prove that 'correct' option thereby disproving the given claim;
• whatever $P(n,m)$ stands for, that 'incorrect' option is at least as strong as that 'correct' option, and to prove the latter (thereby the given claim), it is not necessary to prove the former; that is, we don't need to prove the former.

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P.S. This logic quiz really should have been checked for unintentionally ambiguous sentences (hanging quantifiers) like these:

"There exists an integer $n$ such that $P(n,m)$ is true for all integers $m$"

• [There exists an integer $n$ such that $P(n,m)$ is true] for all integers $m$

or

• There exists an integer $n$ such that [$P(n,m)$ is true for all integers $m$]

I am not sure why this option is incorrect:

There exists an integer m such that $P(n,m)$ is false for all integers n.

• [There exists an integer m such that $P(n,m)$ is false] for all integers $n.$

or

• There exists an integer m such that [$P(n,m)$ is false] for all integers $n.$]