Possible flaw in proof that ∃xG(a,x) ⊨ ∀xG(a,x)

Question

I know that ∀xG(a,x) is not true in the general case where the only premise given is ∃xG(a,x), but using the rules for predicate logic in natural deduction, I've managed to deduce ∃xG(a,x) ⊨ ∀xG(a,x) anyway, but I'm unable to see the flaw(s) in my proof. The proof is:

1 ∃xG(a,x)                    premise
2 x0: G(a,x0) [x/x0]          assumption
      x0:
3        G(a,x0) [x/x0]       Copy 2
4     ∀xG(a,x)                ∀x i 3-3
5 ∀xG(a,x)                    ∃x e 1,2-4

Where am I deducing things incorrectly?

Answer 1

The wrong step in the purported derivation of $\exists x G(a,x) \vdash \forall x G(a,x)$ is the "generalization" used in deriving 4 from 3.

In Natural Deductin the $\forall$I rule: $φ(x)⊢∀xφ(x)$ [or: if $\Gamma \vdash φ[a/x]$, then $\Gamma ⊢ ∀xφ(x)$] has the essential proviso: provided that $x$ is not free in any assumption.

In your derivation, $G(a,x)$ is assumed in step 2 and we cannot apply $\forall$I to derive $∀xG(a,x)$.

Regarding the use of $\exists$E in step 2, it is correct; in Natural Deduction the $∃$E rule licenses the derivation of $ψ$ from $∃xφ(x)$ assuming $φ[a/x]$, provided that $y$ is not free in any other assumption nor in $ψ$.