MAYBE

Problem:
 h(1(1(x1))) -> 1(h(x1))
 1(1(h(b(x1)))) -> 1(1(s(b(x1))))
 1(s(x1)) -> s(1(x1))
 b(s(x1)) -> b(h(x1))
 h(1(b(x1))) -> t(1(1(b(x1))))
 1(t(x1)) -> t(1(1(1(x1))))
 b(t(x1)) -> b(h(x1))

Proof:
 Open