TITLE: If $\Sigma \vdash \phi$ implies $\Sigma \vdash \varphi$ then $\Sigma \vdash \phi \to \varphi$ on propositional logic?
QUESTION [8 upvotes]: My main aim is to prove or disprove that if $\Sigma \vdash \phi$ implies $\Sigma \vdash \varphi$ then $\Sigma \vdash \phi \to \varphi$ where $\Sigma$ donotes a set of sentences in propositional logic.
$\Sigma \vdash \phi$ means there is a deduction from $\Sigma$ where the deduction is a sequence $( \alpha_0 , \dots , \alpha_n)$ with $a_i$ is either in $\Sigma$ or a consequence of MP (that is, for some $j,k<i$ $\alpha_k = \alpha_j \to \alpha_i$ and $\alpha_i$ follows from them) or a tautology.
I'm completely stuck now. I tried to prove it but have no idea on how to bring $\phi$ to a deduction sequence. And I also tried to make a counterexmaple but no simple one I could find. 
Even in my mere intuition, I cannot clearly judge whether it is true or false.
(I also thought of employing Completeness and Soundness Thm..)

REPLY [7 votes]: Take Σ here to be just the axioms of propositional logic, and take $\phi$ to be some propositional atom $p$.  Then Σ ⊢ $\phi$ is false, and thus "Σ ⊢ $\phi$ implies Σ ⊢ $\varphi$" is true for any $\varphi$. Now take $\varphi$ to be some other atom $q$.  $[p \rightarrow q]$ is not a tautology.  If a formula is not a tautology, then it is not provable from Σ, because of the soundness metatheorem.  So, Σ ⊢ $(\phi \rightarrow \varphi)$ is false.  Thus, "if Σ ⊢ ϕ implies Σ ⊢ φ then Σ ⊢ $(\phi \rightarrow \varphi)$" is false in general.
(In fact $p,q$ need not be atoms; they can be taken to be any independent formulas.  That is, formulas such that for each pair of truth values $(b_1,b_2)$, there is at least one valuation for which $p$ and $q$ take values $b_1$ and $b_2$ respectively.)