35.88/10.05	YES
35.88/10.06	proof of /export/starexec/sandbox/benchmark/theBenchmark.xml
35.88/10.06	# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty
35.88/10.06	
35.88/10.06	
35.88/10.06	Termination w.r.t. Q of the given QTRS could be proven:
35.88/10.06	
35.88/10.06	(0) QTRS
35.88/10.06	(1) FlatCCProof [EQUIVALENT, 0 ms]
35.88/10.06	(2) QTRS
35.88/10.06	(3) RootLabelingProof [EQUIVALENT, 1 ms]
35.88/10.06	(4) QTRS
35.88/10.06	(5) QTRSRRRProof [EQUIVALENT, 35 ms]
35.88/10.06	(6) QTRS
35.88/10.06	(7) DependencyPairsProof [EQUIVALENT, 7 ms]
35.88/10.06	(8) QDP
35.88/10.06	(9) DependencyGraphProof [EQUIVALENT, 4 ms]
35.88/10.06	(10) QDP
35.88/10.06	(11) QDPOrderProof [EQUIVALENT, 43 ms]
35.88/10.06	(12) QDP
35.88/10.06	(13) QDPOrderProof [EQUIVALENT, 24 ms]
35.88/10.06	(14) QDP
35.88/10.06	(15) DependencyGraphProof [EQUIVALENT, 0 ms]
35.88/10.06	(16) TRUE
35.88/10.06	
35.88/10.06	
35.88/10.06	----------------------------------------
35.88/10.06	
35.88/10.06	(0)
35.88/10.06	Obligation:
35.88/10.06	Q restricted rewrite system:
35.88/10.06	The TRS R consists of the following rules:
35.88/10.06	
35.88/10.06	   a(x1) -> x1
35.88/10.06	   a(a(x1)) -> a(b(a(b(x1))))
35.88/10.06	   b(b(b(b(x1)))) -> a(x1)
35.88/10.06	
35.88/10.06	Q is empty.
35.88/10.06	
35.88/10.06	----------------------------------------
35.88/10.06	
35.88/10.06	(1) FlatCCProof (EQUIVALENT)
35.88/10.06	We used flat context closure [ROOTLAB]
35.88/10.06	As Q is empty the flat context closure was sound AND complete.
35.88/10.06	
35.88/10.06	----------------------------------------
35.88/10.06	
35.88/10.06	(2)
35.88/10.06	Obligation:
35.88/10.06	Q restricted rewrite system:
35.88/10.06	The TRS R consists of the following rules:
35.88/10.06	
35.88/10.06	   a(a(x1)) -> a(b(a(b(x1))))
35.88/10.06	   a(a(x1)) -> a(x1)
35.88/10.06	   b(a(x1)) -> b(x1)
35.88/10.06	   a(b(b(b(b(x1))))) -> a(a(x1))
35.88/10.06	   b(b(b(b(b(x1))))) -> b(a(x1))
35.88/10.06	
35.88/10.06	Q is empty.
35.88/10.06	
35.88/10.06	----------------------------------------
35.88/10.06	
35.88/10.06	(3) RootLabelingProof (EQUIVALENT)
35.88/10.06	We used plain root labeling [ROOTLAB] with the following heuristic:
35.88/10.06	LabelAll: All function symbols get labeled
35.88/10.06	
35.88/10.06	As Q is empty the root labeling was sound AND complete.
35.88/10.06	
35.88/10.06	----------------------------------------
35.88/10.06	
35.88/10.06	(4)
35.88/10.06	Obligation:
35.88/10.06	Q restricted rewrite system:
35.88/10.06	The TRS R consists of the following rules:
35.88/10.06	
35.88/10.06	   a_{a_1}(a_{a_1}(x1)) -> a_{b_1}(b_{a_1}(a_{b_1}(b_{a_1}(x1))))
35.88/10.06	   a_{a_1}(a_{b_1}(x1)) -> a_{b_1}(b_{a_1}(a_{b_1}(b_{b_1}(x1))))
35.88/10.06	   a_{a_1}(a_{a_1}(x1)) -> a_{a_1}(x1)
35.88/10.06	   a_{a_1}(a_{b_1}(x1)) -> a_{b_1}(x1)
35.88/10.06	   b_{a_1}(a_{a_1}(x1)) -> b_{a_1}(x1)
35.88/10.06	   b_{a_1}(a_{b_1}(x1)) -> b_{b_1}(x1)
35.88/10.06	   a_{b_1}(b_{b_1}(b_{b_1}(b_{b_1}(b_{a_1}(x1))))) -> a_{a_1}(a_{a_1}(x1))
35.88/10.06	   a_{b_1}(b_{b_1}(b_{b_1}(b_{b_1}(b_{b_1}(x1))))) -> a_{a_1}(a_{b_1}(x1))
35.88/10.06	   b_{b_1}(b_{b_1}(b_{b_1}(b_{b_1}(b_{a_1}(x1))))) -> b_{a_1}(a_{a_1}(x1))
35.88/10.06	   b_{b_1}(b_{b_1}(b_{b_1}(b_{b_1}(b_{b_1}(x1))))) -> b_{a_1}(a_{b_1}(x1))
35.88/10.06	
35.88/10.06	Q is empty.
