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Prove or give counter example: $\forall x \exists y(Q(y)\wedge P(x))\vDash \exists y\forall x(Q(y)\wedge P(x))$

I think this statment is true:

$x$ and $y$ are not free, so I can use the following logical equivalences: $$\forall x(\alpha \vee \beta) \equiv \alpha \vee \forall x(\beta)$$ $$\exists x(\alpha \vee \beta) \equiv \alpha \vee \exists x(\beta)$$

$\forall x\exists y(Q(y)\wedge P(x))\Rightarrow\exists y(Q(y)\wedge\forall xP(x)) \Rightarrow \exists y(\forall xP(x)\wedge Q(y))\Rightarrow\forall xP(x)\wedge \exists yQ(y) \Rightarrow\exists y(\forall xP(x)\wedge Q(y))\Rightarrow\exists y( Q(y)\wedge\forall xP(x))\Rightarrow\exists y(\forall x( Q(y)\wedge P(x)))$

$\mathbf{\Rightarrow\exists y\forall x( Q(y)\wedge P(x))}$

$$$$ Did I get it right ? Thank you!

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Your first step:

$$\forall x\exists y(Q(y)\wedge P(x))\Rightarrow\exists y(Q(y)\wedge\forall xP(x))$$

is not an instance of the two equivalence principles you mention, since the $\exists$ is between the $\forall$ and the $\land$

What you can instead is:

$$\forall x\exists y(Q(y)\wedge P(x))\Leftrightarrow \forall x\exists y(P(x)\wedge Q(y))\Leftrightarrow\forall x (P(x)\wedge \exists yQ(y))$$

and go from there

(indeed, note that your steps 3 and 5, as well as 2 and 6, are identical ... that should have told you something went wrong!)

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  • $\begingroup$ Thank you for your answer! How do I continue from here (since you said I'm not using the quivalence principles I mention right)? $\endgroup$ – rose12 Aug 12 '18 at 20:06
  • $\begingroup$ @rose12 commute, bring the $\forall$ in, commute, bring the $\exists$ out, commute, and bring the $\forall$ out $\endgroup$ – Bram28 Aug 12 '18 at 20:10

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