16.72/5.32	YES
16.72/5.33	proof of /export/starexec/sandbox2/benchmark/theBenchmark.xml
16.72/5.33	# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty
16.72/5.33	
16.72/5.33	
16.72/5.33	Termination w.r.t. Q of the given QTRS could be proven:
16.72/5.33	
16.72/5.33	(0) QTRS
16.72/5.33	(1) RootLabelingProof [EQUIVALENT, 0 ms]
16.72/5.33	(2) QTRS
16.72/5.33	(3) DependencyPairsProof [EQUIVALENT, 14 ms]
16.72/5.33	(4) QDP
16.72/5.33	(5) DependencyGraphProof [EQUIVALENT, 0 ms]
16.72/5.33	(6) QDP
16.72/5.33	(7) QDPOrderProof [EQUIVALENT, 128 ms]
16.72/5.33	(8) QDP
16.72/5.33	(9) PisEmptyProof [EQUIVALENT, 0 ms]
16.72/5.33	(10) YES
16.72/5.33	
16.72/5.33	
16.72/5.33	----------------------------------------
16.72/5.33	
16.72/5.33	(0)
16.72/5.33	Obligation:
16.72/5.33	Q restricted rewrite system:
16.72/5.33	The TRS R consists of the following rules:
16.72/5.33	
16.72/5.33	   a(b(b(a(x1)))) -> a(a(a(a(x1))))
16.72/5.33	   b(b(a(a(x1)))) -> b(a(a(a(x1))))
16.72/5.33	   b(b(a(a(x1)))) -> b(a(b(b(x1))))
16.72/5.33	
16.72/5.33	Q is empty.
16.72/5.33	
16.72/5.33	----------------------------------------
16.72/5.33	
16.72/5.33	(1) RootLabelingProof (EQUIVALENT)
16.72/5.33	We used plain root labeling [ROOTLAB] with the following heuristic:
16.72/5.33	LabelAll: All function symbols get labeled
16.72/5.33	
16.72/5.33	As Q is empty the root labeling was sound AND complete.
16.72/5.33	
16.72/5.33	----------------------------------------
16.72/5.33	
16.72/5.33	(2)
16.72/5.33	Obligation:
16.72/5.33	Q restricted rewrite system:
16.72/5.33	The TRS R consists of the following rules:
16.72/5.33	
16.72/5.33	   a_{b_1}(b_{b_1}(b_{a_1}(a_{a_1}(x1)))) -> a_{a_1}(a_{a_1}(a_{a_1}(a_{a_1}(x1))))
16.72/5.33	   a_{b_1}(b_{b_1}(b_{a_1}(a_{b_1}(x1)))) -> a_{a_1}(a_{a_1}(a_{a_1}(a_{b_1}(x1))))
16.72/5.33	   b_{b_1}(b_{a_1}(a_{a_1}(a_{a_1}(x1)))) -> b_{a_1}(a_{a_1}(a_{a_1}(a_{a_1}(x1))))
16.72/5.33	   b_{b_1}(b_{a_1}(a_{a_1}(a_{b_1}(x1)))) -> b_{a_1}(a_{a_1}(a_{a_1}(a_{b_1}(x1))))
16.72/5.33	   b_{b_1}(b_{a_1}(a_{a_1}(a_{a_1}(x1)))) -> b_{a_1}(a_{b_1}(b_{b_1}(b_{a_1}(x1))))
16.72/5.33	   b_{b_1}(b_{a_1}(a_{a_1}(a_{b_1}(x1)))) -> b_{a_1}(a_{b_1}(b_{b_1}(b_{b_1}(x1))))
16.72/5.33	
16.72/5.33	Q is empty.
16.72/5.33	
16.72/5.33	----------------------------------------
16.72/5.33	
16.72/5.33	(3) DependencyPairsProof (EQUIVALENT)
16.72/5.33	Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.
