4.10/1.94	YES
4.10/1.94	proof of /export/starexec/sandbox2/benchmark/theBenchmark.xml
4.10/1.94	# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty
4.10/1.94	
4.10/1.94	
4.10/1.94	Termination w.r.t. Q of the given QTRS could be proven:
4.10/1.94	
4.10/1.94	(0) QTRS
4.10/1.94	(1) DependencyPairsProof [EQUIVALENT, 0 ms]
4.10/1.94	(2) QDP
4.10/1.94	(3) UsableRulesProof [EQUIVALENT, 0 ms]
4.10/1.94	(4) QDP
4.10/1.94	(5) QReductionProof [EQUIVALENT, 0 ms]
4.10/1.94	(6) QDP
4.10/1.94	(7) TransformationProof [EQUIVALENT, 0 ms]
4.10/1.94	(8) QDP
4.10/1.94	(9) DependencyGraphProof [EQUIVALENT, 0 ms]
4.10/1.94	(10) TRUE
4.10/1.94	
4.10/1.94	
4.10/1.94	----------------------------------------
4.10/1.94	
4.10/1.94	(0)
4.10/1.94	Obligation:
4.10/1.94	Q restricted rewrite system:
4.10/1.94	The TRS R consists of the following rules:
4.10/1.94	
4.10/1.94	   h(x, y) -> f(x, y, x)
4.10/1.94	   f(0, 1, x) -> h(x, x)
4.10/1.94	   g(x, y) -> x
4.10/1.94	   g(x, y) -> y
4.10/1.94	
4.10/1.94	The set Q consists of the following terms:
4.10/1.94	
4.10/1.94	   h(x0, x1)
4.10/1.94	   f(0, 1, x0)
4.10/1.94	   g(x0, x1)
4.10/1.94	
4.10/1.94	
4.10/1.94	----------------------------------------
4.10/1.94	
4.10/1.94	(1) DependencyPairsProof (EQUIVALENT)
4.10/1.94	Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.
4.10/1.94	----------------------------------------
4.10/1.94	
4.10/1.94	(2)
4.10/1.94	Obligation:
4.10/1.94	Q DP problem:
4.10/1.94	The TRS P consists of the following rules:
4.10/1.94	
4.10/1.94	   H(x, y) -> F(x, y, x)
4.10/1.94	   F(0, 1, x) -> H(x, x)
4.10/1.94	
4.10/1.94	The TRS R consists of the following rules:
4.10/1.94	
4.10/1.94	   h(x, y) -> f(x, y, x)
4.10/1.94	   f(0, 1, x) -> h(x, x)
4.10/1.94	   g(x, y) -> x
4.10/1.94	   g(x, y) -> y
4.10/1.94	
4.10/1.94	The set Q consists of the following terms:
4.10/1.94	
4.10/1.94	   h(x0, x1)
4.10/1.94	   f(0, 1, x0)
4.10/1.94	   g(x0, x1)
4.10/1.94	
4.10/1.94	We have to consider all minimal (P,Q,R)-chains.
4.10/1.94	----------------------------------------
4.10/1.94	
4.10/1.94	(3) UsableRulesProof (EQUIVALENT)
4.10/1.94	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.
4.10/1.94	----------------------------------------
4.10/1.94	
4.10/1.94	(4)
4.10/1.94	Obligation:
4.10/1.94	Q DP problem:
4.10/1.94	The TRS P consists of the following rules:
4.10/1.94	
4.10/1.94	   H(x, y) -> F(x, y, x)
4.10/1.94	   F(0, 1, x) -> H(x, x)
4.10/1.94	
4.10/1.94	R is empty.
4.10/1.94	The set Q consists of the following terms:
4.10/1.94	
4.10/1.94	   h(x0, x1)
4.10/1.94	   f(0, 1, x0)
4.10/1.94	   g(x0, x1)
4.10/1.94	
4.10/1.94	We have to consider all minimal (P,Q,R)-chains.
4.10/1.94	----------------------------------------
4.10/1.94	
4.10/1.94	(5) QReductionProof (EQUIVALENT)
4.10/1.94	We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].
4.10/1.95	
4.10/1.95	   h(x0, x1)
4.10/1.95	   f(0, 1, x0)
4.10/1.95	   g(x0, x1)
4.10/1.95	
4.10/1.95	
4.10/1.95	----------------------------------------
4.10/1.95	
4.10/1.95	(6)
4.10/1.95	Obligation:
4.10/1.95	Q DP problem:
4.10/1.95	The TRS P consists of the following rules:
4.10/1.95	
4.10/1.95	   H(x, y) -> F(x, y, x)
4.10/1.95	   F(0, 1, x) -> H(x, x)
4.10/1.95	
4.10/1.95	R is empty.
4.10/1.95	Q is empty.
4.10/1.95	We have to consider all minimal (P,Q,R)-chains.
4.10/1.95	----------------------------------------
4.10/1.95	
4.10/1.95	(7) TransformationProof (EQUIVALENT)
4.10/1.95	By instantiating [LPAR04] the rule H(x, y) -> F(x, y, x) we obtained the following new rules [LPAR04]:
4.10/1.95	
4.10/1.95	   (H(z0, z0) -> F(z0, z0, z0),H(z0, z0) -> F(z0, z0, z0))
4.10/1.95	
4.10/1.95	
4.10/1.95	----------------------------------------
4.10/1.95	
4.10/1.95	(8)
4.10/1.95	Obligation:
4.10/1.95	Q DP problem:
4.10/1.95	The TRS P consists of the following rules:
4.10/1.95	
4.10/1.95	   F(0, 1, x) -> H(x, x)
4.10/1.95	   H(z0, z0) -> F(z0, z0, z0)
4.10/1.95	
4.10/1.95	R is empty.
4.10/1.95	Q is empty.
4.10/1.95	We have to consider all minimal (P,Q,R)-chains.
4.10/1.95	----------------------------------------
4.10/1.95	
4.10/1.95	(9) DependencyGraphProof (EQUIVALENT)
4.10/1.95	The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 2 less nodes.
4.10/1.95	----------------------------------------
4.10/1.95	
4.10/1.95	(10)
4.10/1.95	TRUE
4.10/1.96	EOF