20.54/6.19	YES
21.91/6.52	proof of /export/starexec/sandbox/benchmark/theBenchmark.xml
21.91/6.52	# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty
21.91/6.52	
21.91/6.52	
21.91/6.52	Termination w.r.t. Q of the given QTRS could be proven:
21.91/6.52	
21.91/6.52	(0) QTRS
21.91/6.52	(1) QTRS Reverse [EQUIVALENT, 0 ms]
21.91/6.52	(2) QTRS
21.91/6.52	(3) DependencyPairsProof [EQUIVALENT, 5 ms]
21.91/6.52	(4) QDP
21.91/6.52	(5) DependencyGraphProof [EQUIVALENT, 0 ms]
21.91/6.52	(6) QDP
21.91/6.52	(7) MRRProof [EQUIVALENT, 54 ms]
21.91/6.52	(8) QDP
21.91/6.52	(9) QDPOrderProof [EQUIVALENT, 85 ms]
21.91/6.52	(10) QDP
21.91/6.52	(11) PisEmptyProof [EQUIVALENT, 0 ms]
21.91/6.52	(12) YES
21.91/6.52	
21.91/6.52	
21.91/6.52	----------------------------------------
21.91/6.52	
21.91/6.52	(0)
21.91/6.52	Obligation:
21.91/6.52	Q restricted rewrite system:
21.91/6.52	The TRS R consists of the following rules:
21.91/6.52	
21.91/6.52	   a(a(a(a(x1)))) -> b(a(b(b(x1))))
21.91/6.52	   b(b(a(a(x1)))) -> a(a(b(b(x1))))
21.91/6.52	   b(a(b(b(x1)))) -> a(a(b(b(x1))))
21.91/6.52	
21.91/6.52	Q is empty.
21.91/6.52	
21.91/6.52	----------------------------------------
21.91/6.52	
21.91/6.52	(1) QTRS Reverse (EQUIVALENT)
21.91/6.52	We applied the QTRS Reverse Processor [REVERSE].
21.91/6.52	----------------------------------------
21.91/6.52	
21.91/6.52	(2)
21.91/6.52	Obligation:
21.91/6.52	Q restricted rewrite system:
21.91/6.52	The TRS R consists of the following rules:
21.91/6.52	
21.91/6.52	   a(a(a(a(x1)))) -> b(b(a(b(x1))))
21.91/6.52	   a(a(b(b(x1)))) -> b(b(a(a(x1))))
21.91/6.52	   b(b(a(b(x1)))) -> b(b(a(a(x1))))
21.91/6.52	
21.91/6.52	Q is empty.
21.91/6.52	
21.91/6.52	----------------------------------------
21.91/6.52	
21.91/6.52	(3) DependencyPairsProof (EQUIVALENT)
21.91/6.52	Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.
21.91/6.52	----------------------------------------
21.91/6.52	
21.91/6.52	(4)
21.91/6.52	Obligation:
21.91/6.52	Q DP problem:
21.91/6.52	The TRS P consists of the following rules:
21.91/6.52	
21.91/6.52	   A(a(a(a(x1)))) -> B(b(a(b(x1))))
21.91/6.52	   A(a(a(a(x1)))) -> B(a(b(x1)))
21.91/6.52	   A(a(a(a(x1)))) -> A(b(x1))
21.91/6.52	   A(a(a(a(x1)))) -> B(x1)
21.91/6.52	   A(a(b(b(x1)))) -> B(b(a(a(x1))))
21.91/6.52	   A(a(b(b(x1)))) -> B(a(a(x1)))
21.91/6.52	   A(a(b(b(x1)))) -> A(a(x1))
21.91/6.52	   A(a(b(b(x1)))) -> A(x1)
21.91/6.52	   B(b(a(b(x1)))) -> B(b(a(a(x1))))
21.91/6.52	   B(b(a(b(x1)))) -> B(a(a(x1)))
21.91/6.52	   B(b(a(b(x1)))) -> A(a(x1))
21.91/6.52	   B(b(a(b(x1)))) -> A(x1)
21.91/6.52	
21.91/6.52	The TRS R consists of the following rules:
21.91/6.52	
21.91/6.52	   a(a(a(a(x1)))) -> b(b(a(b(x1))))
21.91/6.52	   a(a(b(b(x1)))) -> b(b(a(a(x1))))
21.91/6.52	   b(b(a(b(x1)))) -> b(b(a(a(x1))))
21.91/6.52	
21.91/6.52	Q is empty.
21.91/6.52	We have to consider all minimal (P,Q,R)-chains.
21.91/6.52	----------------------------------------
21.91/6.52	
21.91/6.52	(5) DependencyGraphProof (EQUIVALENT)
21.91/6.52	The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 4 less nodes.
21.91/6.52	----------------------------------------
21.91/6.52	
21.91/6.52	(6)
21.91/6.52	Obligation:
21.91/6.52	Q DP problem:
21.91/6.52	The TRS P consists of the following rules:
21.91/6.52	
21.91/6.52	   B(b(a(b(x1)))) -> B(b(a(a(x1))))
21.91/6.52	   B(b(a(b(x1)))) -> A(a(x1))
21.91/6.52	   A(a(a(a(x1)))) -> B(b(a(b(x1))))
21.91/6.52	   B(b(a(b(x1)))) -> A(x1)
21.91/6.52	   A(a(a(a(x1)))) -> B(x1)
21.91/6.52	   A(a(b(b(x1)))) -> B(b(a(a(x1))))
21.91/6.52	   A(a(b(b(x1)))) -> A(a(x1))
21.91/6.52	   A(a(b(b(x1)))) -> A(x1)
21.91/6.52	
21.91/6.52	The TRS R consists of the following rules:
21.91/6.52	
21.91/6.52	   a(a(a(a(x1)))) -> b(b(a(b(x1))))
21.91/6.52	   a(a(b(b(x1)))) -> b(b(a(a(x1))))
21.91/6.52	   b(b(a(b(x1)))) -> b(b(a(a(x1))))
21.91/6.52	
21.91/6.52	Q is empty.
