while studying memorylessness property of the exponential distribution i came across this:
$$\mathbb P(X>s+t)=\mathbb P(X>t)\cdot \mathbb P(X>s+t|X>t)\tag{1}$$
$$\mathbb P(X>t)\cdot \mathbb P(X>s+t\mid X>t)=\mathbb P(X>t)\cdot \mathbb P(X>s)\tag{2} $$ which implies that $\mathbb P(X>s+t)=\mathbb P(X>s)\cdot \mathbb P(X>t)$.
my problem is that when we let $s=-t$ , then what we get is,
$$1=\mathbb P(X>0)=\mathbb P(X>t-t)=\mathbb P(X>t)\cdot \mathbb P(X>-t)=\mathbb P(X>t)\cdot 1 \neq 1$$
which is a contradiction. Surely i did something wrong but i cant figure out what it is.
edit1:
but if $t>0$ and $s<0$ then $$\mathbb P\{X>s+t\mid X>t\}=\frac{\mathbb P\{X>s+t,X>t\}}{\mathbb P\{X>t\}}\underset{\mathbb P\{X>t\}\neq1}{=}\frac{\mathbb P\{X>t\}}{\mathbb P\{X>t\}}= 1.$$
which implies that (using $(1)$) $\space \space \mathbb P\{X>s+t\} = \mathbb P\{X>t\}*1$
$X$ is the time until the first occurrence in a Poisson process.
$\{X>t+s\}$ thus represents the event of a Poisson arrival not occurring within the interval $(0,t+s]$.
When $s$ is positive the intervals $(0,t]$ and $(t, t+s]$ are disjoint, and the probability that the event does not arrives within the second interval when given that it does not arrive within the first of equal measure to the probability that it does not arrive within $(0,s]$. $$\begin{align}\mathsf P(X\notin(0,t]\land X\notin (t,t+s])~&=~\mathsf P(X\notin(0,t])\cdot\mathsf P(X\notin(t,t+s]\mid X\notin (0,t])\\[1ex]\therefore\quad\mathsf P(X\notin(0,t+s])~&=~\mathsf P(X\notin(0,t)]\cdot\mathsf P(X\notin(0,s])\end{align}$$
However, when $s$ is negative the intervals $(0,t]$ and $(t+s, t]$ are clearly intersecting and thus ....
$$\begin{align}\mathsf P(X\notin(0,t]\land X\notin (t+s,t])~&=~\mathsf P(X\notin(0,t])\cdot\mathsf P(X\notin(t+s,t]\mid X\notin (0,t])\\[2ex]\therefore\quad\mathsf P(X\notin(0,t])~&=~\mathsf P(X\notin(0,t])\cdot 1\qquad\text{unsurprisingly}\end{align}$$