If $x$ does not freely occur in $Φ,$ and $\:Φ, ϕ\: ⊨ ψ,$ then $\:Φ, ∀x\,ϕ\: ⊨ ∀x\,ψ ?$

Question

Below is my proof (I don't know whether it is correct) of $$\text{if }ϕ ⊨ ψ,\text{ then }∀x\,ϕ ⊨ ∀x\,ψ.$$

Given J. Assume that $⟦∀x\,ϕ⟧_J=T$.

Given a in J’s domain, $⟦ψ⟧_(J[x↦a])=T$ because

$⟦ ϕ⟧_(J[x↦a])=T$

$ϕ ⊨ ψ$

So $⟦∀x\,ψ⟧_J=T$. QED.

Given the above result, and that $\:Φ, ϕ\: ⊨ ψ,$ and that $x$ does not freely occur in $Φ,$ what is the procedure to prove that $$Φ, ∀x\,ϕ\: ⊨ ∀x\,ψ ?$$

Answer 1

1. Below is my proof (I don't know whether it is correct) of $$\text{if }ϕ ⊨ ψ,\text{ then }∀x\,ϕ ⊨ ∀x\,ψ.\tag{*}$$

   I suspect that you are using the symbol $⊨$ to mean that $\to$ is true instead of its usual, stronger meaning that $\to$ is a validity (i.e., logically true). In any case, $(*)$ is at least as strong as $$\big(ϕ ⊨ ψ\big)\implies\big(∀x\,ϕ {\implies} ∀x\,ψ\big),$$ which is invalid (i.e., false in some interpretation); so, $(*)$ is also invalid.

2. Given the above result, and that $\:Φ, ϕ\: ⊨ ψ,$ and that $x$ does not freely occur in $Φ,$ what is the procedure to prove that $$Φ, ∀x\,ϕ\: ⊨ ∀x\,ψ ?$$

   Summarising and relabelling for ease of reading: our premises are

   1. $\big(Fx \to Sx\big)\to\big(∀x\,Fx \to ∀x\,Sx\big)$
   2. $\big(P\land Fx\big) \to Sx$
   3. $P \land ∀x\,Fx,$

   and our conclusion is

   • $∀x\,Sx.$

   Here's an informal proof:

   \begin{align} &&\big[\big(Fx \to Sx\big)\to\big(∀x\,Fx \to ∀x\,Sx\big)\big] \land \big[\big(P\land Fx\big) \to Sx\big]\land\big[P \land ∀x\,Fx\big] \\&\equiv &\big[\big(\big(Fx \to Sx\big) \land ∀x\,Fx\big) \to ∀x\,Sx\big] \land \big[\big(\big(P\land Fx\big) \to Sx\big)\land \big(P \land Fx\big)\big]\land\big[ ∀x\,Fx\big] \\&⊨ &\big[\big(\big(Fx \to Sx\big) \land ∀x\,Fx\big) \to ∀x\,Sx\big] \land \big[Sx\big]\land\big[ ∀x\,Fx\big] \\&⊨ &\big[\big(Sx \land ∀x\,Fx\big) \to ∀x\,Sx\big] \land \big[Sx\big]\land\big[ ∀x\,Fx\big] \\&⊨ &∀x\,Sx. \end{align}