8.77/3.06	YES
8.77/3.09	proof of /export/starexec/sandbox/benchmark/theBenchmark.xml
8.77/3.09	# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty
8.77/3.09	
8.77/3.09	
8.77/3.09	Termination w.r.t. Q of the given QTRS could be proven:
8.77/3.09	
8.77/3.09	(0) QTRS
8.77/3.09	(1) QTRSRRRProof [EQUIVALENT, 69 ms]
8.77/3.09	(2) QTRS
8.77/3.09	(3) DependencyPairsProof [EQUIVALENT, 0 ms]
8.77/3.09	(4) QDP
8.77/3.09	(5) DependencyGraphProof [EQUIVALENT, 0 ms]
8.77/3.09	(6) QDP
8.77/3.09	(7) MRRProof [EQUIVALENT, 63 ms]
8.77/3.09	(8) QDP
8.77/3.09	(9) PisEmptyProof [EQUIVALENT, 0 ms]
8.77/3.09	(10) YES
8.77/3.09	
8.77/3.09	
8.77/3.09	----------------------------------------
8.77/3.09	
8.77/3.09	(0)
8.77/3.09	Obligation:
8.77/3.09	Q restricted rewrite system:
8.77/3.09	The TRS R consists of the following rules:
8.77/3.09	
8.77/3.09	   a(x1) -> b(b(x1))
8.77/3.09	   a(b(b(x1))) -> b(b(c(c(c(a(x1))))))
8.77/3.09	   b(b(x1)) -> c(c(c(x1)))
8.77/3.09	   c(c(c(b(b(x1))))) -> a(x1)
8.77/3.09	
8.77/3.09	Q is empty.
8.77/3.09	
8.77/3.09	----------------------------------------
8.77/3.09	
8.77/3.09	(1) QTRSRRRProof (EQUIVALENT)
8.77/3.09	Used ordering:
8.77/3.09	Polynomial interpretation [POLO]:
8.77/3.09	
8.77/3.09	   POL(a(x_1)) = 2 + x_1
8.77/3.09	   POL(b(x_1)) = 1 + x_1
8.77/3.09	   POL(c(x_1)) = x_1
8.77/3.09	With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:
8.77/3.09	
8.77/3.09	   b(b(x1)) -> c(c(c(x1)))
8.77/3.09	
8.77/3.09	
8.77/3.09	
8.77/3.09	
8.77/3.09	----------------------------------------
8.77/3.09	
8.77/3.09	(2)
8.77/3.09	Obligation:
8.77/3.09	Q restricted rewrite system:
8.77/3.09	The TRS R consists of the following rules:
8.77/3.09	
8.77/3.09	   a(x1) -> b(b(x1))
8.77/3.09	   a(b(b(x1))) -> b(b(c(c(c(a(x1))))))
8.77/3.09	   c(c(c(b(b(x1))))) -> a(x1)
8.77/3.09	
8.77/3.09	Q is empty.
8.77/3.09	
8.77/3.09	----------------------------------------
8.77/3.09	
8.77/3.09	(3) DependencyPairsProof (EQUIVALENT)
8.77/3.09	Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.
8.77/3.09	----------------------------------------
8.77/3.09	
8.77/3.09	(4)
8.77/3.09	Obligation:
8.77/3.09	Q DP problem:
8.77/3.09	The TRS P consists of the following rules:
8.77/3.09	
8.77/3.09	   A(b(b(x1))) -> C(c(c(a(x1))))
8.77/3.09	   A(b(b(x1))) -> C(c(a(x1)))
8.77/3.09	   A(b(b(x1))) -> C(a(x1))
8.77/3.09	   A(b(b(x1))) -> A(x1)
8.77/3.09	   C(c(c(b(b(x1))))) -> A(x1)
8.77/3.09	
8.77/3.09	The TRS R consists of the following rules:
8.77/3.09	
8.77/3.09	   a(x1) -> b(b(x1))
8.77/3.09	   a(b(b(x1))) -> b(b(c(c(c(a(x1))))))
8.77/3.09	   c(c(c(b(b(x1))))) -> a(x1)
8.77/3.09	
8.77/3.09	Q is empty.
8.77/3.09	We have to consider all minimal (P,Q,R)-chains.
8.77/3.09	----------------------------------------
8.77/3.09	
8.77/3.09	(5) DependencyGraphProof (EQUIVALENT)
8.77/3.09	The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 2 less nodes.
8.77/3.09	----------------------------------------
8.77/3.09	
8.77/3.09	(6)
8.77/3.09	Obligation:
8.77/3.09	Q DP problem:
8.77/3.09	The TRS P consists of the following rules:
8.77/3.09	
8.77/3.09	   C(c(c(b(b(x1))))) -> A(x1)
8.77/3.09	   A(b(b(x1))) -> C(c(c(a(x1))))
8.77/3.09	   A(b(b(x1))) -> A(x1)
8.77/3.09	
8.77/3.09	The TRS R consists of the following rules:
8.77/3.09	
8.77/3.09	   a(x1) -> b(b(x1))
8.77/3.09	   a(b(b(x1))) -> b(b(c(c(c(a(x1))))))
8.77/3.09	   c(c(c(b(b(x1))))) -> a(x1)
8.77/3.09	
8.77/3.09	Q is empty.
8.77/3.09	We have to consider all minimal (P,Q,R)-chains.
8.77/3.09	----------------------------------------
8.77/3.09	
8.77/3.09	(7) MRRProof (EQUIVALENT)
8.77/3.09	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.
8.77/3.09	
8.77/3.09	Strictly oriented dependency pairs:
8.77/3.09	
8.77/3.09	   C(c(c(b(b(x1))))) -> A(x1)
8.77/3.09	   A(b(b(x1))) -> C(c(c(a(x1))))
8.77/3.09	   A(b(b(x1))) -> A(x1)
8.77/3.09	
8.77/3.09	
8.77/3.09	Used ordering: Polynomial interpretation [POLO]:
8.77/3.09	
8.77/3.09	   POL(A(x_1)) = 1 + x_1
8.77/3.09	   POL(C(x_1)) = x_1
8.77/3.09	   POL(a(x_1)) = 2 + x_1
8.77/3.09	   POL(b(x_1)) = 1 + x_1
8.77/3.09	   POL(c(x_1)) = x_1
8.77/3.09	
8.77/3.09	
8.77/3.09	----------------------------------------
8.77/3.09	
8.77/3.09	(8)
8.77/3.09	Obligation:
8.77/3.09	Q DP problem:
8.77/3.09	P is empty.
8.77/3.09	The TRS R consists of the following rules:
8.77/3.09	
8.77/3.09	   a(x1) -> b(b(x1))
8.77/3.09	   a(b(b(x1))) -> b(b(c(c(c(a(x1))))))
8.77/3.09	   c(c(c(b(b(x1))))) -> a(x1)
8.77/3.09	
8.77/3.09	Q is empty.
8.77/3.09	We have to consider all minimal (P,Q,R)-chains.
8.77/3.09	----------------------------------------
8.77/3.09	
8.77/3.09	(9) PisEmptyProof (EQUIVALENT)
8.77/3.09	The TRS P is empty. Hence, there is no (P,Q,R) chain.
8.77/3.09	----------------------------------------
8.77/3.09	
8.77/3.09	(10)
8.77/3.09	YES
9.18/3.20	EOF