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


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*$\exists x \lnot (\exists y A)$

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

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


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*$\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.
 A: The first two work. Not the third.


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

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

*$\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: $$\lnot\,(\,\lnot (\forall \,x)\exists y\, A ) \quad \equiv \quad\forall x\, \exists y\,A\,$$
A: Since this appears to be homework, I will offer an approach to the solution, rather than a solution. You should consider two questions:


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

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

A: (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 
$A=\{\}$. 
