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I am unsure if i solved those three right. would really appreciate your input to understand if I've done something wrong:

Note: $M$ is a model, $S$ is assignment, $\varphi$ is a formula.

1) if $M \vDash_S \exists x \, \varphi$ then $M \vDash _S \varphi$

2) if $M \vDash_S \varphi$ then $M \vDash_S \forall x \, \varphi$

3) if $M \vDash_S \forall x \varphi$ then $M \vDash_S \varphi$.

What I did:

1) Incorrect. If was written that $M \vDash_S \forall x \, \varphi$ then it would be correct. But if we assume that the left part is true, then there exists a variable $x$ for which it is correct, however it doesn't mean that for every $x$, $M \vDash_S \varphi$ will be true. So incorrect.

2) True. if the left part is true, then the right part is true as its a tautology under model $M$, which means that it is true for every $x$.

3) True. If it is true for every $x$, as written in the left part, I believe we can deduct that it is a tautology under Model $M$ and assignment $S$.

If there's a better way to write the proves, or if I've done a mistake, please correct me so i can learn and improve.

Thank you very much!

Taroccoesbrocco
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BeginningMath
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1 Answers1

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Long comment

The issue is : is $x$ free in $φ$ ?

Consider :

2) if $M,s \vDash φ$, then $M,s \vDash ∀xφ$.

If $x$ is not free in $φ$, we have that $φ \equiv \forall x \varphi$, and thus the property holds.

If instead $x$ is free in $φ$ , the property does not hold.

Consider e.g. as $M$ the structure $\mathbb N$ of natural numbers and as $φ$ the formula $(x=0)$.

With $s : \text {Var} \to \mathbb N$ such that : $s(x)=0$ we have :

$\mathbb N,s \vDash (x=0)$,

but clearly : $\mathbb N,s \nvDash \forall x (x=0)$.

It seems to me that we have an issue regarding the basic semantical definitions.

Regarding :

1) "if we assume that $M,s \vDash ∃xφ$ is true, then there exists a variable $x$ for which it is correct".

No; $M,s \vDash ∃xφ$ means that there is $c \in M$ such that $s'(x)=c$, for some $x$-variant $s'$ of $s$.

"however it doesn't mean that, for every $x$, $M,s \vDash φ$ will be true."

Right, but $M,s \vDash φ$ does not mean that it must hold "for every $x$".

It means that $φ$ is true of the object $d \in M$ such that $s(x)=d$.

Mauro ALLEGRANZA
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