MAYBE
0.07/0.18	MAYBE
0.07/0.18	
0.07/0.18	Problem:
0.07/0.18	 perfectp(0()) -> false()
0.07/0.18	 perfectp(s(x)) -> f(x,s(0()),s(x),s(x))
0.07/0.18	 f(0(),y,0(),u) -> true()
0.07/0.18	 f(0(),y,s(z),u) -> false()
0.07/0.18	 f(s(x),0(),z,u) -> f(x,u,minus(z,s(x)),u)
0.07/0.18	 f(s(x),s(y),z,u) -> if(le(x,y),f(s(x),minus(y,x),z,u),f(x,u,z,u))
0.07/0.18	
0.07/0.18	Proof:
0.07/0.18	 Open
0.07/0.19	EOF