TITLE: If Γ,A∧B⊢Δ is an axiom, then also Γ,A,B⊢Δ is an axiom
QUESTION [2 upvotes]: 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

REPLY [1 votes]: 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.
{"set_name": "stack_exchange", "score": 2, "question_id": 1905275}