23.74/7.04	YES
23.74/7.05	proof of /export/starexec/sandbox2/benchmark/theBenchmark.xml
23.74/7.05	# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty
23.74/7.05	
23.74/7.05	
23.74/7.05	Termination w.r.t. Q of the given QTRS could be proven:
23.74/7.05	
23.74/7.05	(0) QTRS
23.74/7.05	(1) DependencyPairsProof [EQUIVALENT, 27 ms]
23.74/7.05	(2) QDP
23.74/7.05	(3) MRRProof [EQUIVALENT, 71 ms]
23.74/7.05	(4) QDP
23.74/7.05	(5) MRRProof [EQUIVALENT, 13 ms]
23.74/7.05	(6) QDP
23.74/7.05	(7) PisEmptyProof [EQUIVALENT, 0 ms]
23.74/7.05	(8) YES
23.74/7.05	
23.74/7.05	
23.74/7.05	----------------------------------------
23.74/7.05	
23.74/7.05	(0)
23.74/7.05	Obligation:
23.74/7.05	Q restricted rewrite system:
23.74/7.05	The TRS R consists of the following rules:
23.74/7.05	
23.74/7.05	   b(a(a(b(x1)))) -> b(b(b(a(x1))))
23.74/7.05	   b(b(b(b(x1)))) -> a(b(b(b(x1))))
23.74/7.05	   b(a(b(b(x1)))) -> b(a(b(a(x1))))
23.74/7.05	
23.74/7.05	Q is empty.
23.74/7.05	
23.74/7.05	----------------------------------------
23.74/7.05	
23.74/7.05	(1) DependencyPairsProof (EQUIVALENT)
23.74/7.05	Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.
23.74/7.05	----------------------------------------
23.74/7.05	
23.74/7.05	(2)
23.74/7.05	Obligation:
23.74/7.05	Q DP problem:
23.74/7.05	The TRS P consists of the following rules:
23.74/7.05	
23.74/7.05	   B(a(a(b(x1)))) -> B(b(b(a(x1))))
23.74/7.05	   B(a(a(b(x1)))) -> B(b(a(x1)))
23.74/7.05	   B(a(a(b(x1)))) -> B(a(x1))
23.74/7.05	   B(a(b(b(x1)))) -> B(a(b(a(x1))))
23.74/7.05	   B(a(b(b(x1)))) -> B(a(x1))
23.74/7.05	
23.74/7.05	The TRS R consists of the following rules:
23.74/7.05	
23.74/7.05	   b(a(a(b(x1)))) -> b(b(b(a(x1))))
23.74/7.05	   b(b(b(b(x1)))) -> a(b(b(b(x1))))
23.74/7.05	   b(a(b(b(x1)))) -> b(a(b(a(x1))))
23.74/7.05	
23.74/7.05	Q is empty.
23.74/7.05	We have to consider all minimal (P,Q,R)-chains.
23.74/7.05	----------------------------------------
23.74/7.05	
23.74/7.05	(3) MRRProof (EQUIVALENT)
23.74/7.05	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.
23.74/7.05	
23.74/7.05	Strictly oriented dependency pairs:
23.74/7.05	
23.74/7.05	   B(a(a(b(x1)))) -> B(b(a(x1)))
23.74/7.05	   B(a(a(b(x1)))) -> B(a(x1))
23.74/7.05	   B(a(b(b(x1)))) -> B(a(x1))
23.74/7.05	
23.74/7.05	
23.74/7.05	Used ordering: Polynomial interpretation [POLO]:
23.74/7.05	
23.74/7.05	   POL(B(x_1)) = 2*x_1
23.74/7.05	   POL(a(x_1)) = 2 + x_1
23.74/7.05	   POL(b(x_1)) = 2 + x_1
23.74/7.05	
23.74/7.05	
23.74/7.05	----------------------------------------
23.74/7.05	
23.74/7.05	(4)
23.74/7.05	Obligation:
23.74/7.05	Q DP problem:
23.74/7.05	The TRS P consists of the following rules:
23.74/7.05	
23.74/7.05	   B(a(a(b(x1)))) -> B(b(b(a(x1))))
23.74/7.05	   B(a(b(b(x1)))) -> B(a(b(a(x1))))
23.74/7.05	
23.74/7.05	The TRS R consists of the following rules:
23.74/7.05	
23.74/7.05	   b(a(a(b(x1)))) -> b(b(b(a(x1))))
23.74/7.05	   b(b(b(b(x1)))) -> a(b(b(b(x1))))
23.74/7.05	   b(a(b(b(x1)))) -> b(a(b(a(x1))))
23.74/7.05	
23.74/7.05	Q is empty.
23.74/7.05	We have to consider all minimal (P,Q,R)-chains.
23.74/7.05	----------------------------------------
23.74/7.05	
23.74/7.05	(5) MRRProof (EQUIVALENT)
23.74/7.05	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.
23.74/7.05	
23.74/7.05	Strictly oriented dependency pairs:
23.74/7.05	
23.74/7.05	   B(a(a(b(x1)))) -> B(b(b(a(x1))))
23.74/7.05	   B(a(b(b(x1)))) -> B(a(b(a(x1))))
23.74/7.05	
23.74/7.05	Strictly oriented rules of the TRS R:
23.74/7.05	
23.74/7.05	   b(a(a(b(x1)))) -> b(b(b(a(x1))))
23.74/7.05	   b(b(b(b(x1)))) -> a(b(b(b(x1))))
23.74/7.05	   b(a(b(b(x1)))) -> b(a(b(a(x1))))
23.74/7.05	
23.74/7.05	Used ordering: Polynomial interpretation [POLO]:
23.74/7.05	
23.74/7.05	   POL(B(x_1)) = x_1
23.74/7.05	   POL(a(x_1)) = 2*x_1
23.74/7.05	   POL(b(x_1)) = 1 + 2*x_1
23.74/7.05	
23.74/7.05	
23.74/7.05	----------------------------------------
23.74/7.05	
23.74/7.05	(6)
23.74/7.05	Obligation:
23.74/7.05	Q DP problem:
23.74/7.05	P is empty.
23.74/7.05	R is empty.
23.74/7.05	Q is empty.
23.74/7.05	We have to consider all minimal (P,Q,R)-chains.
23.74/7.05	----------------------------------------
23.74/7.05	
23.74/7.05	(7) PisEmptyProof (EQUIVALENT)
23.74/7.05	The TRS P is empty. Hence, there is no (P,Q,R) chain.
23.74/7.05	----------------------------------------
23.74/7.05	
23.74/7.05	(8)
23.74/7.05	YES
24.16/7.09	EOF