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$$\vdash[(\forall x)P(x)]\rightarrow[(\exists x)P(x)]$$ answer:$$\neg P(x)\to\neg P(x)$$$$by QR$$ $$ \neg P(x)\to(\forall x)\neg P(x)$$$$by QR$$ $$(\exists x) \neg P(x)\to(\forall x)\neg P(x)$$ $$\neg((\exists x) \neg P(x))\to\neg((\forall x)\neg P(x))$$ $$[(\forall x)P(x)]\rightarrow[(\exists x)P(x)]$$ could i solve with this way?

zahra
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  • Afaik this is one of the propositions which are false if the domain of the discourse is empty. In any case, it has been asked before around here, so just search on. PS: More tags $\implies$ more people see you asked a question. – Nikolaj-K Dec 02 '14 at 19:34
  • NO - in order to apply $(QR)$ : $\psi \rightarrow \phi \vdash \psi \rightarrow \forall x \phi$, we must have : $x$ not free in $\psi$. – Mauro ALLEGRANZA Dec 02 '14 at 20:18
  • In addition, the last but one step is wrong; the rule is : $p \rightarrow q \vdash \lnot q \rightarrow \lnot p$ – Mauro ALLEGRANZA Dec 02 '14 at 20:23

2 Answers2

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The correct proof, according to

is :

1) $∀xP(x)→P(t)$ --- (Q1) [see page 57]

2) $P(t)→∃xP(x)$ --- (Q2)

3) $∀xP(x) → ∃xP(x)$ --- from 1) and 2) by (PC) [see page 61] using : $p→q,q→r⊢p→r$.

Mauro ALLEGRANZA
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I don't follow your proof, but this result follows trivially if the domain of quantification is non-empty. It may be a bit more clear if you make that domain explicit as follows:

  1. Suppose $\exists x: x\in U$ where $U$ is the domain of quantification.

  2. Suppose $\forall x: [x\in U \implies P(x)]$

  3. $y\in U$ (by existential specification from 1)

  4. $y\in U \implies P(y)$ (by universal specification from 2)

  5. $P(y)$ (by detachment from 3 and 4)

  6. $y\in U \land P(y)$ (from 3 and 5)

  7. $\exists x: [x\in U \land P(x)]$ (by existential generalization from 6)

  8. $\forall x: [x\in U \implies P(x)]\implies \exists x: [x\in U \land P(x)]$ (conclusion from 2 and 7)

Dan Christensen
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