Logic counterexample to argument

Question

$\sim \forall x \; (M(x) \vee W(x))$
$\exists x \; \sim M(x)$

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$\exists x \; \sim W(x)$

This argument is invalid, could any one please provide a counterexample?

The first two lines are the premise, the sentence after the gap is the conclusion

Answer 1

This argument is actually valid. Let $M(x)$ denote "$x$ is a man" and $W(x)$ "$x$ is a woman". The first premise says that it is not the case that everyone is a man or a woman; in other words there is someone who's neither man nor woman. So even without the second premise, the conclusion, which says there is someone who's not a woman, must hold. (Whoever is neither man nor woman must, in particular, not be a woman.)

Answer 2

$\sim \forall x \sim$ is equivalent to $\exists x$ so

$$\sim \forall x \; (M(x) \vee W(x)) = \sim \forall x \; \sim \sim (M(x) \vee W(x)) = \exists x \; \sim(M(x) \vee W(x))$$ $$= \exists x \; (\sim M(x)) \wedge (\sim W(x))$$

Therefore, there is some $x$ for which $M(x)$ and $W(x)$ fail.

In words, if we interpret $M(x)$ to mean "$x$ is a man." and $W(x)$ to mean "$x$ is a woman." The top line means that not everyone is either a man or woman. That is someone has neither gender.

The second line means there is someone who isn't a man.

The third line means there is someone who isn't a woman.

This does follow from the first line since there is someone who isn't either man or woman.

Edit: I am unfamiliar with "fitch". But a proof could go something like what I have above...

$$\sim \forall x \; (M(x) \vee W(x))$$ $$\sim \forall x \; \sim \sim (M(x) \vee W(x)) \qquad \mbox{replace $P$ with $\sim \sim P$}$$ $$\exists x \; \sim(M(x) \vee W(x)) \qquad \sim \forall \sim \mbox{ is equivalent with } \exists $$ $$\exists x \; (\sim M(x)) \wedge (\sim W(x)) \qquad \mbox{De Morgan's law}$$ $$(\sim M(A)) \wedge (\sim W(A)) \qquad \mbox{such an $A$'s existence was asserted.}$$ $$\sim M(A)$$ $$\sim W(A) \qquad \mbox{since both $M(A)$ and $W(A)$ must hold for conjunction to hold.}$$

Answer 3

Here's a Fitch proof (it's ugly and long partly because I had to prove instances of quantifier equivalence and De Morgan laws). I have broken it down into small steps. The premises are (1-2):

To transform (1) to the equivalent formula $\exists x\lnot(M(x) \lor W(x))$, we proceed as follows (3-13):

We then 'De Morgan' (13) into the equivalent formula $\exists x (\lnot M(x) \land \lnot W(x))$ as follows: (14-28):

Finally, we use (28) to get the desired conclusion (29-32):

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Mauro is right of course, none of this is necessary. Bill's and Quinn's arguments are equally satisfactory. You asked for a Fitch derivation, so I typed this up.