Say that a finitely axiomatizable first-order theory $T$ (in a finite language $\Sigma$) is persistently finitely axiomatizable iff the set $T_{\mathsf{w/o=}}$ of equality-free $T$-theorems is finitely axiomatizable in first-order logic without equality.

Building off of this earlier question, I would like to ask:

Is there a finitely axiomatizable relational theory which is not persistently finitely axiomatizable?

Here are the only two relevant facts I currently know:

• One natural attempt for proving a negative answer is the following. Let $$I(x,y)\equiv\bigwedge_{R\in \Sigma} \forall \overline{u},\overline{v}(R(\overline{u},x,\overline{v})\leftrightarrow R(\overline{u},y,\overline{v})).$$ Then we might expect to have $\vdash A\leftrightarrow A_I$ for all first-order sentences $A$, where $A_I$ is gotten from $A$ by replacing "$s=t$" with "$I(s,t)$" throughout. However, this breaks down: consider e.g. $A\equiv \exists x,y(x=y\wedge \neg I(x,y))$. (This counterexample was pointed out by Emil Jerabek.)

• If we allow function symbols we get a positive answer; this was observed by Rodrigo Freire, answering the above-linked original question.

Say that a finitely axiomatizable first-order theory $T$ (in a finite language $\Sigma$) is persistently finitely axiomatizable iff the set $T_{\mathsf{w/o=}}$ of equality-free $T$-theorems is finitely axiomatizable in first-order logic without equality.

Building on this earlier question, I would like to ask:

Is there a finitely axiomatizable relational theory which is not persistently finitely axiomatizable?

Here are the only two relevant facts I currently know:

• One natural attempt for proving a negative answer is the following. Let $$I(x,y)\equiv\bigwedge_{R\in \Sigma} \forall \overline{u},\overline{v}(R(\overline{u},x,\overline{v})\leftrightarrow R(\overline{u},y,\overline{v})).$$ Then we might expect to have $\vdash A\leftrightarrow A_I$ for all first-order sentences $A$, where $A_I$ is gotten from $A$ by replacing "$s=t$" with "$I(s,t)$" throughout. However, this breaks down: consider e.g. $A\equiv \exists x,y(x=y\wedge \neg I(x,y))$. (This counterexample was pointed out by Emil Jerabek.)

• If we allow function symbols we get a positive answer; this was observed by Rodrigo Freire, answering the above-linked original question.