Formula is satisfiable iff its existential closure is satisfiable

Question

How can I prove this statement:

A formula is satisfiable iff its existential closure is satisfiable(the formula has $n$ free variables which are $x_1,\dots,x_n$).

A formula $\varphi$ is satisfiable if and only if there is a structure $\mathcal{M}$ such that $\mathcal{M} \vDash \varphi$ . I applied this definition to $\varphi\left(x_{1}, \ldots, x_{n}\right),$

and say there is a tuple $\vec{a}=(a_{1}, \ldots, a_{n})$ in M satisfying $\varphi$ in $\mathcal{M}$ s.t.

$ \mathcal{M} \vDash \varphi\left(a_{1}, \ldots, a_{n}\right)$

And applying this definition to $\exists x_{1}, \ldots x_{n} \varphi\left(x_{1}, \ldots, x_{n}\right),$

$ \mathcal{M} \vDash \exists x_{1}, \ldots, x_{n} \varphi\left(x_{1}, \ldots, x_{n}\right)$

and after I say these, which steps should I follow?

Answer 1

Welcome to MSE!

Hint:

Recall a formula $\psi$ is satisfiable if and only if in some model $\mathfrak{M} \models \psi$. This is the definition of satisfiability.

So applying this definition to $\varphi(x_1, \ldots, x_n)$, we see this is asking for

• a model $\mathfrak{M}$
• elements $a_1, \ldots, a_n \in \mathfrak{M}$
• $\mathfrak{M} \models \varphi(a_1, \ldots, a_n)$.

And applying this definition to $\exists x_1, \ldots x_n . \varphi(x_1, \ldots, x_n)$, we see we want

• a model $\mathfrak{M}$
• $\mathfrak{M} \models \exists x_1, \ldots, x_n . \varphi(x_1, \ldots, x_n)$

Do you see why these are equivalent conditions? It might be worth reviewing the definition of $\mathfrak{M} \models \exists z . \psi(z)$.

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I hope this helps ^_^