First-order de Morgan's laws for $\nvdash$ This is a continuation of my previous question about $\nvdash$. Assuming that $x$ does not occur free in $\Gamma$, it seems to me that the following two statements in (1) and (2) should hold. These are very similar to first-order de Morgan's laws, but in this case, the negation is not propositional but judgmental.
$$\Gamma\nvdash \forall xA(x)\Longleftrightarrow\ \textit{for some x},\ \Gamma\nvdash A(x)\quad\quad(1) $$
$$\Gamma\nvdash \exists xA(x)\Longleftrightarrow\ \textit{for all x},\ \Gamma\nvdash A(x)\quad\quad(2)$$
But I wonder how to prove both (1) and (2), if possible. Thanks a lot!
 A: We may use the turnstile symbol: $\vdash$ to refer to the derivability relation for a proof system $\text {PS}$ (e.g. Natural Deduction), like in your previous post.
In this case, we read:

$\Gamma \vdash \varphi$

as an assertion about the existence of a derivation of formula $\varphi$ from the set $\Gamma$ of assumptions.
$Γ⊬∀xA(x)$ holds if we have a counter-example to $\forall x A(x)$, i.e. if we have that $\Gamma \cup \{ \lnot A[c/x] \}$ is satisfiable, for some $c$.
But we have that: $\Gamma \cup \{ \lnot \varphi \} \text {  is unsatisfiable iff  } \Gamma \vDash \varphi \text {  iff  } \Gamma \vdash \varphi$.
Thus: $\Gamma \cup \{ \lnot A[c/x] \}$ is satisfiable, for some $c$, iff not-$(\Gamma \vdash A[c/x])$.

Buy see also the post: Judgment-level negation of $\nvdash$.
Following the discussion in this post, we can consider the proof-theoretical approach specific of calculus of sequents: $\Gamma \vdash A$.
This type of calculus is aimed at the analysis of formal deductions, and it is expressed through rules.
See e.g. Kosta Došen, Logical constants as punctuation marks, NDJFL, 1989, page 367, for the rules regarding quantifiers.
The rules are "double-lined", i.e. they are invertible, meaning that we can read them both ways: one going downward and
the other upward.
A: You need to be careful that in the statements, $x$ would be forced to refer to some fixed term, which can only be built up from atomic variables by applying function symbols in the first-order language you're considering.  And even if you can find some model in which some formula's negation is satisfiable, it might not necessarily be possible to express the witness to satisfiability in terms of the assignments of the atomic variables.
For example, let us consider the "restricted language of groups" where the only function symbol is the binary operation $\cdot$; and the axioms $GroupAxioms$ include $\exists e \forall y (e \cdot y = y \land y \cdot e = y)$, etc.  Now, for every term $x$ in terms of atomic variables,
$$GroupAxioms \not\vdash \forall y (x \cdot y = y).$$
To see this: for example, if we consider the model $\mathbb{Z}$ and the assignment where every atomic variable is sent to $1$, then by induction, every term built up from these atomic variables, using only $\cdot$ (interpreted as integer addition), has an interpretation which is strictly positive and therefore cannot get us to the identity of the group.  On the other hand, clearly
$$GroupAxioms \vdash \exists x \forall y (x \cdot y = y)$$
which gives a counterexample to the reverse direction in (2).  (The forward direction in (2), on the other hand, is true if we incorporate the appropriate restrictions on overlaps between free variables in $\Gamma$ and in the term $x$ - this is an easy consequence of the ${\exists}I$ proof rule, that if $\Gamma \vdash A(\tau)$ for some term $\tau$, then $\Gamma \vdash \exists x \, A(x)$.)
Likewise, for (1), the reverse implication is true as a result of the ${\forall}E$ proof rule (i.e. instantiation of universal quantifier: if $\Gamma \vdash \forall x \, A(x)$ and $\tau$ is a term, then $\Gamma \vdash A(\tau)$).  Here, the forward implication in (1) is also true: suppose $\Gamma \not\vdash \forall x \, A(x)$.  If we choose as the term $x$ an atomic variable which does not appear in any of the group axioms or in $A$, then $\Gamma \vdash A(x)$ implies $\Gamma \vdash \forall x\, A(x)$ by the ${\forall}I$ proof rule, given a contradiction; thus, we have found $x$ such that $\Gamma \not\vdash A(x)$.
