NO

Problem:
 a(x1) -> x1
 a(b(x1)) -> b(a(c(b(c(x1)))))
 b(x1) -> a(x1)
 c(c(x1)) -> x1

Proof:
 Unfolding Processor:
  loop length: 11
  terms:
   a(b(c(b(x33527))))
   b(a(c(b(c(c(b(x33527)))))))
   a(a(c(b(c(c(b(x33527)))))))
   a(a(c(a(c(c(b(x33527)))))))
   a(a(c(a(b(x33527)))))
   a(a(c(b(a(c(b(c(x33527))))))))
   a(a(c(a(a(c(b(c(x33527))))))))
   a(a(c(a(c(b(c(x33527)))))))
   a(a(c(c(b(c(x33527))))))
   a(a(b(c(x33527))))
   a(b(a(c(b(c(c(x33527)))))))
  context: []
  substitution:
   x33527 -> c(c(x33527))
  Qed