Consider a simple random walk $\bf x$ with ${\bf x}(0) = 0$, let $R$ be the first return time (return to 0), and let $\#(R,x)$ be the number of visits of $x$ before time $R$ (WLOG assume $x>0$, so in the calculation we only consider random walks that stay above zero before first return). Please help explain why the following highlighted equality holds.
If you have conditioned on $x(n) = x$ and $x_i > 0$ for $i = 1,\ldots, n$, then the new process translated by $x$ must have the property that $$ \begin{cases} x - x(i) < x, \qquad i = 1,\ldots,n-1, \\ x - x(n) = 0. \end{cases} $$
Otherwise you would have $$ x-x(i) \geq x \mbox{ for some } i \in 1,\ldots,n-1 \implies x(i) \leq 0. $$ Now take that same sequence of steps, but starting from $n$ and you have a sequence such that $x_0 = 0$ and $x_n = x$ and for all $x_i$ you have $x_i < x$ which is the desired property. In terms of the sample path of this sequence the $i$th step is the $(n - i)$th of the original sequence, for example you might have the mapping $$ (L,L,R,L,L,L,R) \mapsto (R,L,L,L,R,L,L) $$ but the probability remains invariant under this relabelling and so you get the desired claim. Attached a quick picture which hopefully supports my argument