MAYBE

Problem:
 a(b(a(x))) -> b(a(b(x)))
 b(a(b(x))) -> a(b(a(x)))
 a(c(a(x))) -> c(a(c(x)))
 c(a(c(x))) -> a(c(a(x)))
 b(c(x)) -> c(b(x))
 c(b(x)) -> b(c(x))
 p() -> a(p())
 p() -> b(p())
 p() -> c(p())

Proof:
 Open