Implication Introduction in reverse way In Gentzen system, there is an inference rule such that one can deduce $\Gamma \to \Delta, \mathfrak{A} \supset \mathfrak{B}$  from  $\Gamma, \mathfrak{A} \to \Delta, \mathfrak{B}$.
Can we, in reverse way, deduce  $\Gamma, \mathfrak{A} \to \Delta, \mathfrak{B}$ from $\Gamma \to \Delta, \mathfrak{A} \supset \mathfrak{B}$? More precisely, 


*

*In Gentzen sequent calculus, is there an inference rule of the form below?
\begin{align}
  \frac{\Gamma \to \Delta, \mathfrak{A} \supset \mathfrak{B}}{\Gamma, \mathfrak{A} \to \Delta, \mathfrak{B}}*
\end{align}

*In Gentzen sequent calculus, is there a derivation of $\Gamma, \to \Delta,$ from the assumption $\Gamma \to \Delta, \supset $? In other words, is the rule ($*$) derivable in Genztzen sequent calculus?
 A: No; in Sequent Calculus you do not "unpack" complex formulae but always build them up from their "components".
The rules for the conditional connective $\supset$ are:
\begin{align}
{\cfrac{C, \Gamma \to \Delta, D}{\Gamma \to \Delta, C \supset D} \supset \text {-right}}
\end{align}
\begin{align}
{\cfrac{\Gamma \to \Delta, C \ \ \ \ \ \ \ D, \Pi \to \Lambda}{C \supset D, \Gamma, \Pi \to \Delta, \Lambda} \supset \text {-left}}
\end{align}

An important feature of Sequent Calculus is that the rules are invertible, i.e. we can use them "bottom-up" in a proof-search procedure. In this case, what you are asking is nothing else than $\supset \text {-right}$ read bottom-up.
A: Mauro Allegranza's reply correctly answers Point 1 of the question. But the answer to Point 2 of the question is positive. Indeed, the rule $\supset$-right is reversible, which means that the inference rule 
\begin{align}
  \frac{\Gamma \to \Delta, \mathfrak{A} \supset \mathfrak{B}}{\Gamma, \mathfrak{A} \to \Delta, \mathfrak{B}}*
\end{align}
(i.e. the bottom-up version of the rule $\supset$-right) is derivable in Gentzen sequent calculus: ($*$) is not an inference rule of original Gentzen sequent calculus but it can be "simulated" in Gentzen sequent calculus, i.e. there exists a derivation of $\Gamma, \mathfrak{A} \to \Delta, \mathfrak{B}$ from the assumption $\Gamma \to \Delta, \mathfrak{A} \supset \mathfrak{B}$, for instance the following
\begin{align}\tag{1}
\dfrac{\Gamma \to \Delta, \mathfrak{A} \supset \mathfrak{B} \qquad \dfrac{\dfrac{}{\mathfrak{A} \to \mathfrak{A}}\text{ax} \qquad \dfrac{}{\mathfrak{B} \to \mathfrak{B}}\text{ax}}{\mathfrak{A},\mathfrak{A} \supset \mathfrak{B} \to \mathfrak{B}}\supset\!\!\text{-right} }{\Gamma, \mathfrak{A \to \Delta, \mathfrak{B}}}\text{cut}
\end{align}
This entails that if the sequent $\Gamma \to \Delta, \mathfrak{A} \supset \mathfrak{B}$ is derivable, then $\Gamma, \mathfrak{A} \to \Delta, \mathfrak{B}$ is derivable: given a derivation $\pi$ of $\Gamma \to \Delta, \mathfrak{A} \supset \mathfrak{B}$, the composition of $\pi$ with the derivation $(1)$ yields a derivation of $\Gamma, \mathfrak{A} \to \Delta, \mathfrak{B}$.
