Correct Hilbert Style proof for $\vdash (A\vee B) \equiv (A \vee C)$?

Question

Is this appropriate Hilbert Style proof for $\vdash (A\vee B) \equiv (A \vee C)$?

(1) $A\vee (B \equiv C ) \equiv A \vee B \equiv A \vee C$ (by the Axiom of Distributivity of $\vee$ over $\equiv$)

QED?

Answer 1

$\def\getsto{\leftrightarrow} (A\vee(B\getsto C))\getsto((A\vee B)\getsto(A\vee C))$ is a provable tautology in classical logic.

However, $A\vee(B\getsto C)$ is not, so you cannot use the above to prove $(A\vee B)\getsto(A\vee C)$ except conditionally.$$A\vee(B\getsto C)\vdash (A\vee B)\getsto(A\vee C)$$

Do you have a way to justify a premise of $A\vee(B\getsto C)$?

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$(A\vee(B\equiv C))\equiv(A\vee B)\equiv(A\vee C)$ is a tautology, and well, may be used as an axiom for a Hilbert-esque proof system.

However, $A\vee(B\equiv C)$ is not, so you cannot use the above to prove $(A\vee B)\equiv(A\vee C)$ except conditionally. $$A\vee(B\equiv C)\vdash (A\vee B)\equiv(A\vee C)$$