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$$\forall x\exists y \bigl(P(x)\to P(y)\bigr)\to\forall x\exists y\bigl(P(x)\to(y)\bigr)$$

Here's what I have so far, but I think it's wrong: $$\begin{align*} &\neg\Bigl( \forall x\exists y\bigl(P(x)\to P(y)\bigr) \to \forall x\exists y\bigl( P(x)\to (y)\bigr)\Bigr) &&\text{implication}\\ &\neg\forall x\exists y \bigl( P(x)\to P(y)\bigr)\lor \forall x\exists y \bigl( P(x)\to (y)\bigr) &&\text{implication}\\ &\forall x\exists y\bigl( P(x)\to(y)\bigr) \lor \neg\forall x\exists y\bigl( P(x)\to P(y)\bigr) \equiv \text{TRUE} &&\text{negation} \end{align*}$$

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4  
In the first displayed formula, and the second, and the third, what does $(y)$ mean? It is certainly non-standard notation, or a typo. –  André Nicolas Jan 17 '12 at 16:40
    
To be honest, I'm not sure, the P(y) represents a proposition with parameter y, I have no idea what the (y) by itself means, which is part of the problem I'm having. –  kylex Jan 17 '12 at 16:43
1  
Note that $\neg(P\to Q)\equiv P\land \neg Q$. Instead, you have $\neg(P\to Q)\equiv \neg P\lor Q$, which is incorrect. If your second line was meant to be $\neg(\neg P\lor Q)$, then you are missing a negation, and the negation of that would be of the form $P\land \neg Q$. –  Arturo Magidin Jan 17 '12 at 16:48

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