3.95/1.88	YES
3.95/1.89	proof of /export/starexec/sandbox2/benchmark/theBenchmark.xml
3.95/1.89	# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty
3.95/1.89	
3.95/1.89	
3.95/1.89	Termination w.r.t. Q of the given QTRS could be proven:
3.95/1.89	
3.95/1.89	(0) QTRS
3.95/1.89	(1) DependencyPairsProof [EQUIVALENT, 0 ms]
3.95/1.89	(2) QDP
3.95/1.89	(3) UsableRulesProof [EQUIVALENT, 0 ms]
3.95/1.89	(4) QDP
3.95/1.89	(5) QReductionProof [EQUIVALENT, 0 ms]
3.95/1.89	(6) QDP
3.95/1.89	(7) TransformationProof [EQUIVALENT, 0 ms]
3.95/1.89	(8) QDP
3.95/1.89	(9) DependencyGraphProof [EQUIVALENT, 0 ms]
3.95/1.89	(10) TRUE
3.95/1.89	
3.95/1.89	
3.95/1.89	----------------------------------------
3.95/1.89	
3.95/1.89	(0)
3.95/1.89	Obligation:
3.95/1.89	Q restricted rewrite system:
3.95/1.89	The TRS R consists of the following rules:
3.95/1.89	
3.95/1.89	   f(0, 1, X) -> h(X, X)
3.95/1.89	   h(0, X) -> f(0, X, X)
3.95/1.89	   g(X, Y) -> X
3.95/1.89	   g(X, Y) -> Y
3.95/1.89	
3.95/1.89	The set Q consists of the following terms:
3.95/1.89	
3.95/1.89	   f(0, 1, x0)
3.95/1.89	   h(0, x0)
3.95/1.89	   g(x0, x1)
3.95/1.89	
3.95/1.89	
3.95/1.89	----------------------------------------
3.95/1.89	
3.95/1.89	(1) DependencyPairsProof (EQUIVALENT)
3.95/1.89	Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.
3.95/1.89	----------------------------------------
3.95/1.89	
3.95/1.89	(2)
3.95/1.89	Obligation:
3.95/1.89	Q DP problem:
3.95/1.89	The TRS P consists of the following rules:
3.95/1.89	
3.95/1.89	   F(0, 1, X) -> H(X, X)
3.95/1.89	   H(0, X) -> F(0, X, X)
3.95/1.89	
3.95/1.89	The TRS R consists of the following rules:
3.95/1.89	
3.95/1.89	   f(0, 1, X) -> h(X, X)
3.95/1.89	   h(0, X) -> f(0, X, X)
3.95/1.89	   g(X, Y) -> X
3.95/1.89	   g(X, Y) -> Y
3.95/1.89	
3.95/1.89	The set Q consists of the following terms:
3.95/1.89	
3.95/1.89	   f(0, 1, x0)
3.95/1.89	   h(0, x0)
3.95/1.89	   g(x0, x1)
3.95/1.89	
3.95/1.89	We have to consider all minimal (P,Q,R)-chains.
3.95/1.89	----------------------------------------
3.95/1.89	
3.95/1.89	(3) UsableRulesProof (EQUIVALENT)
3.95/1.89	As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.
3.95/1.89	----------------------------------------
3.95/1.89	
3.95/1.89	(4)
3.95/1.89	Obligation:
3.95/1.89	Q DP problem:
3.95/1.89	The TRS P consists of the following rules:
3.95/1.89	
3.95/1.89	   F(0, 1, X) -> H(X, X)
3.95/1.89	   H(0, X) -> F(0, X, X)
3.95/1.89	
3.95/1.89	R is empty.
3.95/1.89	The set Q consists of the following terms:
3.95/1.89	
3.95/1.89	   f(0, 1, x0)
3.95/1.89	   h(0, x0)
3.95/1.89	   g(x0, x1)
3.95/1.89	
3.95/1.89	We have to consider all minimal (P,Q,R)-chains.
3.95/1.89	----------------------------------------
3.95/1.89	
3.95/1.89	(5) QReductionProof (EQUIVALENT)
3.95/1.89	We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].
3.95/1.89	
3.95/1.89	   f(0, 1, x0)
3.95/1.89	   h(0, x0)
3.95/1.89	   g(x0, x1)
3.95/1.89	
3.95/1.89	
3.95/1.89	----------------------------------------
3.95/1.89	
3.95/1.89	(6)
3.95/1.89	Obligation:
3.95/1.89	Q DP problem:
3.95/1.89	The TRS P consists of the following rules:
3.95/1.89	
3.95/1.89	   F(0, 1, X) -> H(X, X)
3.95/1.89	   H(0, X) -> F(0, X, X)
3.95/1.89	
3.95/1.89	R is empty.
3.95/1.89	Q is empty.
3.95/1.89	We have to consider all minimal (P,Q,R)-chains.
3.95/1.89	----------------------------------------
3.95/1.89	
3.95/1.89	(7) TransformationProof (EQUIVALENT)
3.95/1.89	By instantiating [LPAR04] the rule F(0, 1, X) -> H(X, X) we obtained the following new rules [LPAR04]:
3.95/1.89	
3.95/1.89	   (F(0, 1, 1) -> H(1, 1),F(0, 1, 1) -> H(1, 1))
3.95/1.89	
3.95/1.89	
3.95/1.89	----------------------------------------
3.95/1.89	
3.95/1.89	(8)
3.95/1.89	Obligation:
3.95/1.89	Q DP problem:
3.95/1.89	The TRS P consists of the following rules:
3.95/1.89	
3.95/1.89	   H(0, X) -> F(0, X, X)
3.95/1.89	   F(0, 1, 1) -> H(1, 1)
3.95/1.89	
3.95/1.89	R is empty.
3.95/1.89	Q is empty.
3.95/1.89	We have to consider all minimal (P,Q,R)-chains.
3.95/1.89	----------------------------------------
3.95/1.89	
3.95/1.89	(9) DependencyGraphProof (EQUIVALENT)
3.95/1.89	The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 2 less nodes.
3.95/1.89	----------------------------------------
3.95/1.89	
3.95/1.89	(10)
3.95/1.89	TRUE
4.05/1.91	EOF