1.20/0.70	YES
1.20/0.70	
1.20/0.70	Problem:
1.20/0.70	 f(s(x1)) -> s(s(f(p(s(x1)))))
1.20/0.70	 f(0(x1)) -> 0(x1)
1.20/0.70	 p(s(x1)) -> x1
1.20/0.70	
1.20/0.70	Proof:
1.20/0.70	 Bounds Processor:
1.20/0.70	  bound: 0
1.20/0.70	  enrichment: match
1.20/0.70	  automaton:
1.20/0.70	   final states: {2,7,1}
1.20/0.70	   transitions:
1.20/0.70	    f40() -> 2*
1.20/0.70	    s0(5) -> 6*
1.20/0.70	    s0(2) -> 3*
1.20/0.70	    s0(6) -> 1*
1.20/0.70	    f0(4) -> 5*
1.20/0.70	    p0(3) -> 4*
1.20/0.70	    00(2) -> 7*
1.20/0.70	    1 -> 5*
1.20/0.70	    2 -> 4*
1.20/0.70	    7 -> 5*
1.20/0.70	  problem:
1.20/0.70	   
1.20/0.70	  Qed
1.63/0.70	EOF