I have this:
Two sets of formulas $\Sigma$ and $\Gamma$ in which for each formula $\delta,\Sigma\vdash\delta\iff\Gamma\vdash\delta$. Supposing $\delta1,\ldots,\delta n$ is a deduction of the formula $\delta$ from the set $\Sigma$ then $\delta1,\ldots,\delta n$ is also a deduction of $\delta$ from the set $\Gamma$.
So my doubt is if $\delta$ is only derivable from $\Sigma$ when $\Gamma\vdash\delta$ then how can you from $\Sigma$ deduce $\delta$ which is said later on in the question, is this false? Is there any way a set $\Sigma$ needs another set $\Gamma$ to derive $\delta$ but then it can also deduce $\delta$? I am not getting this question very well.
Since $\delta \in \Sigma$ implies $\Sigma \vdash \delta$ we know that every element of $\Sigma$ is a theorem of $\Gamma$ and vice versa. From this we can deduce that $\Sigma$ and $\Gamma$ will have the same theorems. However, a proof of $\delta_{1}$ from $\Gamma$ might require multiple steps. In general $\delta_{1}, \cdots, \delta_{n}$ will not be a deduction from $\Gamma$.