MAYBE

Problem:
 acka(y(),n) -> s(n)
 acka(a(m),y()) -> acka(m,s(0()))
 acka(a(m),a(n)) -> acka(m,ackb(s(m),n))
 ackb(z(),n) -> s(n)
 ackb(b(m),z()) -> ackb(m,s(0()))
 ackb(b(m),b(n)) -> acka(m,ackb(s(m),n))
 s(n) -> a(n)
 s(n) -> b(n)
 0() -> y()
 0() -> z()

Proof:
 Open