35.88/10.06	
35.88/10.06	----------------------------------------
35.88/10.06	
35.88/10.06	(5) QTRSRRRProof (EQUIVALENT)
35.88/10.06	Used ordering:
35.88/10.06	Polynomial interpretation [POLO]:
35.88/10.06	
35.88/10.06	   POL(a_{a_1}(x_1)) = 2 + x_1
35.88/10.06	   POL(a_{b_1}(x_1)) = x_1
35.88/10.06	   POL(b_{a_1}(x_1)) = 1 + x_1
35.88/10.06	   POL(b_{b_1}(x_1)) = 1 + x_1
35.88/10.06	With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:
35.88/10.06	
35.88/10.06	   a_{a_1}(a_{a_1}(x1)) -> a_{b_1}(b_{a_1}(a_{b_1}(b_{a_1}(x1))))
35.88/10.06	   a_{a_1}(a_{a_1}(x1)) -> a_{a_1}(x1)
35.88/10.06	   a_{a_1}(a_{b_1}(x1)) -> a_{b_1}(x1)
35.88/10.06	   b_{a_1}(a_{a_1}(x1)) -> b_{a_1}(x1)
35.88/10.06	   a_{b_1}(b_{b_1}(b_{b_1}(b_{b_1}(b_{b_1}(x1))))) -> a_{a_1}(a_{b_1}(x1))
35.88/10.06	   b_{b_1}(b_{b_1}(b_{b_1}(b_{b_1}(b_{a_1}(x1))))) -> b_{a_1}(a_{a_1}(x1))
35.88/10.06	   b_{b_1}(b_{b_1}(b_{b_1}(b_{b_1}(b_{b_1}(x1))))) -> b_{a_1}(a_{b_1}(x1))
35.88/10.06	
35.88/10.06	
35.88/10.06	
35.88/10.06	
35.88/10.06	----------------------------------------
35.88/10.06	
35.88/10.06	(6)
35.88/10.06	Obligation:
35.88/10.06	Q restricted rewrite system:
35.88/10.06	The TRS R consists of the following rules:
35.88/10.06	
35.88/10.06	   a_{a_1}(a_{b_1}(x1)) -> a_{b_1}(b_{a_1}(a_{b_1}(b_{b_1}(x1))))
35.88/10.06	   b_{a_1}(a_{b_1}(x1)) -> b_{b_1}(x1)
35.88/10.06	   a_{b_1}(b_{b_1}(b_{b_1}(b_{b_1}(b_{a_1}(x1))))) -> a_{a_1}(a_{a_1}(x1))
35.88/10.06	
35.88/10.06	Q is empty.
35.88/10.06	
35.88/10.06	----------------------------------------
35.88/10.06	
35.88/10.06	(7) DependencyPairsProof (EQUIVALENT)
35.88/10.06	Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.
35.88/10.06	----------------------------------------
35.88/10.06	
35.88/10.06	(8)
35.88/10.06	Obligation:
35.88/10.06	Q DP problem:
35.88/10.06	The TRS P consists of the following rules:
35.88/10.06	
35.88/10.06	   A_{A_1}(a_{b_1}(x1)) -> A_{B_1}(b_{a_1}(a_{b_1}(b_{b_1}(x1))))
35.88/10.06	   A_{A_1}(a_{b_1}(x1)) -> B_{A_1}(a_{b_1}(b_{b_1}(x1)))
35.88/10.06	   A_{A_1}(a_{b_1}(x1)) -> A_{B_1}(b_{b_1}(x1))
35.88/10.06	   A_{B_1}(b_{b_1}(b_{b_1}(b_{b_1}(b_{a_1}(x1))))) -> A_{A_1}(a_{a_1}(x1))
35.88/10.06	   A_{B_1}(b_{b_1}(b_{b_1}(b_{b_1}(b_{a_1}(x1))))) -> A_{A_1}(x1)
35.88/10.06	
35.88/10.06	The TRS R consists of the following rules:
35.88/10.06	
35.88/10.06	   a_{a_1}(a_{b_1}(x1)) -> a_{b_1}(b_{a_1}(a_{b_1}(b_{b_1}(x1))))
35.88/10.06	   b_{a_1}(a_{b_1}(x1)) -> b_{b_1}(x1)
35.88/10.06	   a_{b_1}(b_{b_1}(b_{b_1}(b_{b_1}(b_{a_1}(x1))))) -> a_{a_1}(a_{a_1}(x1))
35.88/10.06	
35.88/10.06	Q is empty.
