Equivalent expressions for $\lnot (\forall x)\exists y A $

Question

I need equivalent expressions for: $\lnot (\forall x)\exists y A$

1. $\exists x \lnot (\exists y A)$

2. $\exists x \lnot (\exists y) A$ The same as (1) I think

3. $\exists x \exists y \lnot A $

Which one is OK? And what would be the result for the negation: $\lnot(\lnot (\forall x)\exists y A)$?

1. $\forall x \lnot (\exists y A) $

Is that OK?

In fact, what I want is $\lnot (\forall x)\exists y A$ without any negated quantifier, if this is possible.

Answer 1

The first two work. Not the third.

1. $\lnot (\forall x)\exists y A \equiv \exists x \lnot (\exists y A)\quad$ YES!

2. $\lnot (\forall x)\exists y A \equiv \exists x \lnot (\exists y) A \quad ?\quad$ The same as (1): More or less, but stick with $(1)$

3. $\lnot (\forall x)\exists y A \not\equiv \exists x \exists y \lnot A $

Then we have from $(1)$: $$\lnot (\forall x)\,\exists y\, A \quad \equiv \quad\exists x \,\lnot (\exists y \,A) \quad \equiv \quad \exists x \,\forall y \,(\lnot A)$$

Now: $$\lnot\,(\,\lnot (\forall \,x)\exists y\, A ) \quad \equiv \quad\forall x\, \exists y\,A\,$$

Answer 2

Since this appears to be homework, I will offer an approach to the solution, rather than a solution. You should consider two questions:

1. How can you rewrite $\lnot \forall x\,A$ in terms of $\exists$?
2. How can you rewrite $\lnot \exists x\,A$ in terms of $\forall$?

Answer 3

(1) and (2) in your answer are the same and are the answer.

You can also write $\lnot\exists$ as $\nexists$.

If you pin down $A$ as a relation from domain $X$ to range $Y$, so that $A\colon X \to Y$, then you can write $A=\{\}$.