If Γ,A∧B⊢Δ is an axiom, then also Γ,A,B⊢Δ is an axiom

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Context

I am studying sequent calculus, and I am trying to understand the proof that the rule L∧ introducing

"$\land $" on the left: ${\displaystyle \quad {\cfrac {\Gamma ,A,B\vdash \Delta }{\Gamma ,A\land B\vdash \Delta }}}$

is invertible, where invertible means that as soon as I have a derivation of height $n$ of the conclusion, I also have a derivation of the same height of the premise. Here I am considering sequents for classical logic, and I define an axiom as a sequent $\Gamma \vdash \Delta $ , such that $\Gamma$ and $\Delta$ are finite multisets sharing at least one formula. The proof is by induction on the height of the derivation, and starts with the case $n=0$.

Problem:

The proof is easy and follows inductively from the fact that, if

(1)$\displaystyle \quad \Gamma ,A\land B\vdash \Delta$ is an axiom,

then also (2) $\displaystyle \quad \Gamma ,A,B\vdash \Delta$ is an axiom.

My reasoning is that it can be the case that $\Gamma$ and $\Delta$ don't share any formula, and $A \land B$ is in $\Delta$, so that (1) is an axiom, but in this case I don't understand how to prove rigorously why (2) should be.

Thanks for reading

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Probably it is because you work in sequent calculus, which only admits axioms in form $\Gamma\vdash\Delta$, where $\Gamma$ and $\Delta$ share an atomic formula. I would recommend a book by Troelstra and Schwichteberg: Basic proof theory. Here this system is called $\mathrm{G}3$, see Definition 3.5.1.