Why does the inference rule of negation moves a term to the other side of the turnstile the excellent tutorial http://logitext.mit.edu/tutorial presents 
the negation inference rule like so: 
\begin{align}
 \frac{\Gamma \vdash A, \Delta}{\Gamma, \lnot A \vdash \Delta} \lnot_L
\end{align}
and 
\begin{align}
 \frac{\Gamma, A \vdash \Delta}{\Gamma \vdash \lnot A, \Delta} \lnot_R
\end{align}
It's elegant and reminds of algebra but I can't convince myself of the first example truthiness. 
The proposition itself seems impossible A and not A implies A? 
Isn't the above a contradiction? 
 A: They must be understood "semantically": intuitively, a sequent $A_1,\ldots,A_m \vdash B_1,\ldots, B_n$ means: if $A_1 \land \ldots \land A_m$, then $B_1 \lor \ldots \lor B_n$.

A sequent $\Gamma \vdash \Delta$ is satisfied in an interpretation $\mathfrak I$ if either some formula in $\Gamma$ is not satisfied by $\mathfrak I$, or some 
  formula in $\Delta$ is satisfied by $\mathfrak I$. 
A sequent is valid if it is satisfied in every interpretation. 

The inference rules must be sound, i.e. they must derive true conclusion from true premises.
Consider now the first rule: 

$(\lnot \text{left}) \ \ \ \ \dfrac{\Gamma \vdash \Delta, A}{\lnot A, \Gamma \vdash\Delta};$

forgetting about the contexts ($\Gamma$ and $\Delta$), if the upper sequent is true in $\mathfrak I$, this means that $A$ is true, and thus $\lnot A$ is false.
The same for $(\lnot \text{right}).$
A: The "design intent" of the classical sequent calculus is that the sequent $$A_1,\ldots,A_m\vdash B_1,\ldots,B_n$$
should be derivable if and only if
$$ \neg A_1 \lor \cdots \lor \neg A_m \lor B_1 \lor \cdots \lor B_n $$
is logically valid.
Viewed in this light, passing between $\Gamma,A\vdash \Delta$ and $\Gamma\vdash \neg A,\Delta$ does not change the correctness condition at all, so it is reasonable that they should be derivable from each other.
And passing between $\Gamma\vdash B,\Delta$ and $\Gamma,\neg B\vdash \Delta$ doesn't change the meaning of the correctness criterion, if we accept that $\neg\neg B$ has the same meaning as $B$ itself.