16.72/5.33	----------------------------------------
16.72/5.33	
16.72/5.33	(4)
16.72/5.33	Obligation:
16.72/5.33	Q DP problem:
16.72/5.33	The TRS P consists of the following rules:
16.72/5.33	
16.72/5.33	   B_{B_1}(b_{a_1}(a_{a_1}(a_{a_1}(x1)))) -> A_{B_1}(b_{b_1}(b_{a_1}(x1)))
16.72/5.33	   B_{B_1}(b_{a_1}(a_{a_1}(a_{a_1}(x1)))) -> B_{B_1}(b_{a_1}(x1))
16.72/5.33	   B_{B_1}(b_{a_1}(a_{a_1}(a_{b_1}(x1)))) -> A_{B_1}(b_{b_1}(b_{b_1}(x1)))
16.72/5.33	   B_{B_1}(b_{a_1}(a_{a_1}(a_{b_1}(x1)))) -> B_{B_1}(b_{b_1}(x1))
16.72/5.33	   B_{B_1}(b_{a_1}(a_{a_1}(a_{b_1}(x1)))) -> B_{B_1}(x1)
16.72/5.33	
16.72/5.33	The TRS R consists of the following rules:
16.72/5.33	
16.72/5.33	   a_{b_1}(b_{b_1}(b_{a_1}(a_{a_1}(x1)))) -> a_{a_1}(a_{a_1}(a_{a_1}(a_{a_1}(x1))))
16.72/5.33	   a_{b_1}(b_{b_1}(b_{a_1}(a_{b_1}(x1)))) -> a_{a_1}(a_{a_1}(a_{a_1}(a_{b_1}(x1))))
16.72/5.33	   b_{b_1}(b_{a_1}(a_{a_1}(a_{a_1}(x1)))) -> b_{a_1}(a_{a_1}(a_{a_1}(a_{a_1}(x1))))
16.72/5.33	   b_{b_1}(b_{a_1}(a_{a_1}(a_{b_1}(x1)))) -> b_{a_1}(a_{a_1}(a_{a_1}(a_{b_1}(x1))))
16.72/5.33	   b_{b_1}(b_{a_1}(a_{a_1}(a_{a_1}(x1)))) -> b_{a_1}(a_{b_1}(b_{b_1}(b_{a_1}(x1))))
16.72/5.33	   b_{b_1}(b_{a_1}(a_{a_1}(a_{b_1}(x1)))) -> b_{a_1}(a_{b_1}(b_{b_1}(b_{b_1}(x1))))
16.72/5.33	
16.72/5.33	Q is empty.
16.72/5.33	We have to consider all minimal (P,Q,R)-chains.
16.72/5.33	----------------------------------------
16.72/5.33	
16.72/5.33	(5) DependencyGraphProof (EQUIVALENT)
16.72/5.33	The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 2 less nodes.
16.72/5.33	----------------------------------------
16.72/5.33	
16.72/5.33	(6)
16.72/5.33	Obligation:
16.72/5.33	Q DP problem:
16.72/5.33	The TRS P consists of the following rules:
16.72/5.33	
16.72/5.33	   B_{B_1}(b_{a_1}(a_{a_1}(a_{b_1}(x1)))) -> B_{B_1}(b_{b_1}(x1))
16.72/5.33	   B_{B_1}(b_{a_1}(a_{a_1}(a_{a_1}(x1)))) -> B_{B_1}(b_{a_1}(x1))
16.72/5.33	   B_{B_1}(b_{a_1}(a_{a_1}(a_{b_1}(x1)))) -> B_{B_1}(x1)
16.72/5.33	
16.72/5.33	The TRS R consists of the following rules:
16.72/5.33	
16.72/5.33	   a_{b_1}(b_{b_1}(b_{a_1}(a_{a_1}(x1)))) -> a_{a_1}(a_{a_1}(a_{a_1}(a_{a_1}(x1))))
16.72/5.33	   a_{b_1}(b_{b_1}(b_{a_1}(a_{b_1}(x1)))) -> a_{a_1}(a_{a_1}(a_{a_1}(a_{b_1}(x1))))
16.72/5.33	   b_{b_1}(b_{a_1}(a_{a_1}(a_{a_1}(x1)))) -> b_{a_1}(a_{a_1}(a_{a_1}(a_{a_1}(x1))))
16.72/5.33	   b_{b_1}(b_{a_1}(a_{a_1}(a_{b_1}(x1)))) -> b_{a_1}(a_{a_1}(a_{a_1}(a_{b_1}(x1))))
16.72/5.33	   b_{b_1}(b_{a_1}(a_{a_1}(a_{a_1}(x1)))) -> b_{a_1}(a_{b_1}(b_{b_1}(b_{a_1}(x1))))
16.72/5.33	   b_{b_1}(b_{a_1}(a_{a_1}(a_{b_1}(x1)))) -> b_{a_1}(a_{b_1}(b_{b_1}(b_{b_1}(x1))))
16.72/5.33	
16.72/5.33	Q is empty.
16.72/5.33	We have to consider all minimal (P,Q,R)-chains.
16.72/5.33	----------------------------------------
16.72/5.33	
16.72/5.33	(7) QDPOrderProof (EQUIVALENT)
16.72/5.33	We use the reduction pair processor [LPAR04,JAR06].
16.72/5.33	
16.72/5.33	
16.72/5.33	The following pairs can be oriented strictly and are deleted.
