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Assume that $P$ is a 1-place predicate. We define the set of formulas $\Gamma$ like this: $$\Gamma =\{\lnot \forall x_0 P(x_0),P(x_0),P(x_1),P(x_2),...\}.$$

Is $\Gamma $ consistent?

My answer is no, and my deduction is that it is obvious that $\Gamma \vdash \forall x P(x)$ and also $\Gamma \vdash \lnot\forall x P(x)$ so $\Gamma \vdash \bot$, so it is inconsistent, but how can I deduce the part that $\Gamma \vdash \forall x P(x)$? I think it's obvious but I can't prove it. Any help will be appreciated.

mrp
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MAh2014
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  • Using natural deduction, assume $\neg\forall x P(x)$, then there exists $x_0$ such that $\neg P(x_0)$, but you can find $P(x_0)$ in $\Gamma$, so you've found a contradiction. – Javi May 04 '17 at 15:09
  • "it is obvious that $Γ⊢∀xP(x)$..." can you write the derivation ? – Mauro ALLEGRANZA May 04 '17 at 15:27
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    what are $x_0,x_1,x_2,\ldots$ ? – mercio May 04 '17 at 15:31
  • As $P(x_i)$ is a member of $\Gamma$ for all $i$ so the result will be derived naturally ,but this is exactly my problem,how to deduce this result? – MAh2014 May 04 '17 at 15:31
  • We need some details... what about the def of consistency ? Is it "equivalent" to satisfiable ? – Mauro ALLEGRANZA May 04 '17 at 15:36
  • @MauroALLEGRANZA Yes it is. – MAh2014 May 04 '17 at 15:46
  • @Javi Could you please tell the deduction of "..¬∀xP(x) , then there exists x0 such that ¬P(x0).."? – MAh2014 May 04 '17 at 16:43
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    $\Gamma$ is satisfiable; consider a domain $D= { 0,1 }$ and interpret $P$ with ${ 0 }$. Then consider a variable assignment function $s$ such that $s(x_i)=0$ for every $i$. Clearly $D,s \vDash \Gamma$. – Mauro ALLEGRANZA May 05 '17 at 06:00
  • @MauroALLEGRANZA but P is a one place predicate,not a function. – MAh2014 May 05 '17 at 11:44
  • And thus ? The "meaning" assigned to $P$ is a subset of the domain, and this is the way we have to interpret unary predicates. – Mauro ALLEGRANZA May 05 '17 at 12:17
  • Techincal question/comment: In $\neg \forall x_0 P(x_0)$ the $x_0$ is used as a variable, but in $P(x_0)$ it is used as a constant ... Which is it? Are $x_0, x_1, ...$ all constants? And if so, would you be ok with changing $\neg \forall x_0 P(x_0)$ to $\neg \forall x P(x)$? – Bram28 May 05 '17 at 13:41
  • @Bram28 I was thinking about this too,I'm not sure if we can do this,and $x_i$ are individual variables. – MAh2014 May 05 '17 at 14:31
  • @MAh2014 But if $x_i$ is a variable, then $P(x_i)$ is not a statement to which we can assign a truth-value. Hence, $\Gamma$ is not a set of sentences of which we can determine its consistency. So, for your question to make sense, we really need to regard all $x_i$ to be constants ... but we'll also need to change $\forall x_0 P(x_0)$ to $\forall x P(x)$ – Bram28 May 05 '17 at 15:51
  • @MauroALLEGRANZA I don't have any idea about Bram28 question,would you please help? – MAh2014 May 06 '17 at 13:58
  • @MauroALLEGRANZA I can't understand what is $s$,we can't assign any thing to individual variables?! – MAh2014 May 08 '17 at 12:47

1 Answers1

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The only reasonable interpretation of consistency of a set of formulae is that one cannot prove a contradiction from them, which is then equivalent to satisfiability, which means existence of a model, which must assign every free variable to an object. It is certainly possible to satisfy $Γ$ as given, since one could construct a structure with just two objects, such that one of them satisfies $P$ and the other does not, and in this structure assign all free variables $x_0,x_1,...$ to the one object that satisfies $P$.

If you're not familiar with this terminology, refer to Wolfgang Rautenberg's Concise Introduction to Logic or Stephen Simpson's notes as linked from here.

user21820
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