Adding $\tfrac{}{\Gamma \varphi}$ (for a fixed non-correct $\Gamma \varphi$) to the rules of the sequent calculus, can one now derive every sequent?

Question

One knows that the sequent calculus over the set of sequents $\Gamma \varphi$ is correct and complete, meaning that the derivable sequents are precisely the correct ones.

However, adding just one axiom $\tfrac{}{\Gamma \varphi}$ (where $\Gamma \varphi$ is not correct) to the rules of the sequent calculus, can one now derive every sequent?

To prove or disprove this, it might be important to know that I am talking about the specific sequent calculus introduced in "Mathematical Logic" by Ebbinghaus, Thomas and Flum.

Answer 1

The answer to my question is: not in general. Thanks to Bram28 for giving me an idea in his answer.

Assuming $\Gamma \varphi$ only contains sentences (!), one can proof the following by induction over the enlarged sequent calculus: If $\Delta \psi$ can be derived in the enlarged calculus, then for all interpretations ℑ: If (If ℑ$\models \Gamma$, then ℑ$\models \varphi$), then (If ℑ$\models \Delta$, then ℑ$\models \psi$). The proof is analogous to the one showing correctness of the sequent calculus. $\Gamma \varphi$ only containing sentences is needed to avoid problems with free variables while dealing with the rule ($\exists$A) (see Ebbinghaus for the rule).

Now we use a counterexample similarly to Bram28's approach: Take $\Gamma=\emptyset$ and a sentence $\varphi$ that is not valid (making the sequent $\Gamma \varphi$ not correct), but let it also be satisfiable. For example, use $\varphi=\exists x \exists y \lnot x \equiv y$ where $x$ and $y$ are different variables (i.e. "the universe contains at least two elements", which is not always true, but it can be).

If the sequent $\Delta\psi=\emptyset\lnot\varphi$ was derivable, the above result yields: For all interpretations ℑ: If (ℑ$\models\varphi$), then (ℑ$\models\lnot\varphi$), i.e. $\varphi\models\lnot\varphi$. This is false because $\varphi$ is satisfiable.

Answer 2

Not in general, no.

Consider adding the sequent $\{ \} P$ where $P$ is a specific logic statement that is not a contradiction.

Then all statements that you can derive (as a sequent $\{ \} \phi$) are going to be either tautologies, or statements that are logically implied by $\phi$... which means you cannot derive the sequent $\{ \} \neg P$

The only time that adding a specific sequent would allow you to infer all sequents is when the sequent you add is equivalent to a contradiction, e.g. if you add the sequent $\{ \} P \land \neg P$ for some specific $P$.