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I figured that implications statements can be written as:

P implies Q meaning that if P then Q

And I thought the negation of if-then statements can be written in the format: p and not q

Hence I thought that the negation would be: "xy is even and x or y is not even"

But I am not sure about the answer and would appreciate it if anyone could help.

Sammy
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  • In english, x is even or odd, but y is not parity of x unless its even. But excluding 1, -1, and 0 for any argument. Don't know if a negative number less than 0 can be considered even or odd though? – GetSet Oct 10 '20 at 02:55
  • Your answer is correct. – user400188 Oct 10 '20 at 03:01

3 Answers3

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Presumably this is quantified over $x$ and $y$, and $x$ and $y$ are integers. So the statement would read

"For all $x \in \mathbb{Z}$, for all $y \in \mathbb{Z}$, if $xy$ is even, then $x$ is even or $y$ is even."

You can write this as sentence in first order logic like: $$\forall (x \in \mathbb{Z})\forall (y \in \mathbb{Z})[Even (xy) \implies Even(x) \vee Even(y) ]$$ You can specify what $Even(n)$ means more precisely, but that is not the relevant issue right now. Then, to negate this, just put a negation sign in front: $$\neg \forall (x \in \mathbb{Z})\forall (y \in \mathbb{Z})[Even (xy) \implies Even(x) \vee Even(y) ]$$ So, you could rightfully declare that the negation is...

"It is not the case that for all $x \in \mathbb{Z}$, for all $y \in \mathbb{Z}$, if $xy$ is even, then $x$ is even or $y$ is even."

Not very helpful, right? We can put it in a different form by following the rules for moving negation signs through quantifiers $$\neg \forall (x \in \mathbb{Z})\forall (y \in \mathbb{Z})[Even (xy) \implies Even(x) \vee Even(y) ]$$ $$ \exists(x \in \mathbb{Z})\neg \forall (y \in \mathbb{Z})[Even (xy) \implies Even(x) \vee Even(y) ]$$ $$\exists (x \in \mathbb{Z})\exists (y \in \mathbb{Z})\neg [Even (xy) \implies Even(x) \vee Even(y) ]$$ "There exists $x \in \mathbb{Z}$ and a $y \in \mathbb{Z}$ such that it is not the case that if $xy$ is even, then $x$ or $y$ is even."

But we can do better and negate the conditional "if-then" if we want: $$\exists (x \in \mathbb{Z})\exists (y \in \mathbb{Z})[Even (xy) \wedge \neg [Even(x) \vee Even(y)] ]$$ "There exists $x \in \mathbb{Z}$ and a $y \in \mathbb{Z}$ such that $xy$ is even, and it is not the case that $x$ or $y$ is even." $$\exists (x \in \mathbb{Z})\exists (y \in \mathbb{Z})[Even (xy) \wedge [\neg Even(x) \wedge \neg Even(y)] ]$$ "There exists $x \in \mathbb{Z}$ and a $y \in \mathbb{Z}$ such that $xy$ is even, and $x$ is not even and $y$ is not even."

Jbag1212
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  • It should be noted that it is ambiguous whether x and y are universally quantified or whether they refer to something previously identified. – DanielV Oct 10 '20 at 03:31
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The negation is $xy$ is even and $x$ and $y$ are odd.

That's because, as you noted, $\lnot (p\implies q)$ is $ p\land \lnot q$.

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The negation would simply be "xy is even does not imply that x or y is even." The best way to go about writing negations is to add "not" before your q.

The way you wrote it doesn't really have a p or a q. There is no conclusion, but rather two variables: "xy is even", and "x or y is not even".

Hope this helps!

Crimson
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  • It took me a while to understand what you meant, but I think you are right. The answer is better phrased with "$x$ or $y$ are not even", as opposed to is, which is singular. – user400188 Oct 10 '20 at 03:05