MAYBE

Problem:
 s(b(x1)) -> b(s(s(s(x1))))
 s(b(s(x1))) -> b(t(x1))
 t(b(x1)) -> b(s(x1))
 t(b(s(x1))) -> u(t(b(x1)))
 b(u(x1)) -> b(s(x1))
 t(s(x1)) -> t(t(x1))
 t(u(x1)) -> u(t(x1))
 s(u(x1)) -> s(s(x1))

Proof:
 Open