20.53/6.11	YES
20.66/6.20	proof of /export/starexec/sandbox/benchmark/theBenchmark.xml
20.66/6.20	# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty
20.66/6.20	
20.66/6.20	
20.66/6.20	Termination w.r.t. Q of the given QTRS could be proven:
20.66/6.20	
20.66/6.20	(0) QTRS
20.66/6.20	(1) DependencyPairsProof [EQUIVALENT, 19 ms]
20.66/6.20	(2) QDP
20.66/6.20	(3) DependencyGraphProof [EQUIVALENT, 1 ms]
20.66/6.20	(4) QDP
20.66/6.20	(5) QDPOrderProof [EQUIVALENT, 90 ms]
20.66/6.20	(6) QDP
20.66/6.20	(7) PisEmptyProof [EQUIVALENT, 0 ms]
20.66/6.20	(8) YES
20.66/6.20	
20.66/6.20	
20.66/6.20	----------------------------------------
20.66/6.20	
20.66/6.20	(0)
20.66/6.20	Obligation:
20.66/6.20	Q restricted rewrite system:
20.66/6.20	The TRS R consists of the following rules:
20.66/6.20	
20.66/6.20	   a(x1) -> x1
20.66/6.20	   a(b(x1)) -> c(b(b(a(a(x1)))))
20.66/6.20	   b(x1) -> c(x1)
20.66/6.20	   c(c(x1)) -> x1
20.66/6.20	
20.66/6.20	Q is empty.
20.66/6.20	
20.66/6.20	----------------------------------------
20.66/6.20	
20.66/6.20	(1) DependencyPairsProof (EQUIVALENT)
20.66/6.20	Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.
20.66/6.20	----------------------------------------
20.66/6.20	
20.66/6.20	(2)
20.66/6.20	Obligation:
20.66/6.20	Q DP problem:
20.66/6.20	The TRS P consists of the following rules:
20.66/6.20	
20.66/6.20	   A(b(x1)) -> C(b(b(a(a(x1)))))
20.66/6.20	   A(b(x1)) -> B(b(a(a(x1))))
20.66/6.20	   A(b(x1)) -> B(a(a(x1)))
20.66/6.20	   A(b(x1)) -> A(a(x1))
20.66/6.20	   A(b(x1)) -> A(x1)
20.66/6.20	   B(x1) -> C(x1)
20.66/6.20	
20.66/6.20	The TRS R consists of the following rules:
20.66/6.20	
20.66/6.20	   a(x1) -> x1
20.66/6.20	   a(b(x1)) -> c(b(b(a(a(x1)))))
20.66/6.20	   b(x1) -> c(x1)
20.66/6.20	   c(c(x1)) -> x1
20.66/6.20	
20.66/6.20	Q is empty.
20.66/6.20	We have to consider all minimal (P,Q,R)-chains.
20.66/6.20	----------------------------------------
20.66/6.20	
20.66/6.20	(3) DependencyGraphProof (EQUIVALENT)
20.66/6.20	The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 4 less nodes.
20.66/6.20	----------------------------------------
20.66/6.20	
20.66/6.20	(4)
20.66/6.20	Obligation:
20.66/6.20	Q DP problem:
20.66/6.20	The TRS P consists of the following rules:
20.66/6.20	
20.66/6.20	   A(b(x1)) -> A(x1)
20.66/6.20	   A(b(x1)) -> A(a(x1))
20.66/6.20	
20.66/6.20	The TRS R consists of the following rules:
20.66/6.20	
20.66/6.20	   a(x1) -> x1
20.66/6.20	   a(b(x1)) -> c(b(b(a(a(x1)))))
20.66/6.20	   b(x1) -> c(x1)
20.66/6.20	   c(c(x1)) -> x1
20.66/6.20	
20.66/6.20	Q is empty.
20.66/6.20	We have to consider all minimal (P,Q,R)-chains.
20.66/6.20	----------------------------------------
20.66/6.20	
20.66/6.20	(5) QDPOrderProof (EQUIVALENT)
20.66/6.20	We use the reduction pair processor [LPAR04,JAR06].
20.66/6.20	
20.66/6.20	
20.66/6.20	The following pairs can be oriented strictly and are deleted.
20.66/6.20	
20.66/6.20	   A(b(x1)) -> A(x1)
20.66/6.20	   A(b(x1)) -> A(a(x1))
20.66/6.20	The remaining pairs can at least be oriented weakly.
20.66/6.20	Used ordering:  Matrix interpretation [MATRO] with arctic natural numbers [ARCTIC]:
20.66/6.20	
20.66/6.20	   <<<
20.66/6.20	 POL(A(x_1)) =  	[[-I]] 	 +  	[[0A, -I, -I]] 	* 	x_1
20.66/6.20	>>>
20.66/6.20	
20.66/6.20	   <<<
20.66/6.20	 POL(b(x_1)) =  	[[0A], [-I], [-I]] 	 +  	[[1A, 0A, 0A], [0A, -I, -I], [0A, 0A, 0A]] 	* 	x_1
20.66/6.20	>>>
20.66/6.20	
20.66/6.20	   <<<
20.66/6.20	 POL(a(x_1)) =  	[[-I], [0A], [-I]] 	 +  	[[0A, -I, -I], [1A, 0A, 0A], [1A, 0A, 0A]] 	* 	x_1
20.66/6.20	>>>
20.66/6.20	
20.66/6.20	   <<<
20.66/6.20	 POL(c(x_1)) =  	[[0A], [-I], [-I]] 	 +  	[[-I, 0A, 0A], [0A, -I, -I], [0A, 0A, 0A]] 	* 	x_1
20.66/6.20	>>>
20.66/6.20	
20.66/6.20	
20.66/6.20	The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:
20.66/6.20	
20.66/6.20	   a(x1) -> x1
20.66/6.20	   a(b(x1)) -> c(b(b(a(a(x1)))))
20.66/6.20	   b(x1) -> c(x1)
20.66/6.20	   c(c(x1)) -> x1
20.66/6.20	
20.66/6.20	
20.66/6.20	----------------------------------------
20.66/6.20	
20.66/6.20	(6)
20.66/6.20	Obligation:
20.66/6.20	Q DP problem:
20.66/6.20	P is empty.
20.66/6.20	The TRS R consists of the following rules:
20.66/6.20	
20.66/6.20	   a(x1) -> x1
20.66/6.20	   a(b(x1)) -> c(b(b(a(a(x1)))))
20.66/6.20	   b(x1) -> c(x1)
20.66/6.20	   c(c(x1)) -> x1
20.66/6.20	
20.66/6.20	Q is empty.
20.66/6.20	We have to consider all minimal (P,Q,R)-chains.
20.66/6.20	----------------------------------------
20.66/6.20	
20.66/6.20	(7) PisEmptyProof (EQUIVALENT)
20.66/6.20	The TRS P is empty. Hence, there is no (P,Q,R) chain.
20.66/6.20	----------------------------------------
20.66/6.20	
20.66/6.20	(8)
20.66/6.20	YES
20.99/6.28	EOF