If $\phi(A)$, and $A\vdash B$, does $\phi(B)$ hold?

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Simple logic question concerning substitution, where $A$ and $B$ are propositions, and $\phi(A)$ is a formula with one or more instances of the propositional variable $A$,

If $\phi(A)$, and $A\vdash B$, does $\phi(B)$ hold?

This has probably been asked before, but I'm not sure what to search to get an answer.

Obviously, if $\phi(A)$, $A\vdash B$ and $B\vdash A$, then it's safe to assert $\phi(B)$, but I'm unsure if $B\vdash A$ is necessary.

My intuition so far is that $A\vdash B$ is insufficient, but I'm unable to come up with a counterexample.

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Certainly not. Take $\phi(C) = \neg C$. Let $A = \bot$ and $B = \top$. Then $\vdash \phi(A)$, and $A \vdash B$. But $\vdash \phi(B)$ fails.