On Shoenfield's Mathematical Logic, Chapter 3 Q2 $
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I need help in the following exercise by Shoenfield's Mathematical Logic (Chapter 3, Problem 2):

Let $T$ be a theory with no nonlogical axioms. For every formula $\A$ of $T$, let $\A^*$ be the formula obtained from $\A$ by omitting all quantifiers and replacing each term by a new constant $\mathbf{e}$. Show that if $\vdash_T\A$, then $\A^*$ is a tautological consequence of formulas of the form $\a = \a$. Conclude that there is no formula $\A$ such that $\vdash_T\A$ and $\vdash_T\lnot\A$.

I'm aware that, by the definition of tautological consequence, we would like to show that if we have a truth valuation $V$ such that $V(\a = \a) = \mathbf{T}$ for some collection of formulas $\a$'s, then $V(\A^*) = \mathbf{T}$. However, I'm not sure how we can arrive at such a conclusion from $V(\a = \a)$. In fact, I'm not even sure how $\a = \a$ is related to $\A^*$. Furthermore, I'm not sure how the second part of the question is related to the first.
Any advice/help on this problem is appreciated.
 A: I think that the proof must be by induction on the derivation $\vdash_T \mathbf{A}$.
Consider that $T$ has no non-logical axioms but can have non-logical symbols. Silly example: only a non-logical binary predicate $E(x,y)$.
With this language, atoms must be: $E(x,y)$ and $x=y$. [We can write $\in$ in place of $E$ and consider "pure" set theory, i.e. the mathematical theory in the first-order language of sets without mathematical axioms.]
Thus, what can we prove with only logical axioms and rules?
Base step: all "$^*$-transform" of logical axioms must be tautological consequences of of formulas of the form $\mathbf{a}=\mathbf{a}$.
Obvious for propositional axioms $\lnot \mathbf{A} \lor \mathbf{A}$ (they are tautologies) and quantifier axioms: $\mathbf{A}_{x}[\mathbf{a}] \to \exists x\mathbf{A}$, whose transform will be: $\mathbf{A}^*_{x}[\mathbf{e}] \to \mathbf{A}^*_{x}[\mathbf{e}]$ (again a tautology).
Also the transform ofr an equality axiom: $x = y \to (E(x,z) \to E(y,z))$, will be a tautology.
The only axioms that are not tautologies are the identity axioms: $\mathbf{x}=\mathbf{x}$.
The inductive step is straightforward: rules are sound with respect to tautological consequence (see the Tautology Theorem).
The only case to consider is the $\exists$-Introduction rule.
But again, if in $T$ we apply it to derive $\exists x \mathbf{A} \to \mathbf{B}$ from $\mathbf{A} \to \mathbf{B}$ ($x$ not free in  $\mathbf{B}$), the corresponding formula will be $\mathbf{A}^*_{x}[\mathbf{e}] \to \mathbf{B}^*$.
And this is a tautological consequence of $\mathbf{e}=\mathbf{e}$, if $\mathbf{A}^*_{x}[\mathbf{e}] \to \mathbf{B}^*$ is, using equality axiom:

$\mathbf{e}=\mathbf{e} \to ((\mathbf{A}^*_{x}[\mathbf{e}] \to \mathbf{B}^*) \to (\mathbf{A}^*_{x}[\mathbf{e}] \to \mathbf{B}^*)).$


The second part is straightforward; assume that for some formula $\mathbf{A}$ we have both $\vdash_T \mathbf{A}$ and $\vdash_T \lnot \mathbf{A}$.
Then, using the previous results, we have that both $\mathbf{A}^*$ and $\lnot \mathbf{A}^*$ are tautological consequences of identity axioms, contradicting their validity.
