One will need a copy of Feller's text (3rd edition) to answer this question.
The proof I'm having difficulty with is Theorem 1, pages 84-85.
When he discusses the r=1 case, he says ... "To the latter section we apply the result for r=0 but interchanging the roles of plus and minus. We conclude that the number of paths of length 2n-2v starting at (2v,-2) and not crossing the level -1 equals the number of paths from (2v,-2) to (2n+1,-3)."
My question is, how does Feller obtain (2n+1,-3)? Related is, what is meant (exactly) by interchanging the roles of plus and minus?
Note, I do believe that he is using the reflection lemma as he does in the r=0 case.
Consider any path, $P_1$, from $(2v,-2)$ of length $2n-2v$ that doesn't cross the line $-1$. Reflect it in the line $-1$ and shift it left $2v$ units. Then we have a path, $P_2$, starting from $(0,0)$ of length $2n-2v$ that doesn't cross the line $-1$. Note that "interchanging the roles of plus and minus" refers to the effect of this reflection on the individual steps in the path. Then to $P_2$ we can apply the result for $r=0$ as follows:
\begin{eqnarray*} && \dfrac{\text{#Paths from $(2v,-2)$ of length $2n-2v$ that don't cross the line $-1$}}{\text{#Paths of length $2n-2v$}} \\ && \\ &&\qquad = P(\text{A path from $(2v,-2)$ of length $2n-2v$ doesn't cross the line $-1$}) \\ && \\ &&\qquad = 2p_{2n-2v+1,1} \qquad\qquad\qquad\qquad\qquad\qquad\text{(applying result for $r=0$)} \\ && \\ &&\qquad = 2p_{2n-2v+1,-1} \qquad\qquad\qquad\qquad\qquad\qquad\text{(by symmetry)} \\ && \\ &&\qquad = 2\times \dfrac{\text{#Paths from $(2v,-2)$ to $(2n+1,-3)$}}{\text{#Paths of length $2n-2v+1$}} \\ && \\ &&\qquad = \dfrac{\text{#Paths from $(2v,-2)$ to $(2n+1,-3)$}}{\text{#Paths of length $2n-2v$}}. \\ && \\ && \text{Equating numerators, we have:} \\ && \\ && \text{#Paths from $(2v,-2)$ of length $2n-2v$ that don't cross the line $-1$} \\ &&\qquad = \text{#Paths from $(2v,-2)$ to $(2n+1,-3)$}. \\ \end{eqnarray*}