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I encountered an expression during my homework.

$\exists x (Qx\rightarrow \forall x Qx)$ where $Q$ was a unary predicate symbol. However, I'm not sure if this sentence was "legal". Surely, it could be built up from wffs by inductive steps.Thus the syntax was correct. However, I'm using Enderton's textbook so $\exists x\alpha$ was translated as there was some $z\in|B|$ such that $\alpha[s(x|z)]$ was satisfied, where $|B|$ was the universe.

So the translation seemed to be strange, as there seemed to be an expansion where $\exists x\forall x Qx$ for the second term.

My question was that

  1. How to deal with the case such that $\forall x(\exists x \alpha)$, $\exists x(\forall x \alpha)$, $\forall x(\forall x \alpha)$, or $\exists x(\exists x \alpha)$...

2 Can we or would it be better to write the original expression as $\exists x (Qx\rightarrow \forall y Qy)$?

1 Answers1

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I'm put off by the fact that $x$ is bound twice -- you can't bind a nonfree variable! It seems that the statement is meant to say $\exists x (Qx\rightarrow \forall y Qy)$.

Andrew Tindall
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  • I was wondering about the same question, but the original expression was in fact correct as wffs, since it was built up by automic formulas and the logical symbols. So we should be able to translate it. I even thought about rewrite $\exists$ back to $\forall$. But the direct explanation I came up with was such that the inner $\forall x$ reassigned $s(x|a)$ to $s(x|b)(x|a)$(operate from right to left) for every $b\in |B|$. –  Nov 03 '17 at 19:05