16.72/5.33	
16.72/5.33	   B_{B_1}(b_{a_1}(a_{a_1}(a_{b_1}(x1)))) -> B_{B_1}(b_{b_1}(x1))
16.72/5.33	   B_{B_1}(b_{a_1}(a_{a_1}(a_{a_1}(x1)))) -> B_{B_1}(b_{a_1}(x1))
16.72/5.33	   B_{B_1}(b_{a_1}(a_{a_1}(a_{b_1}(x1)))) -> B_{B_1}(x1)
16.72/5.33	The remaining pairs can at least be oriented weakly.
16.72/5.33	Used ordering:  Polynomial Order [NEGPOLO,POLO] with Interpretation:
16.72/5.33	
16.72/5.33	POL( B_{B_1}_1(x_1) ) = max{0, 2x_1 - 2}
16.72/5.33	POL( b_{b_1}_1(x_1) ) = x_1 + 1
16.72/5.33	POL( b_{a_1}_1(x_1) ) = x_1
16.72/5.33	POL( a_{a_1}_1(x_1) ) = x_1 + 1
16.72/5.33	POL( a_{b_1}_1(x_1) ) = x_1 + 2
16.72/5.33	
16.72/5.33	The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:
16.72/5.33	
16.72/5.33	   b_{b_1}(b_{a_1}(a_{a_1}(a_{a_1}(x1)))) -> b_{a_1}(a_{a_1}(a_{a_1}(a_{a_1}(x1))))
16.72/5.33	   b_{b_1}(b_{a_1}(a_{a_1}(a_{b_1}(x1)))) -> b_{a_1}(a_{a_1}(a_{a_1}(a_{b_1}(x1))))
16.72/5.33	   b_{b_1}(b_{a_1}(a_{a_1}(a_{a_1}(x1)))) -> b_{a_1}(a_{b_1}(b_{b_1}(b_{a_1}(x1))))
16.72/5.33	   b_{b_1}(b_{a_1}(a_{a_1}(a_{b_1}(x1)))) -> b_{a_1}(a_{b_1}(b_{b_1}(b_{b_1}(x1))))
16.72/5.33	   a_{b_1}(b_{b_1}(b_{a_1}(a_{a_1}(x1)))) -> a_{a_1}(a_{a_1}(a_{a_1}(a_{a_1}(x1))))
16.72/5.33	   a_{b_1}(b_{b_1}(b_{a_1}(a_{b_1}(x1)))) -> a_{a_1}(a_{a_1}(a_{a_1}(a_{b_1}(x1))))
16.72/5.33	
16.72/5.33	
16.72/5.33	----------------------------------------
16.72/5.33	
16.72/5.33	(8)
16.72/5.33	Obligation:
16.72/5.33	Q DP problem:
16.72/5.33	P is empty.
16.72/5.33	The TRS R consists of the following rules:
16.72/5.33	
16.72/5.33	   a_{b_1}(b_{b_1}(b_{a_1}(a_{a_1}(x1)))) -> a_{a_1}(a_{a_1}(a_{a_1}(a_{a_1}(x1))))
16.72/5.33	   a_{b_1}(b_{b_1}(b_{a_1}(a_{b_1}(x1)))) -> a_{a_1}(a_{a_1}(a_{a_1}(a_{b_1}(x1))))
16.72/5.33	   b_{b_1}(b_{a_1}(a_{a_1}(a_{a_1}(x1)))) -> b_{a_1}(a_{a_1}(a_{a_1}(a_{a_1}(x1))))
16.72/5.33	   b_{b_1}(b_{a_1}(a_{a_1}(a_{b_1}(x1)))) -> b_{a_1}(a_{a_1}(a_{a_1}(a_{b_1}(x1))))
16.72/5.33	   b_{b_1}(b_{a_1}(a_{a_1}(a_{a_1}(x1)))) -> b_{a_1}(a_{b_1}(b_{b_1}(b_{a_1}(x1))))
16.72/5.33	   b_{b_1}(b_{a_1}(a_{a_1}(a_{b_1}(x1)))) -> b_{a_1}(a_{b_1}(b_{b_1}(b_{b_1}(x1))))
16.72/5.33	
16.72/5.33	Q is empty.
16.72/5.33	We have to consider all minimal (P,Q,R)-chains.
16.72/5.33	----------------------------------------
16.72/5.33	
16.72/5.33	(9) PisEmptyProof (EQUIVALENT)
16.72/5.33	The TRS P is empty. Hence, there is no (P,Q,R) chain.
16.72/5.33	----------------------------------------
16.72/5.33	
16.72/5.33	(10)
16.72/5.33	YES
17.25/5.43	EOF