These statements came up when my friend and I wanted to prove a proposition by using proof by contradiction. We seemed to negate the statement differently.
My negation of the statement was of the form $$(\exists x \neg P(x)) \wedge(\exists y \neg Q(y))\tag A$$ and his was $$\exists x,y \neg P(x) \wedge \neg Q(y).\tag {B1}$$
Note here that the only difference is the quantifier. It seems intuitive to us that these statements are equivalent, but negating both of these give us $$(\forall x P(x)) \vee (\forall y Q(y))\tag{NA}$$
and $$\forall x,y P(x) \vee Q(y).\tag{NB1}$$
For example, consider
"All humans are male or all humans are female"
and
"All humans are either male or female."
Which seems to be two really different statements. What went wrong?
For context, I have not been into predicate logic for a while now.
UPDATE
To add more context, the statement we were trying to prove is (see here)
Either $xy=y$ for all $x,y\in S$ or $xy=x$ for all $x,y\in S$
Which seemed very different to
For all $x,y\in S$ either $xy = y$ or $xy = x$.
In the first statement, one of the properties needs to hold for all $x,y$. While in the second one, it need not be, since they can be mixed. I already understand the difference between these statements, thanks to the comment of Mauro ALLEGRANZA. The latter is of the form $\forall x P(x) \vee Q(x)$ rather than NB1, hence not equivalent.
My friend's negation of the first statement is
There exist $x,y,z,w \in S$ such that $xy \neq x$ and $wz \neq z,$
while I think it should be
There exist $x,y\in S$ such that $xy \neq x$ and there exist $w,z\in S$ such that $wz \neq z$.
Which one is correct? Are they both wrong? Since it seems they are most likely equivalent, I think they are both correct.
Best Answer
Please note that neither your original Question nor your Addendum-Update is an accurate translation of the linked exercise. Here, I shall address your Addendum-Update (which really is a New Question)—which may have only a tenuous link to the motivating linked exercise.
Here is the translation: $$\forall\, x,y\in S\:\,P(x,y) \;\;\lor\;\; \forall\, w,z\in S\:\,Q(w,z).\tag1$$
Here is the translation: $$\forall\, x,y\in S\,\bigg(P(x,y)\lor Q(x,y)\bigg).\tag2$$
You are correct that $$(1)\not\equiv(2).$$
No it is not. Incidentally, please understand that $$\bigg(\forall x P(x)\bigg) \vee Q(x)\quad\text{ and }\quad\forall x\bigg( P(x) \vee Q(x)\bigg)$$ have different meanings, and that your suggestion is the same as the former, not the latter. In any case, none of these three is equivalent to the correct translation $(2).$
For ease of reading, here again is the first statement: $$\forall\, x,y\in S\:\,P(x,y) \;\;\lor\;\; \forall\, w,z\in S\:\,Q(w,z).\tag1$$
And here is its negation: $$\exists\, x,y\in S\:\,\lnot P(x,y) \;\;\land\;\; \exists\, w,z\in S\:\,\lnot Q(w,z)\tag{N1}$$ $$\exists\, x,y\in S\:\,xy\neq x \;\;\land\;\; \exists\, w,z\in S\:\,wz\neq z\tag{N1}$$ $$\exists\, x,y\in S\:\, \bigg(xy\neq x \;\;\land\;\; \exists\, w,z\in S\:\,wz\neq z \bigg)\tag{N1}$$ $$\exists\, x,y,w,z\in S\,\bigg(xy\neq x \land wz\neq z\bigg)\tag{N1}$$
Since your negation attempt is exactly the second line, while your friend's is exactly the fourth line, you are both correct!