Given a clone $\mathcal{C}$ over $\{\top,\perp\}$, let $\mathsf{FOL}^\mathcal{C}$ be the version of first-order logic with connectives from $\mathcal{C}$ in place of the usual Booleans. Given a clone $\mathcal{C}$ and a structure $\mathfrak{M}$, even if $\mathsf{FOL}^\mathcal{C}$ is weaker than $\mathsf{FOL}$ it may still be the case that $\mathsf{FOL}$ and $\mathsf{FOL}^\mathcal{C}$ "coincide on $\mathfrak{M}$" in the sense that every $\mathsf{FOL}$-definable relation on $\mathfrak{M}$ is $\mathsf{FOL}^\mathcal{C}$-definable (with parameters allowed throughout). Call clones with this latter property "$\mathfrak{M}$-strong," and let $[\mathfrak{M}]$ be the intersection of all $\mathfrak{M}$-strong clones; intuitively, $[\mathfrak{M}]$ is the clone of connectives which $\mathfrak{M}$ can't "simplify" in any useful way.
Question: Need $[\mathfrak{M}]$ always be $\mathfrak{M}$-strong?
I see no reason why this should be the case; however, I also don't immediately see how to produce a counterexample. (Note that an affirmative answer to this question would yield a negative answer to this earlier question of mine.)
