237.48/61.08	YES
237.75/61.13	proof of /export/starexec/sandbox/benchmark/theBenchmark.xml
237.75/61.13	# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty
237.75/61.13	
237.75/61.13	
237.75/61.13	Termination w.r.t. Q of the given QTRS could be proven:
237.75/61.13	
237.75/61.13	(0) QTRS
237.75/61.13	(1) DependencyPairsProof [EQUIVALENT, 0 ms]
237.75/61.13	(2) QDP
237.75/61.13	(3) QDPOrderProof [EQUIVALENT, 38 ms]
237.75/61.13	(4) QDP
237.75/61.13	(5) DependencyGraphProof [EQUIVALENT, 0 ms]
237.75/61.13	(6) QDP
237.75/61.13	(7) QDPOrderProof [EQUIVALENT, 480 ms]
237.75/61.13	(8) QDP
237.75/61.13	(9) PisEmptyProof [EQUIVALENT, 0 ms]
237.75/61.13	(10) YES
237.75/61.13	
237.75/61.13	
237.75/61.13	----------------------------------------
237.75/61.13	
237.75/61.13	(0)
237.75/61.13	Obligation:
237.75/61.13	Q restricted rewrite system:
237.75/61.13	The TRS R consists of the following rules:
237.75/61.13	
237.75/61.13	   0(0(0(0(x1)))) -> 0(1(1(1(x1))))
237.75/61.13	   1(0(0(1(x1)))) -> 0(0(0(0(x1))))
237.75/61.13	
237.75/61.13	Q is empty.
237.75/61.13	
237.75/61.13	----------------------------------------
237.75/61.13	
237.75/61.13	(1) DependencyPairsProof (EQUIVALENT)
237.75/61.13	Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.
237.75/61.13	----------------------------------------
237.75/61.13	
237.75/61.13	(2)
237.75/61.13	Obligation:
237.75/61.13	Q DP problem:
237.75/61.13	The TRS P consists of the following rules:
237.75/61.13	
237.75/61.13	   0^1(0(0(0(x1)))) -> 0^1(1(1(1(x1))))
237.75/61.13	   0^1(0(0(0(x1)))) -> 1^1(1(1(x1)))
237.75/61.13	   0^1(0(0(0(x1)))) -> 1^1(1(x1))
237.75/61.13	   0^1(0(0(0(x1)))) -> 1^1(x1)
237.75/61.13	   1^1(0(0(1(x1)))) -> 0^1(0(0(0(x1))))
237.75/61.13	   1^1(0(0(1(x1)))) -> 0^1(0(0(x1)))
237.75/61.13	   1^1(0(0(1(x1)))) -> 0^1(0(x1))
237.75/61.13	   1^1(0(0(1(x1)))) -> 0^1(x1)
237.75/61.13	
237.75/61.13	The TRS R consists of the following rules:
237.75/61.13	
237.75/61.13	   0(0(0(0(x1)))) -> 0(1(1(1(x1))))
237.75/61.13	   1(0(0(1(x1)))) -> 0(0(0(0(x1))))
237.75/61.13	
237.75/61.13	Q is empty.
237.75/61.13	We have to consider all minimal (P,Q,R)-chains.
237.75/61.13	----------------------------------------
237.75/61.13	
237.75/61.13	(3) QDPOrderProof (EQUIVALENT)
237.75/61.13	We use the reduction pair processor [LPAR04,JAR06].
237.75/61.13	
237.75/61.13	
237.75/61.13	The following pairs can be oriented strictly and are deleted.
237.75/61.13	
237.75/61.13	   0^1(0(0(0(x1)))) -> 1^1(1(1(x1)))
237.75/61.13	   0^1(0(0(0(x1)))) -> 1^1(1(x1))
237.75/61.13	   0^1(0(0(0(x1)))) -> 1^1(x1)
237.75/61.13	   1^1(0(0(1(x1)))) -> 0^1(0(0(x1)))
237.75/61.13	   1^1(0(0(1(x1)))) -> 0^1(0(x1))
237.75/61.13	   1^1(0(0(1(x1)))) -> 0^1(x1)
237.75/61.13	The remaining pairs can at least be oriented weakly.
237.75/61.13	Used ordering:  Polynomial interpretation [POLO]:
237.75/61.13	
237.75/61.13	   POL(0(x_1)) = 1 + x_1
237.75/61.13	   POL(0^1(x_1)) = 1 + x_1
237.75/61.13	   POL(1(x_1)) = 1 + x_1
237.75/61.13	   POL(1^1(x_1)) = 1 + x_1
237.75/61.13	
237.75/61.13	The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:
237.75/61.13	
237.75/61.13	   1(0(0(1(x1)))) -> 0(0(0(0(x1))))
237.75/61.13	   0(0(0(0(x1)))) -> 0(1(1(1(x1))))
237.75/61.13	
237.75/61.13	
237.75/61.13	----------------------------------------
237.75/61.13	
237.75/61.13	(4)
237.75/61.13	Obligation:
237.75/61.13	Q DP problem:
237.75/61.13	The TRS P consists of the following rules:
237.75/61.13	
237.75/61.13	   0^1(0(0(0(x1)))) -> 0^1(1(1(1(x1))))
237.75/61.13	   1^1(0(0(1(x1)))) -> 0^1(0(0(0(x1))))
237.75/61.13	
237.75/61.13	The TRS R consists of the following rules:
237.75/61.13	
237.75/61.13	   0(0(0(0(x1)))) -> 0(1(1(1(x1))))
237.75/61.13	   1(0(0(1(x1)))) -> 0(0(0(0(x1))))
237.75/61.13	
237.75/61.13	Q is empty.
