19.03/5.79	YES
21.32/6.75	proof of /export/starexec/sandbox2/benchmark/theBenchmark.xml
21.32/6.75	# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty
21.32/6.75	
21.32/6.75	
21.32/6.75	Termination w.r.t. Q of the given QTRS could be proven:
21.32/6.75	
21.32/6.75	(0) QTRS
21.32/6.75	(1) QTRS Reverse [EQUIVALENT, 0 ms]
21.32/6.75	(2) QTRS
21.32/6.75	(3) DependencyPairsProof [EQUIVALENT, 25 ms]
21.32/6.75	(4) QDP
21.32/6.75	(5) DependencyGraphProof [EQUIVALENT, 5 ms]
21.32/6.75	(6) QDP
21.32/6.75	(7) QDPOrderProof [EQUIVALENT, 62 ms]
21.32/6.75	(8) QDP
21.32/6.75	(9) PisEmptyProof [EQUIVALENT, 0 ms]
21.32/6.75	(10) YES
21.32/6.75	
21.32/6.75	
21.32/6.75	----------------------------------------
21.32/6.75	
21.32/6.75	(0)
21.32/6.75	Obligation:
21.32/6.75	Q restricted rewrite system:
21.32/6.75	The TRS R consists of the following rules:
21.32/6.75	
21.32/6.75	   b(a(b(a(x1)))) -> b(b(a(a(x1))))
21.32/6.75	   a(a(a(b(x1)))) -> a(b(a(a(x1))))
21.32/6.75	   a(a(b(a(x1)))) -> b(b(a(a(x1))))
21.32/6.75	   a(b(b(b(x1)))) -> b(a(b(a(x1))))
21.32/6.75	
21.32/6.75	Q is empty.
21.32/6.75	
21.32/6.75	----------------------------------------
21.32/6.75	
21.32/6.75	(1) QTRS Reverse (EQUIVALENT)
21.32/6.75	We applied the QTRS Reverse Processor [REVERSE].
21.32/6.75	----------------------------------------
21.32/6.75	
21.32/6.75	(2)
21.32/6.75	Obligation:
21.32/6.75	Q restricted rewrite system:
21.32/6.75	The TRS R consists of the following rules:
21.32/6.75	
21.32/6.75	   a(b(a(b(x1)))) -> a(a(b(b(x1))))
21.32/6.75	   b(a(a(a(x1)))) -> a(a(b(a(x1))))
21.32/6.75	   a(b(a(a(x1)))) -> a(a(b(b(x1))))
21.32/6.75	   b(b(b(a(x1)))) -> a(b(a(b(x1))))
21.32/6.75	
21.32/6.75	Q is empty.
21.32/6.75	
21.32/6.75	----------------------------------------
21.32/6.75	
21.32/6.75	(3) DependencyPairsProof (EQUIVALENT)
21.32/6.75	Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.
21.32/6.75	----------------------------------------
21.32/6.75	
21.32/6.75	(4)
21.32/6.75	Obligation:
21.32/6.75	Q DP problem:
21.32/6.75	The TRS P consists of the following rules:
21.32/6.75	
21.32/6.75	   A(b(a(b(x1)))) -> A(a(b(b(x1))))
21.32/6.75	   A(b(a(b(x1)))) -> A(b(b(x1)))
21.32/6.75	   A(b(a(b(x1)))) -> B(b(x1))
21.32/6.75	   B(a(a(a(x1)))) -> A(a(b(a(x1))))
21.32/6.75	   B(a(a(a(x1)))) -> A(b(a(x1)))
21.32/6.75	   B(a(a(a(x1)))) -> B(a(x1))
21.32/6.75	   A(b(a(a(x1)))) -> A(a(b(b(x1))))
21.32/6.75	   A(b(a(a(x1)))) -> A(b(b(x1)))
21.32/6.75	   A(b(a(a(x1)))) -> B(b(x1))
21.32/6.75	   A(b(a(a(x1)))) -> B(x1)
21.32/6.75	   B(b(b(a(x1)))) -> A(b(a(b(x1))))
21.32/6.75	   B(b(b(a(x1)))) -> B(a(b(x1)))
21.32/6.75	   B(b(b(a(x1)))) -> A(b(x1))
21.32/6.75	   B(b(b(a(x1)))) -> B(x1)
21.32/6.75	
21.32/6.75	The TRS R consists of the following rules:
21.32/6.75	
21.32/6.75	   a(b(a(b(x1)))) -> a(a(b(b(x1))))
21.32/6.75	   b(a(a(a(x1)))) -> a(a(b(a(x1))))
21.32/6.75	   a(b(a(a(x1)))) -> a(a(b(b(x1))))
21.32/6.75	   b(b(b(a(x1)))) -> a(b(a(b(x1))))
21.32/6.75	
21.32/6.75	Q is empty.
21.32/6.75	We have to consider all minimal (P,Q,R)-chains.
21.32/6.75	----------------------------------------
21.32/6.75	
21.32/6.75	(5) DependencyGraphProof (EQUIVALENT)
21.32/6.75	The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 3 less nodes.
