Poisson process expected value Let {${N(t), t \geq0}$} be a Poisson process with rate $\lambda$. Let $S_n$ denote the time of the $nth$ event. What is $E[N(4)-N(2)|N(1)=3]$?

(Note: $E[X]$ is the expected value or mean). 

I know that generally $E[N(t)]= \lambda t$, where $t$ is time. Since this is a Poisson process, wouldn't it not matter that $N(1)=3$ since the increments are independent? If so, then would it be correct to claim that $E[N(4)-N(2)|N(1)=3]$= $E[N(4)-N(2)]$=$2 \lambda$?
I don't feel I can contribute any more by writing out an answer, except:

yes you are correct. It follows from independent increment. that $N(4)-N(2)$ is independent of $N(1)-N(0)=N(1)$. therefore, is is just $E[N(4)-N(2)]=2\lambda$
Actually, wouldn't the answer be $2\lambda+3$? $2\lambda$ is the expected occurrences from time $2$ to time $4$ $(N(4)-N(2))$ and you are guaranteed $3$ from the condition.