How to prove the following property of implication?

The question with which I am struggling is the following,

Question. Let $$\Gamma$$ be a consistent set of wffs of propositional calculus (see the axioms and rule of inference below). Let $$\alpha,\beta$$ be two wffs. If $$\Gamma\nvdash\alpha$$ and $$\Gamma\nvdash\beta$$, is it possible to have $$\Gamma\vdash\alpha\to\beta$$?

The axioms and rules of inference are (here $$P,Q$$ and $$S$$ are arbitrary formulas),

$$\color{crimson}{\text{Axiom 1.}}\ P\to (Q\to P)$$

$$\color{crimson}{\text{Axiom 2.}}\ (S\to (P\to Q))\to((S\to P)\to (S\to Q))$$

$$\color{crimson}{\text{Axiom 3.}}\ (\neg Q\to\neg P)\to(P\to Q)$$

$$\color{crimson}{\text{Rule of Inference.}}$$ Modus Ponens.

My Attempt

I tried by assuming on the contrary that $$\Gamma\nvdash\alpha\to\beta$$ and then extending $$\Gamma$$ to a maximal consistent set $$\Delta$$ such that $$\Delta\nvdash\alpha\to\beta$$. I have also noted that $$\Delta\vdash\alpha$$ and $$\Delta\nvdash\beta$$ as well. But I can't find any contradiction.

Yes, it is possible to have $$\Gamma \vdash \alpha \to \beta$$, for suitable formulas $$\alpha, \beta$$.
Indeed, since $$\Gamma$$ is consistent, there exists a formula $$\alpha$$ such that $$\Gamma \not\vdash \alpha$$. Take $$\beta = \alpha$$. Thus, $$\Gamma \not\vdash \beta$$ but $$\Gamma \vdash \alpha \to \beta$$, as you can see here for a derivation in Hilbert system.
Note that the statement "if $$\Gamma$$ is consistent and $$\alpha,\beta$$ are formulas such that $$\Gamma \not\vdash \alpha$$ and $$\Gamma \not\vdash \beta$$, then $$\Gamma \vdash \alpha \to \beta$$" is false for arbitrary formulas $$\alpha, \beta$$ and arbitrary $$\Gamma$$. For instance, if $$\alpha, \beta$$ are two distinct propositional variables and $$\Gamma$$ is the empty set of formulas, then $$\Gamma$$ is consistent, $$\Gamma \not\vdash \alpha$$ and $$\Gamma \not\vdash \beta$$ but $$\Gamma \not\vdash \alpha \to \beta$$.
• What properties of Γ would ensure that for arbitrary formulas $α,β$ that if $Γ⊬α$ and $Γ⊬β$ then $Γ\vdash α→β$ as well? – user170039 Mar 19 '20 at 4:47
• I think if $\Gamma$ is maximal consistent then we have the required thing. – user170039 Mar 19 '20 at 5:58
• @user170039 - Exactly! Do you have a proof? Hint: if $\Gamma$ is maximally consistent then it is complete (i.e. for every formula $\alpha$, either $\Gamma \vdash \alpha$ or $\Gamma \vdash \lnot \alpha$). – Taroccoesbrocco Mar 19 '20 at 7:53
• Yep. I do have a proof. If $\Gamma$ is maximal consistent and $\Gamma\nvdash\alpha$ then we have $\alpha\notin\Gamma$. Hence $\Gamma\cup\{\alpha\}$ is inconsistent. Hence we have $\Gamma\cup\{\alpha\}\vdash\beta$. Applying DT $\Gamma\vdash\alpha\to\beta$. – user170039 Mar 19 '20 at 8:26