21.32/6.75	----------------------------------------
21.32/6.75	
21.32/6.75	(6)
21.32/6.75	Obligation:
21.32/6.75	Q DP problem:
21.32/6.75	The TRS P consists of the following rules:
21.32/6.75	
21.32/6.75	   A(b(a(b(x1)))) -> B(b(x1))
21.32/6.75	   B(a(a(a(x1)))) -> A(b(a(x1)))
21.32/6.75	   A(b(a(b(x1)))) -> A(b(b(x1)))
21.32/6.75	   A(b(a(a(x1)))) -> A(b(b(x1)))
21.32/6.75	   A(b(a(a(x1)))) -> B(b(x1))
21.32/6.75	   B(a(a(a(x1)))) -> B(a(x1))
21.32/6.75	   B(b(b(a(x1)))) -> A(b(a(b(x1))))
21.32/6.75	   A(b(a(a(x1)))) -> B(x1)
21.32/6.75	   B(b(b(a(x1)))) -> B(a(b(x1)))
21.32/6.75	   B(b(b(a(x1)))) -> A(b(x1))
21.32/6.75	   B(b(b(a(x1)))) -> B(x1)
21.32/6.75	
21.32/6.75	The TRS R consists of the following rules:
21.32/6.75	
21.32/6.75	   a(b(a(b(x1)))) -> a(a(b(b(x1))))
21.32/6.75	   b(a(a(a(x1)))) -> a(a(b(a(x1))))
21.32/6.75	   a(b(a(a(x1)))) -> a(a(b(b(x1))))
21.32/6.75	   b(b(b(a(x1)))) -> a(b(a(b(x1))))
21.32/6.75	
21.32/6.75	Q is empty.
21.32/6.75	We have to consider all minimal (P,Q,R)-chains.
21.32/6.75	----------------------------------------
21.32/6.75	
21.32/6.75	(7) QDPOrderProof (EQUIVALENT)
21.32/6.75	We use the reduction pair processor [LPAR04,JAR06].
21.32/6.75	
21.32/6.75	
21.32/6.75	The following pairs can be oriented strictly and are deleted.
21.32/6.75	
21.32/6.75	   A(b(a(b(x1)))) -> B(b(x1))
21.32/6.75	   B(a(a(a(x1)))) -> A(b(a(x1)))
21.32/6.75	   A(b(a(b(x1)))) -> A(b(b(x1)))
21.32/6.75	   A(b(a(a(x1)))) -> A(b(b(x1)))
21.32/6.75	   A(b(a(a(x1)))) -> B(b(x1))
21.32/6.75	   B(a(a(a(x1)))) -> B(a(x1))
21.32/6.75	   B(b(b(a(x1)))) -> A(b(a(b(x1))))
21.32/6.75	   A(b(a(a(x1)))) -> B(x1)
21.32/6.75	   B(b(b(a(x1)))) -> B(a(b(x1)))
21.32/6.75	   B(b(b(a(x1)))) -> A(b(x1))
21.32/6.75	   B(b(b(a(x1)))) -> B(x1)
21.32/6.75	The remaining pairs can at least be oriented weakly.
21.32/6.75	Used ordering:  Polynomial interpretation [POLO]:
21.32/6.75	
21.32/6.75	   POL(A(x_1)) = x_1
21.32/6.75	   POL(B(x_1)) = 1 + x_1
21.32/6.75	   POL(a(x_1)) = 1 + x_1
21.32/6.75	   POL(b(x_1)) = 1 + x_1
21.32/6.75	
21.32/6.75	The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:
21.32/6.75	
21.32/6.75	   b(a(a(a(x1)))) -> a(a(b(a(x1))))
21.32/6.75	   b(b(b(a(x1)))) -> a(b(a(b(x1))))
21.32/6.75	   a(b(a(a(x1)))) -> a(a(b(b(x1))))
21.32/6.75	   a(b(a(b(x1)))) -> a(a(b(b(x1))))
21.32/6.75	
21.32/6.75	
21.32/6.75	----------------------------------------
21.32/6.75	
21.32/6.75	(8)
21.32/6.75	Obligation:
21.32/6.75	Q DP problem:
21.32/6.75	P is empty.
21.32/6.75	The TRS R consists of the following rules:
21.32/6.75	
21.32/6.75	   a(b(a(b(x1)))) -> a(a(b(b(x1))))
21.32/6.75	   b(a(a(a(x1)))) -> a(a(b(a(x1))))
21.32/6.75	   a(b(a(a(x1)))) -> a(a(b(b(x1))))
21.32/6.75	   b(b(b(a(x1)))) -> a(b(a(b(x1))))
21.32/6.75	
21.32/6.75	Q is empty.
21.32/6.75	We have to consider all minimal (P,Q,R)-chains.
21.32/6.75	----------------------------------------
21.32/6.75	
21.32/6.75	(9) PisEmptyProof (EQUIVALENT)
21.32/6.75	The TRS P is empty. Hence, there is no (P,Q,R) chain.
21.32/6.75	----------------------------------------
21.32/6.75	
21.32/6.75	(10)
21.32/6.75	YES
21.99/7.78	EOF