# Computing the number of renewals

My question is from the book Introduction to Probability Models, 10th edition, by Sheldon Ross. Page 463, $\S$7.8.

There is a paragraph in the book talking about the computation of renewal function:

Let $Y$ be an exponential random variable having rate $\lambda$, and suppose that $Y$ is independent of the renewal process $\{N(t), t \geq 0\}$. We start by determining $E[N(Y)]$, the expected number of renewals by the random time $Y$. To do so, we first condition on $X_1$, the time of the first renewal. This yields $$E[N(Y)] = \int_0^\infty E[N(Y)|X_1=x]f(x)dx$$ where $f$ is the interarrival density. To determine $E[N(Y)|X_1 =x]$, we now condition on whether or not $Y$ exceeds $x$. Now, if $Y < x$, then as the first renewal occurs at time $x$, it follows that the number of renewals by time $Y$ is equal to $0$. On the other hand, if we are given that x < Y, then the number of renewals by time $Y$ will equal $1$ (the one at $x$) plus the number of additional renewals between $x$ and $Y$. But by the memoryless property of exponential random variables, it follows that, given that $Y > x$, the amount by which it exceeds $x$ is also exponentialwith rate $\lambda$, and so given that $Y > x$ the number of renewals between $x$ and $Y$ will have the same distribution as $N(Y)$. Hence, $$E[N(Y)|X_1 = x, Y < x] = 0$$ $$E[N(Y)|X_1 = x, Y > x] = 1 + \color{red}{E[N(Y)]}$$

My question is about the last item. While I think the last item should be $E[N(Y-x)]$ instead of $\color{red}{E[N(Y)]}$ in this case.

Even though the author provides an explanation before the equations, I find it confusing. Can anybody help me explain why the final item would be $\color{red}{E[N(Y)]}$?