21.91/6.52	We have to consider all minimal (P,Q,R)-chains.
21.91/6.52	----------------------------------------
21.91/6.52	
21.91/6.52	(7) MRRProof (EQUIVALENT)
21.91/6.52	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.
21.91/6.52	
21.91/6.52	Strictly oriented dependency pairs:
21.91/6.52	
21.91/6.52	   B(b(a(b(x1)))) -> A(a(x1))
21.91/6.52	   A(a(a(a(x1)))) -> B(b(a(b(x1))))
21.91/6.52	   B(b(a(b(x1)))) -> A(x1)
21.91/6.52	   A(a(a(a(x1)))) -> B(x1)
21.91/6.52	   A(a(b(b(x1)))) -> B(b(a(a(x1))))
21.91/6.52	   A(a(b(b(x1)))) -> A(a(x1))
21.91/6.52	   A(a(b(b(x1)))) -> A(x1)
21.91/6.52	
21.91/6.52	
21.91/6.52	Used ordering: Polynomial interpretation [POLO]:
21.91/6.52	
21.91/6.52	   POL(A(x_1)) = 2 + x_1
21.91/6.52	   POL(B(x_1)) = x_1
21.91/6.52	   POL(a(x_1)) = 2 + x_1
21.91/6.52	   POL(b(x_1)) = 2 + x_1
21.91/6.52	
21.91/6.52	
21.91/6.52	----------------------------------------
21.91/6.52	
21.91/6.52	(8)
21.91/6.52	Obligation:
21.91/6.52	Q DP problem:
21.91/6.52	The TRS P consists of the following rules:
21.91/6.52	
21.91/6.52	   B(b(a(b(x1)))) -> B(b(a(a(x1))))
21.91/6.52	
21.91/6.52	The TRS R consists of the following rules:
21.91/6.52	
21.91/6.52	   a(a(a(a(x1)))) -> b(b(a(b(x1))))
21.91/6.52	   a(a(b(b(x1)))) -> b(b(a(a(x1))))
21.91/6.52	   b(b(a(b(x1)))) -> b(b(a(a(x1))))
21.91/6.52	
21.91/6.52	Q is empty.
21.91/6.52	We have to consider all minimal (P,Q,R)-chains.
21.91/6.52	----------------------------------------
21.91/6.52	
21.91/6.52	(9) QDPOrderProof (EQUIVALENT)
21.91/6.52	We use the reduction pair processor [LPAR04,JAR06].
21.91/6.52	
21.91/6.52	
21.91/6.52	The following pairs can be oriented strictly and are deleted.
21.91/6.52	
21.91/6.52	   B(b(a(b(x1)))) -> B(b(a(a(x1))))
21.91/6.52	The remaining pairs can at least be oriented weakly.
21.91/6.52	Used ordering:  Matrix interpretation [MATRO] with arctic natural numbers [ARCTIC]:
21.91/6.52	
21.91/6.52	   <<<
21.91/6.52	 POL(B(x_1)) =  	[[-I]] 	 +  	[[0A, -I, 0A]] 	* 	x_1
21.91/6.52	>>>
21.91/6.52	
21.91/6.52	   <<<
21.91/6.52	 POL(b(x_1)) =  	[[0A], [0A], [0A]] 	 +  	[[-I, -I, -I], [1A, -I, -I], [0A, -I, -I]] 	* 	x_1
21.91/6.52	>>>
21.91/6.52	
21.91/6.52	   <<<
21.91/6.52	 POL(a(x_1)) =  	[[0A], [-I], [-I]] 	 +  	[[-I, 1A, 0A], [0A, -I, -I], [0A, -I, -I]] 	* 	x_1
21.91/6.52	>>>
21.91/6.52	
21.91/6.52	
21.91/6.52	The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:
21.91/6.52	
21.91/6.52	   a(a(a(a(x1)))) -> b(b(a(b(x1))))
21.91/6.52	   a(a(b(b(x1)))) -> b(b(a(a(x1))))
21.91/6.52	   b(b(a(b(x1)))) -> b(b(a(a(x1))))
21.91/6.52	
21.91/6.52	
21.91/6.52	----------------------------------------
21.91/6.52	
21.91/6.52	(10)
21.91/6.52	Obligation:
21.91/6.52	Q DP problem:
21.91/6.52	P is empty.
21.91/6.52	The TRS R consists of the following rules:
21.91/6.52	
21.91/6.52	   a(a(a(a(x1)))) -> b(b(a(b(x1))))
21.91/6.52	   a(a(b(b(x1)))) -> b(b(a(a(x1))))
21.91/6.52	   b(b(a(b(x1)))) -> b(b(a(a(x1))))
21.91/6.52	
21.91/6.52	Q is empty.
21.91/6.52	We have to consider all minimal (P,Q,R)-chains.
21.91/6.52	----------------------------------------
21.91/6.52	
21.91/6.52	(11) PisEmptyProof (EQUIVALENT)
21.91/6.52	The TRS P is empty. Hence, there is no (P,Q,R) chain.
21.91/6.52	----------------------------------------
21.91/6.52	
21.91/6.52	(12)
21.91/6.52	YES
22.05/10.34	EOF