21.86/6.45	YES
22.17/6.50	proof of /export/starexec/sandbox/benchmark/theBenchmark.xml
22.17/6.50	# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty
22.17/6.50	
22.17/6.50	
22.17/6.50	Termination w.r.t. Q of the given QTRS could be proven:
22.17/6.50	
22.17/6.50	(0) QTRS
22.17/6.50	(1) DependencyPairsProof [EQUIVALENT, 24 ms]
22.17/6.50	(2) QDP
22.17/6.50	(3) MRRProof [EQUIVALENT, 62 ms]
22.17/6.50	(4) QDP
22.17/6.50	(5) QDPOrderProof [EQUIVALENT, 74 ms]
22.17/6.50	(6) QDP
22.17/6.50	(7) PisEmptyProof [EQUIVALENT, 0 ms]
22.17/6.50	(8) YES
22.17/6.50	
22.17/6.50	
22.17/6.50	----------------------------------------
22.17/6.50	
22.17/6.50	(0)
22.17/6.50	Obligation:
22.17/6.50	Q restricted rewrite system:
22.17/6.50	The TRS R consists of the following rules:
22.17/6.50	
22.17/6.50	   0(0(0(0(x1)))) -> 1(0(0(1(x1))))
22.17/6.50	   0(1(0(1(x1)))) -> 0(0(1(0(x1))))
22.17/6.50	
22.17/6.50	Q is empty.
22.17/6.50	
22.17/6.50	----------------------------------------
22.17/6.50	
22.17/6.50	(1) DependencyPairsProof (EQUIVALENT)
22.17/6.50	Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.
22.17/6.50	----------------------------------------
22.17/6.50	
22.17/6.50	(2)
22.17/6.50	Obligation:
22.17/6.50	Q DP problem:
22.17/6.50	The TRS P consists of the following rules:
22.17/6.50	
22.17/6.50	   0^1(0(0(0(x1)))) -> 0^1(0(1(x1)))
22.17/6.50	   0^1(0(0(0(x1)))) -> 0^1(1(x1))
22.17/6.50	   0^1(1(0(1(x1)))) -> 0^1(0(1(0(x1))))
22.17/6.50	   0^1(1(0(1(x1)))) -> 0^1(1(0(x1)))
22.17/6.50	   0^1(1(0(1(x1)))) -> 0^1(x1)
22.17/6.50	
22.17/6.50	The TRS R consists of the following rules:
22.17/6.50	
22.17/6.50	   0(0(0(0(x1)))) -> 1(0(0(1(x1))))
22.17/6.50	   0(1(0(1(x1)))) -> 0(0(1(0(x1))))
22.17/6.50	
22.17/6.50	Q is empty.
22.17/6.50	We have to consider all minimal (P,Q,R)-chains.
22.17/6.50	----------------------------------------
22.17/6.50	
22.17/6.50	(3) MRRProof (EQUIVALENT)
22.17/6.50	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.
22.17/6.50	
22.17/6.50	Strictly oriented dependency pairs:
22.17/6.50	
22.17/6.50	   0^1(0(0(0(x1)))) -> 0^1(0(1(x1)))
22.17/6.50	   0^1(0(0(0(x1)))) -> 0^1(1(x1))
22.17/6.50	   0^1(1(0(1(x1)))) -> 0^1(1(0(x1)))
22.17/6.50	   0^1(1(0(1(x1)))) -> 0^1(x1)
22.17/6.50	
22.17/6.50	
22.17/6.50	Used ordering: Polynomial interpretation [POLO]:
22.17/6.50	
22.17/6.50	   POL(0(x_1)) = 1 + x_1
22.17/6.50	   POL(0^1(x_1)) = x_1
22.17/6.50	   POL(1(x_1)) = 1 + x_1
22.17/6.50	
22.17/6.50	
22.17/6.50	----------------------------------------
22.17/6.50	
22.17/6.50	(4)
22.17/6.50	Obligation:
22.17/6.50	Q DP problem:
22.17/6.50	The TRS P consists of the following rules:
22.17/6.50	
22.17/6.50	   0^1(1(0(1(x1)))) -> 0^1(0(1(0(x1))))
22.17/6.50	
22.17/6.50	The TRS R consists of the following rules:
22.17/6.50	
22.17/6.50	   0(0(0(0(x1)))) -> 1(0(0(1(x1))))
22.17/6.50	   0(1(0(1(x1)))) -> 0(0(1(0(x1))))
22.17/6.50	
22.17/6.50	Q is empty.
22.17/6.50	We have to consider all minimal (P,Q,R)-chains.
22.17/6.50	----------------------------------------
22.17/6.50	
22.17/6.50	(5) QDPOrderProof (EQUIVALENT)
22.17/6.50	We use the reduction pair processor [LPAR04,JAR06].
22.17/6.50	
22.17/6.50	
22.17/6.50	The following pairs can be oriented strictly and are deleted.
22.17/6.50	
22.17/6.50	   0^1(1(0(1(x1)))) -> 0^1(0(1(0(x1))))
22.17/6.50	The remaining pairs can at least be oriented weakly.
22.17/6.50	Used ordering:  Matrix interpretation [MATRO] with arctic natural numbers [ARCTIC]:
22.17/6.50	
22.17/6.50	   <<<
22.17/6.50	 POL(0^1(x_1)) =  	[[-I]] 	 +  	[[0A, -I, -I]] 	* 	x_1
22.17/6.50	>>>
22.17/6.50	
22.17/6.50	   <<<
22.17/6.50	 POL(1(x_1)) =  	[[0A], [0A], [0A]] 	 +  	[[1A, 0A, 1A], [0A, 0A, 1A], [-I, -I, -I]] 	* 	x_1
22.17/6.50	>>>
22.17/6.50	
22.17/6.50	   <<<
22.17/6.50	 POL(0(x_1)) =  	[[0A], [-I], [-I]] 	 +  	[[-I, 0A, 0A], [-I, 1A, 1A], [1A, 0A, 0A]] 	* 	x_1
22.17/6.50	>>>
22.17/6.50	
22.17/6.50	
22.17/6.50	The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:
22.17/6.50	
22.17/6.50	   0(0(0(0(x1)))) -> 1(0(0(1(x1))))
22.17/6.50	   0(1(0(1(x1)))) -> 0(0(1(0(x1))))
22.17/6.50	
22.17/6.50	
22.17/6.50	----------------------------------------
22.17/6.50	
22.17/6.50	(6)
22.17/6.50	Obligation:
22.17/6.50	Q DP problem:
22.17/6.50	P is empty.
22.17/6.50	The TRS R consists of the following rules:
22.17/6.50	
22.17/6.50	   0(0(0(0(x1)))) -> 1(0(0(1(x1))))
22.17/6.50	   0(1(0(1(x1)))) -> 0(0(1(0(x1))))
22.17/6.50	
22.17/6.50	Q is empty.
22.17/6.50	We have to consider all minimal (P,Q,R)-chains.
22.17/6.50	----------------------------------------
22.17/6.50	
22.17/6.50	(7) PisEmptyProof (EQUIVALENT)
22.17/6.50	The TRS P is empty. Hence, there is no (P,Q,R) chain.
22.17/6.50	----------------------------------------
22.17/6.50	
22.17/6.50	(8)
22.17/6.50	YES
22.35/6.58	EOF