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$\begingroup$

Can someone tell me if im doing this correct?!?! im a little confused on pulling the quantifiers out to the left after i remove the "->" and distribute the negation.

$\exists xP(x)\wedge \exists yQ(y) \rightarrow \exists z(P(z) \wedge Q(z))$

$\equiv \lnot (\exists xP(x)\wedge \exists yQ(y)) \lor \exists z(P(z) \wedge Q(z))$

$\equiv \exists z\lnot (\exists xP(x)\wedge \exists yQ(y)) \lor P(z) \wedge Q(z)$

$\equiv \exists z(\lnot\exists xP(x)\lor \lnot\exists yQ(y)) \lor P(z) \wedge Q(z)$

$\equiv \exists z(\forall x \lnot P(x)\lor \forall y\lnot Q(y)) \lor P(z) \wedge Q(z)$

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    You will find it is easier to get help if you post something readable. Type it out. – 2017-02-21
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    im unsure how to get the symbols into the text, sorry im new at this – 2017-02-21
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    This may help, [Mathjax: a Basic Tutorial and Quick Reference](http://meta.math.stackexchange.com/questions/5020/mathjax-basic-tutorial-and-quick-reference) – 2017-02-21
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    ok i fixed it now – 2017-02-21
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    On the first line you have $... \to \exists z\,\big(P(\color{blue}{\,x\,})\land Q(z)\big)$ . Is that supposed to be a *free* $x$ or a $z$. It mysteriously jumps to being a $z$ on the third line, – 2017-02-22
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    yes let me fix thanks – 2017-02-22

1 Answers 1

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That seems okay, although you should keep the bracketing around the predicate for $\exists z$.

$\begin{align} &\exists x\,P(x)\land \exists y\,Q(y) ~\to~ \exists z\,(P(z) \wedge Q(z)) \\[1ex] \equiv ~& \lnot \big(\exists x\,P(x)\land \exists y\,Q(y)\big) ~\lor~ \exists z\,\big(P(z) \land Q(z)\big) && \text{Implication Equivalence} \\[1ex] \equiv ~& \exists z\,\Bigl(\lnot \bigl(\exists x\,P(x)\land \exists y\,Q(y)\bigr) \lor \bigl(P(z) \wedge Q(z)\bigr)\Bigr) && \text{quantifier distribution} \\[1ex] \equiv ~& \exists z\,\Bigl(\bigl(\lnot\exists x\,P(x)\lor \lnot\exists y\,Q(y)\bigr) \lor \bigl(P(z) \wedge Q(z)\bigr)\Bigr) && \text{de Morgan's Laws} \\[1ex] \equiv ~& \exists z~\Bigl(\forall x\,\lnot P(x)\lor \forall y\,\lnot Q(y) \lor \bigl(P(z) \wedge Q(z)\bigr)\Bigr) && \text{quantifier duality} \\[3ex] \equiv ~& \forall x\,\forall y\,\exists z~\Bigl(\lnot P(x)\lor \lnot Q(y) \lor \bigl(P(z) \wedge Q(z)\bigr)\Bigr) && \text{quantifier distribution} \\[3ex] \end{align}$

Which is DNF, to get CNF just distribute the predicate.

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    how do you get the other quantifiers to the left? – 2017-02-22
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    Same rule. Since each of their bound variables are free in the other clauses you can distribute the quantifiers safely. – 2017-02-22
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    last question, can you distribute the first two terms into the third term? – 2017-02-22