Reflexion principle equivalence of statements

Question

$B_t$ is a Brownian motion, $S_t$ is defined as $$ S_t := \sup_{0 \leq s \leq t} B_s $$ I want to show that $$ P(S_t\geq b , B_t \leq a) = P(B_t \geq 2b-a) $$ for $a \leq b$ implies $$ P(S_t \geq a) = 2 P (B_t \geq a).$$ I have tried posing $b=a$, however this leads to $$ P(S_t\geq a) = P(S_t\geq a , B_t \leq a) = P(B_t \geq a) $$ Which is different from the second statement. One of those four must be wrong:

1. My one-line conclusion
2. The first statement (page 2 of https://www.ceremade.dauphine.fr/~mischler/Enseignements/ProcContM1/PrincipeReflexion.pdf)
3. The second statement (page 2 of https://ocw.mit.edu/courses/sloan-school-of-management/15-070j-advanced-stochastic-processes-fall-2013/lecture-notes/MIT15_070JF13_Lec7.pdf)
4. Mathematics (we are still unsure whether mathematics is consistent, right?).

Answer 1

Your (my) one-line conclusion is wrong: $$ P(S_t \geq a) = P(S_t \geq a, B_t > a) + P(S_t \geq a, B_t\leq a).$$ I don't know why on earth you (I) assumed that $P(S_t \geq a, B_t > a) = 0$.

Since $B_t > a$ implies $S_t >a$, then $$ P(S_t \geq a, B_t > a) = P(B_t \geq a).$$ Hence $$ P(S_t \geq a) = P(S_t \geq a, B_t > a) + P(S_t \geq a, B_t\leq a) = 2 P(B_t \geq a). $$ The consistency of mathematics is saved... at least for now...