Usually I see that in the structural operational semantics representation for the while loop, the program state don't change:
$(while \> B \> do \>S, \sigma) \rightarrow (if \>B \> then \>S; (while \> B \> do \>S) \> else \> SKIP, \sigma)$
For me, this not intuitive, if the state don't change (i.e. the status of the memory stays the same) then $B$ will continue to stay true and the program will never terminate.
Can anyone please explain why the state don't change in this rule?
The state can change in subsequent reduction steps because on the right hand side of
$$ \langle while\ B\ do\ S, \sigma \rangle \quad\rightarrow\quad \langle if\ B\ then\ ( {\color{red}{S}};\ while\ B\ do\ S )\ else\ skip, \sigma\rangle $$
the $while$-loop is guarded (preceeded) by $S$. The computation of $S$ may change the state so that the condition $B$ can evaluate to $\mathsf{false}$.
In programming language semantics, the notion of program state is not a vague philosophical notion, but a very precise mathematical one. A state $s$ in this small-step operational semantics is a partial function
$$ s : \mathbf{Var} \hookrightarrow \mathbb{Z} $$
that records the values of the variables. So if $s\, x = v$, then variable $x$ has value $v$. The state is necessarily a partial function, since it only makes sense to record the values of variables that actually occur.
The unfolding axiom
$$ \langle \texttt{while}\; b\; \texttt{do}\; S,s \rangle \Rightarrow \langle \texttt{if}\; b\; \texttt{then}\; S; \texttt{while}\; b\; \texttt{do}\; S \; \texttt{else skip},s \rangle$$
is simply telling us that we unfold a while-loop into a conditional statement, one of whose branches contains the loop. No variables will change their value because of this, and for this reason the state does not change.
The state $\sigma$ does not change when we consider $B$ to decide whether to perform one iteration of the loop, but it can change later when we run the body $S$. And so, the next time we consider $B$, there can be a change of $\sigma$.
