Show that $\exists v_{0} (\phi \rightarrow \forall v_{0} \phi)$ is valid.

So what I have done is that I start with the assumption that it is not valid and trying to get a inconsistency.

$s \not\in \text{Sat}_{M} (\exists v_{0} (\phi \rightarrow \forall v_{0} \phi)) $

$s \not\in \text{Sat}_{M} (\neg \forall v_{0} \neg(\phi \rightarrow \forall v_{0} \phi))$

$s \not\in {^NM} \setminus \text{Sat}_{M} (\forall v_{0} \neg(\phi \rightarrow \forall v_{0} \phi))$

$s \in \text{Sat}_{M} (\forall v_{0} \neg(\phi \rightarrow \forall v_{0} \phi))$

$s \in A_{0}(\text{Sat}_{M} (\neg(\phi \rightarrow \forall v_{0} \phi)))$

$s(a/0) \in \text{Sat}_{M} (\neg(\phi \rightarrow \forall v_{0} \phi))$ for all $a\in M$

$s(a/0) \in {^NM} \setminus \text{Sat}_{M} (\phi \rightarrow \forall v_{0} \phi)$ for all $a\in M$

$s(a/0) \not\in \text{Sat}_{M} (\phi \rightarrow \forall v_{0} \phi)$ for all $a\in M$

$s(a/0) \not\in ({^NM} \setminus \text{Sat}_{M} (\phi)) \cup (\text{Sat}_{M}(\forall v_{0} \phi))$ for all $a\in M$

From here I get two conditions (not sure if the right word).


$s(a/0) \in \text{Sat}_{M} (\phi)$ for all $a\in M$



$s(a/0) \not\in \text{Sat}_{M}(\forall v_{0} \phi)$ for all $a\in M$

$s(a/0) \not\in A_{0}(\text{Sat}_{M}(\phi))$ for all $a\in M$

$s(a/0)s(b/0) \not\in \text{Sat}_{M}(\phi)$ for all $a\in M$ and for some $b \in M$

Now since we have that for all $a \in M$, $s(a/0) \in \text{Sat}_{M} (\phi)$ and $s(a/0) \not\in \text{Sat}_{M} (\phi)$ we have a inconsistency.

This is what I've done but I have a feeling in the back of my head that there is something wrong in my solution.

  • $\begingroup$ This is known as The Drinker's Paradox. There is someone in the bar, that if they take a sip, everyone is taking a sip. $\endgroup$ – Asaf Karagila Mar 6 '17 at 15:04
  • $\begingroup$ @AsafKaragila That is true. However while I can find lot of different kinds of proofs for this I was wondering if what I have done here is correct or not. Couldn't find similiar style solutions anywhere and that is why I am asking this. $\endgroup$ – E.K. Mar 6 '17 at 15:27
  • $\begingroup$ Admittedly, I have no idea what half the symbols in your proof mean. So I can't really give my opinion on the topic. $\endgroup$ – Asaf Karagila Mar 6 '17 at 15:30
  • $\begingroup$ Up to the "separation" of the argument in point 1. and 2. it looks fine to me... You are saying (correct me if I'm worng) that the variable assignment function $s(a/0)$ satisfy $\phi$ and does not satisfy $\forall v_0 \phi$, for all $a \in M$. $\endgroup$ – Mauro ALLEGRANZA Mar 6 '17 at 15:34
  • $\begingroup$ @MauroALLEGRANZA Yes, that is correct. $\endgroup$ – E.K. Mar 6 '17 at 15:40

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