237.75/61.13	We have to consider all minimal (P,Q,R)-chains.
237.75/61.13	----------------------------------------
237.75/61.13	
237.75/61.13	(5) DependencyGraphProof (EQUIVALENT)
237.75/61.13	The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 1 less node.
237.75/61.13	----------------------------------------
237.75/61.13	
237.75/61.13	(6)
237.75/61.13	Obligation:
237.75/61.13	Q DP problem:
237.75/61.13	The TRS P consists of the following rules:
237.75/61.13	
237.75/61.13	   0^1(0(0(0(x1)))) -> 0^1(1(1(1(x1))))
237.75/61.13	
237.75/61.13	The TRS R consists of the following rules:
237.75/61.13	
237.75/61.13	   0(0(0(0(x1)))) -> 0(1(1(1(x1))))
237.75/61.13	   1(0(0(1(x1)))) -> 0(0(0(0(x1))))
237.75/61.13	
237.75/61.13	Q is empty.
237.75/61.13	We have to consider all minimal (P,Q,R)-chains.
237.75/61.14	----------------------------------------
237.75/61.14	
237.75/61.14	(7) QDPOrderProof (EQUIVALENT)
237.75/61.14	We use the reduction pair processor [LPAR04,JAR06].
237.75/61.14	
237.75/61.14	
237.75/61.14	The following pairs can be oriented strictly and are deleted.
237.75/61.14	
237.75/61.14	   0^1(0(0(0(x1)))) -> 0^1(1(1(1(x1))))
237.75/61.14	The remaining pairs can at least be oriented weakly.
237.75/61.14	Used ordering:  Matrix interpretation [MATRO] with arctic natural numbers [ARCTIC]:
237.75/61.14	
237.75/61.14	   <<<
237.75/61.14	 POL(0^1(x_1)) =  	[[-I]] 	 +  	[[0A, -I, -I, 0A, 0A]] 	* 	x_1
237.75/61.14	>>>
237.75/61.14	
237.75/61.14	   <<<
237.75/61.14	 POL(0(x_1)) =  	[[0A], [0A], [0A], [-I], [0A]] 	 +  	[[-I, 0A, 0A, 0A, 0A], [-I, 0A, -I, 0A, 0A], [-I, 1A, -I, 0A, 0A], [-I, 0A, -I, 0A, 0A], [0A, 0A, -I, 0A, 0A]] 	* 	x_1
237.75/61.14	>>>
237.75/61.14	
237.75/61.14	   <<<
237.75/61.14	 POL(1(x_1)) =  	[[0A], [0A], [1A], [0A], [0A]] 	 +  	[[-I, 0A, -I, 0A, 0A], [-I, -I, -I, 0A, 0A], [1A, 1A, 0A, 1A, 1A], [-I, 0A, -I, 0A, 0A], [-I, 0A, -I, 0A, 0A]] 	* 	x_1
237.75/61.14	>>>
237.75/61.14	
237.75/61.14	
237.75/61.14	The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:
237.75/61.14	
237.75/61.14	   1(0(0(1(x1)))) -> 0(0(0(0(x1))))
237.75/61.14	   0(0(0(0(x1)))) -> 0(1(1(1(x1))))
237.75/61.14	
237.75/61.14	
237.75/61.14	----------------------------------------
237.75/61.14	
237.75/61.14	(8)
237.75/61.14	Obligation:
237.75/61.14	Q DP problem:
237.75/61.14	P is empty.
237.75/61.14	The TRS R consists of the following rules:
237.75/61.14	
237.75/61.14	   0(0(0(0(x1)))) -> 0(1(1(1(x1))))
237.75/61.14	   1(0(0(1(x1)))) -> 0(0(0(0(x1))))
237.75/61.14	
237.75/61.14	Q is empty.
237.75/61.14	We have to consider all minimal (P,Q,R)-chains.
237.75/61.14	----------------------------------------
237.75/61.14	
237.75/61.14	(9) PisEmptyProof (EQUIVALENT)
237.75/61.14	The TRS P is empty. Hence, there is no (P,Q,R) chain.
237.75/61.14	----------------------------------------
237.75/61.14	
237.75/61.14	(10)
237.75/61.14	YES
237.75/61.19	EOF