Is "$(\exists x \neg P(x) \wedge(\exists y \neg Q(y))$" equivalent with "$\exists x,y \neg P(x) \wedge \neg Q(y)$?"

Question

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.

Answer 1

Please note that this Answer addresses only the portion of the OP before the Addendum-Update.

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$$(\exists x \neg P(x)) \wedge(\exists y \neg Q(y))\tag A$$ $$\exists x \neg P(x) \wedge\exists y \neg Q(y)\tag {A rewrite}$$ $$\exists x,y \neg P(x) \wedge \neg Q(y)\tag {B1}$$ $$\bigg(\exists x,y \neg P(x)\bigg) \wedge \neg Q(z)\tag {B1 rewrite}$$ $$\exists x,y\bigg( \neg P(x) \wedge \neg Q(y)\bigg)\tag {B2}$$ $\text A$$\equiv$$\text {B2}$$\not\equiv$$\text B1.$

Translation of $\text{B2}:$ Some dog is not a canine and some lion is not a cat.

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$$(\forall x P(x)) \vee (\forall y Q(y))\tag{NA}$$ $$\forall x P(x) \vee \forall y Q(y)\tag{NA rewrite}$$ $$\forall x,y P(x) \vee Q(y)\tag{NB1}$$ $$\bigg(\forall x,y P(x)\bigg) \vee Q(z)\tag{NB1 rewrite}$$ $$\forall x,y \bigg(P(x) \vee Q(y)\bigg)\tag{NB2}$$ $\text {NA}\equiv \text{NB2}\not\equiv \text{NB1}.$

Translation of $\text{NB2}:$ Every dog is a canine or every lion is a cat.

Answer 2

UPDATE To add more context, the statement we were trying to prove is (see here)

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.

Either $xy=y$ for all $x,y\in S$ or $xy=x$ for all $x,y\in S$

Here is the translation: $$\forall\, x,y\in S\:\,P(x,y) \;\;\lor\;\; \forall\, w,z\in S\:\,Q(w,z).\tag1$$

Which seemed very different to

For all $x,y\in S$ either $xy = y$ or $xy = x$.

Here is the translation: $$\forall\, x,y\in S\,\bigg(P(x,y)\lor Q(x,y)\bigg).\tag2$$

You are correct that $$(1)\not\equiv(2).$$

The latter is of the form $$\forall x P(x) \vee Q(x)$$

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 $(2).$

My friend's negation of the first statement is

There exist $x,y,z,w \in S$ such that $xy \neq x$ and $zw \neq w,$

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.

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!

Answer 3

Saying $$(\lnot \forall x) \phi(x) \; \land \; (\lnot \forall y) \psi(y)$$ Is equivalent to $$(\lnot \forall x) (\lnot \forall y)(\phi(x) \; \land \; \psi(y))$$ Is only true, if $\{y\} \subseteq$ free($\phi$) is not true. If $y$ was free in $\phi$ then quantifying over it would not make it equivalent, and the same thing goes with $x$ and $\psi$. Also, there are cases that use quantifier elimination to find an equivalent free formula for a quantified formula too (this is applicable to all formulae in Prenex Normal Form [i.e. even if it is not the case that they're equivalent, then it can be resoled]). Therefore, they're equivalent (as you can observe that the bounded variables don't change).