Let $W_t$ be the Brownian motion starting at $0$. Consider the following random variables. $M_t = \sup_{0\leq s \leq t} W_s$ and $|W_t|$.
We first calculate
$$\Bbb{P}(|W_t|>a ) = \Bbb{P}(W_t >a) + \Bbb{P}(W_t< -a)$$
Fact: The random variable $M_t - W_t$ has the same distribution of $|W_t|$.
I was thinking, is there a way of proving that
$$\Bbb{P}(M_t - W_t >a) = \Bbb{P}(W_t >a) + \Bbb{P}(W_t <-a) $$
using the reflexion principle?
Attempt: The event $$[M_t-W_t >a] = [M_t-W_t >a, W_t\geq - a] \cup [M_t-W_t >a, W_t< - a] = [M_t-W_t >a, W_t\geq - a] \cup [ W_t< - a]. $$
What I am trying to do is to prove (using some sort of reflexion principle) that $$\Bbb{P}(M_t-W_t >a, W_t\geq - a) = \Bbb{P}(W_t >a)$$
Edit:
The reflexion principle is given by $$P^0[W_t \leq a \mid \mathcal{F}_{T_b}] = P^0[W_t \geq 2b-a \mid \mathcal{F}_{T_b}] \text{ on } \{T_b \leq t\} $$
where $T_b = \inf \{t \geq 0 , W_t = b\}$