MAYBE Problem: b(a(b(x))) -> b(a(a(a(b(x))))) b(a(a(a(b(a(a(b(x)))))))) -> b(a(a(b(a(a(b(a(a(a(b(b(x)))))))))))) b(a(a(a(b(a(a(a(b(x))))))))) -> b(x) b(a(a(a(b(b(b(x))))))) -> b(b(b(a(a(a(b(x))))))) b(a(a(b(b(x))))) -> b(x) b(b(a(a(b(x))))) -> b(x) b(a(a(a(b(a(b(x))))))) -> b(x) b(a(b(a(a(a(b(x))))))) -> b(x) Proof: Open