 Alternating split of a Poisson process

Alternately routing a rate-$r$ Poisson process produces non-Poisson streams with common interarrival law Erlang$(2,r)$. The odd-numbered stream has an initial exponential delay, while the even-numbered stream is an ordinary renewal process. Their equilibrium delay survivor is $e^{-rt}(1+rt/2)$. The probability of no odd-stream point in $(s,s+t]$ is $e^{-rt}[1+rt(1-e^{-2rs})/2]$; for the even stream replace the minus by a plus. Conditioning on the parity of $N(s)$ proves these formulas.