21.19/6.36	YES
21.52/6.39	proof of /export/starexec/sandbox2/benchmark/theBenchmark.xml
21.52/6.39	# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty
21.52/6.39	
21.52/6.39	
21.52/6.39	Termination w.r.t. Q of the given QTRS could be proven:
21.52/6.39	
21.52/6.39	(0) QTRS
21.52/6.39	(1) QTRSRRRProof [EQUIVALENT, 37 ms]
21.52/6.39	(2) QTRS
21.52/6.39	(3) DependencyPairsProof [EQUIVALENT, 24 ms]
21.52/6.39	(4) QDP
21.52/6.39	(5) MRRProof [EQUIVALENT, 7 ms]
21.52/6.39	(6) QDP
21.52/6.39	(7) PisEmptyProof [EQUIVALENT, 0 ms]
21.52/6.39	(8) YES
21.52/6.39	
21.52/6.39	
21.52/6.39	----------------------------------------
21.52/6.39	
21.52/6.39	(0)
21.52/6.39	Obligation:
21.52/6.39	Q restricted rewrite system:
21.52/6.39	The TRS R consists of the following rules:
21.52/6.39	
21.52/6.39	   a(x1) -> b(x1)
21.52/6.39	   a(b(x1)) -> b(c(a(x1)))
21.52/6.39	   b(x1) -> c(x1)
21.52/6.39	   c(b(x1)) -> a(x1)
21.52/6.39	
21.52/6.39	Q is empty.
21.52/6.39	
21.52/6.39	----------------------------------------
21.52/6.39	
21.52/6.39	(1) QTRSRRRProof (EQUIVALENT)
21.52/6.39	Used ordering:
21.52/6.39	Polynomial interpretation [POLO]:
21.52/6.39	
21.52/6.39	   POL(a(x_1)) = 1 + x_1
21.52/6.39	   POL(b(x_1)) = 1 + x_1
21.52/6.39	   POL(c(x_1)) = x_1
21.52/6.39	With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:
21.52/6.39	
21.52/6.39	   b(x1) -> c(x1)
21.52/6.39	
21.52/6.39	
21.52/6.39	
21.52/6.39	
21.52/6.39	----------------------------------------
21.52/6.39	
21.52/6.39	(2)
21.52/6.39	Obligation:
21.52/6.39	Q restricted rewrite system:
21.52/6.39	The TRS R consists of the following rules:
21.52/6.39	
21.52/6.39	   a(x1) -> b(x1)
21.52/6.39	   a(b(x1)) -> b(c(a(x1)))
21.52/6.39	   c(b(x1)) -> a(x1)
21.52/6.39	
21.52/6.39	Q is empty.
21.52/6.39	
21.52/6.39	----------------------------------------
21.52/6.39	
21.52/6.39	(3) DependencyPairsProof (EQUIVALENT)
21.52/6.39	Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.
21.52/6.39	----------------------------------------
21.52/6.39	
21.52/6.39	(4)
21.52/6.39	Obligation:
21.52/6.39	Q DP problem:
21.52/6.39	The TRS P consists of the following rules:
21.52/6.39	
21.52/6.39	   A(b(x1)) -> C(a(x1))
21.52/6.39	   A(b(x1)) -> A(x1)
21.52/6.39	   C(b(x1)) -> A(x1)
21.52/6.39	
21.52/6.39	The TRS R consists of the following rules:
21.52/6.39	
21.52/6.39	   a(x1) -> b(x1)
21.52/6.39	   a(b(x1)) -> b(c(a(x1)))
21.52/6.39	   c(b(x1)) -> a(x1)
21.52/6.39	
21.52/6.39	Q is empty.
21.52/6.39	We have to consider all minimal (P,Q,R)-chains.
21.52/6.39	----------------------------------------
21.52/6.39	
21.52/6.39	(5) MRRProof (EQUIVALENT)
21.52/6.39	By using the rule removal processor [LPAR04] with the following ordering, at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented.
21.52/6.39	
21.52/6.39	Strictly oriented dependency pairs:
21.52/6.39	
21.52/6.39	   A(b(x1)) -> C(a(x1))
21.52/6.39	   A(b(x1)) -> A(x1)
21.52/6.39	   C(b(x1)) -> A(x1)
21.52/6.39	
21.52/6.39	
21.52/6.39	Used ordering: Polynomial interpretation [POLO]:
21.52/6.39	
21.52/6.39	   POL(A(x_1)) = 3 + 2*x_1
21.52/6.39	   POL(C(x_1)) = 2*x_1
21.52/6.39	   POL(a(x_1)) = 2 + x_1
21.52/6.39	   POL(b(x_1)) = 2 + x_1
21.52/6.39	   POL(c(x_1)) = x_1
21.52/6.39	
21.52/6.39	
21.52/6.39	----------------------------------------
21.52/6.39	
21.52/6.39	(6)
21.52/6.39	Obligation:
21.52/6.39	Q DP problem:
21.52/6.39	P is empty.
21.52/6.39	The TRS R consists of the following rules:
21.52/6.39	
21.52/6.39	   a(x1) -> b(x1)
21.52/6.39	   a(b(x1)) -> b(c(a(x1)))
21.52/6.39	   c(b(x1)) -> a(x1)
21.52/6.39	
21.52/6.39	Q is empty.
21.52/6.39	We have to consider all minimal (P,Q,R)-chains.
21.52/6.39	----------------------------------------
21.52/6.39	
21.52/6.39	(7) PisEmptyProof (EQUIVALENT)
21.52/6.39	The TRS P is empty. Hence, there is no (P,Q,R) chain.
21.52/6.39	----------------------------------------
21.52/6.39	
21.52/6.39	(8)
21.52/6.39	YES
21.72/6.51	EOF