NO
0.07/0.18	NO
0.07/0.18	
0.07/0.18	Problem:
0.07/0.18	 U11(tt()) -> N
0.07/0.18	 U21(tt()) -> s(plus(N,M))
0.07/0.18	 and(tt()) -> X
0.07/0.18	 isNat() -> tt()
0.07/0.18	 isNat() -> and(isNat())
0.07/0.18	 isNat() -> isNat()
0.07/0.18	 plus(N,0()) -> U11(isNat())
0.07/0.18	 plus(N,s(M)) -> U21(and(isNat()))
0.07/0.18	
0.07/0.18	Proof:
0.07/0.18	 Fresh Variable Processor: loop length: 1
0.07/0.18	                           terms:
0.07/0.18	                            U11(tt())
0.07/0.18	                           context: []
0.07/0.18	                           substitution:
0.07/0.18	                            N -> U11(tt())
0.07/0.18	  Qed
0.07/0.18	EOF