3.45/1.67	YES
3.45/1.68	proof of /export/starexec/sandbox/benchmark/theBenchmark.xml
3.45/1.68	# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty
3.45/1.68	
3.45/1.68	
3.45/1.68	Termination w.r.t. Q of the given QTRS could be proven:
3.45/1.68	
3.45/1.68	(0) QTRS
3.45/1.68	(1) DependencyPairsProof [EQUIVALENT, 0 ms]
3.45/1.68	(2) QDP
3.45/1.68	(3) DependencyGraphProof [EQUIVALENT, 0 ms]
3.45/1.68	(4) TRUE
3.45/1.68	
3.45/1.68	
3.45/1.68	----------------------------------------
3.45/1.68	
3.45/1.68	(0)
3.45/1.68	Obligation:
3.45/1.68	Q restricted rewrite system:
3.45/1.68	The TRS R consists of the following rules:
3.45/1.68	
3.45/1.68	   f(s(x)) -> f(g(x, x))
3.45/1.68	   g(0, 1) -> s(0)
3.45/1.68	   0 -> 1
3.45/1.68	
3.45/1.68	The set Q consists of the following terms:
3.45/1.68	
3.45/1.68	   f(s(x0))
3.45/1.68	   0
3.45/1.68	
3.45/1.68	
3.45/1.68	----------------------------------------
3.45/1.68	
3.45/1.68	(1) DependencyPairsProof (EQUIVALENT)
3.45/1.68	Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.
3.45/1.68	----------------------------------------
3.45/1.68	
3.45/1.68	(2)
3.45/1.68	Obligation:
3.45/1.68	Q DP problem:
3.45/1.68	The TRS P consists of the following rules:
3.45/1.68	
3.45/1.68	   F(s(x)) -> F(g(x, x))
3.45/1.68	   F(s(x)) -> G(x, x)
3.45/1.68	
3.45/1.68	The TRS R consists of the following rules:
3.45/1.68	
3.45/1.68	   f(s(x)) -> f(g(x, x))
3.45/1.68	   g(0, 1) -> s(0)
3.45/1.68	   0 -> 1
3.45/1.68	
3.45/1.68	The set Q consists of the following terms:
3.45/1.68	
3.45/1.68	   f(s(x0))
3.45/1.68	   0
3.45/1.68	
3.45/1.68	We have to consider all minimal (P,Q,R)-chains.
3.45/1.68	----------------------------------------
3.45/1.68	
3.45/1.68	(3) DependencyGraphProof (EQUIVALENT)
3.45/1.68	The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 2 less nodes.
3.45/1.68	----------------------------------------
3.45/1.68	
3.45/1.68	(4)
3.45/1.68	TRUE
3.52/1.69	EOF