35.88/10.06	We have to consider all minimal (P,Q,R)-chains.
35.88/10.06	----------------------------------------
35.88/10.06	
35.88/10.06	(9) DependencyGraphProof (EQUIVALENT)
35.88/10.06	The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 1 less node.
35.88/10.06	----------------------------------------
35.88/10.06	
35.88/10.06	(10)
35.88/10.06	Obligation:
35.88/10.06	Q DP problem:
35.88/10.06	The TRS P consists of the following rules:
35.88/10.06	
35.88/10.06	   A_{B_1}(b_{b_1}(b_{b_1}(b_{b_1}(b_{a_1}(x1))))) -> A_{A_1}(a_{a_1}(x1))
35.88/10.06	   A_{A_1}(a_{b_1}(x1)) -> A_{B_1}(b_{a_1}(a_{b_1}(b_{b_1}(x1))))
35.88/10.06	   A_{B_1}(b_{b_1}(b_{b_1}(b_{b_1}(b_{a_1}(x1))))) -> A_{A_1}(x1)
35.88/10.06	   A_{A_1}(a_{b_1}(x1)) -> A_{B_1}(b_{b_1}(x1))
35.88/10.06	
35.88/10.06	The TRS R consists of the following rules:
35.88/10.06	
35.88/10.06	   a_{a_1}(a_{b_1}(x1)) -> a_{b_1}(b_{a_1}(a_{b_1}(b_{b_1}(x1))))
35.88/10.06	   b_{a_1}(a_{b_1}(x1)) -> b_{b_1}(x1)
35.88/10.06	   a_{b_1}(b_{b_1}(b_{b_1}(b_{b_1}(b_{a_1}(x1))))) -> a_{a_1}(a_{a_1}(x1))
35.88/10.06	
35.88/10.06	Q is empty.
35.88/10.06	We have to consider all minimal (P,Q,R)-chains.
35.88/10.06	----------------------------------------
35.88/10.06	
35.88/10.06	(11) QDPOrderProof (EQUIVALENT)
35.88/10.06	We use the reduction pair processor [LPAR04,JAR06].
35.88/10.06	
35.88/10.06	
35.88/10.06	The following pairs can be oriented strictly and are deleted.
35.88/10.06	
35.88/10.06	   A_{B_1}(b_{b_1}(b_{b_1}(b_{b_1}(b_{a_1}(x1))))) -> A_{A_1}(x1)
35.88/10.06	   A_{A_1}(a_{b_1}(x1)) -> A_{B_1}(b_{b_1}(x1))
35.88/10.06	The remaining pairs can at least be oriented weakly.
35.88/10.06	Used ordering:  Polynomial Order [NEGPOLO,POLO] with Interpretation:
35.88/10.06	
35.88/10.06	POL( A_{A_1}_1(x_1) ) = max{0, 2x_1 - 2}
35.88/10.06	POL( A_{B_1}_1(x_1) ) = max{0, 2x_1 - 2}
35.88/10.06	POL( b_{a_1}_1(x_1) ) = max{0, x_1 - 1}
35.88/10.06	POL( a_{b_1}_1(x_1) ) = x_1 + 2
35.88/10.06	POL( b_{b_1}_1(x_1) ) = x_1 + 1
35.88/10.06	POL( a_{a_1}_1(x_1) ) = x_1 + 2
35.88/10.06	
35.88/10.06	The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:
35.88/10.06	
35.88/10.06	   a_{b_1}(b_{b_1}(b_{b_1}(b_{b_1}(b_{a_1}(x1))))) -> a_{a_1}(a_{a_1}(x1))
35.88/10.06	   a_{a_1}(a_{b_1}(x1)) -> a_{b_1}(b_{a_1}(a_{b_1}(b_{b_1}(x1))))
35.88/10.06	   b_{a_1}(a_{b_1}(x1)) -> b_{b_1}(x1)
35.88/10.06	
35.88/10.06	
35.88/10.06	----------------------------------------
35.88/10.06	
35.88/10.06	(12)
35.88/10.06	Obligation:
35.88/10.06	Q DP problem:
35.88/10.06	The TRS P consists of the following rules:
35.88/10.06	
35.88/10.06	   A_{B_1}(b_{b_1}(b_{b_1}(b_{b_1}(b_{a_1}(x1))))) -> A_{A_1}(a_{a_1}(x1))
35.88/10.06	   A_{A_1}(a_{b_1}(x1)) -> A_{B_1}(b_{a_1}(a_{b_1}(b_{b_1}(x1))))
35.88/10.06	
35.88/10.06	The TRS R consists of the following rules:
35.88/10.06	
35.88/10.06	   a_{a_1}(a_{b_1}(x1)) -> a_{b_1}(b_{a_1}(a_{b_1}(b_{b_1}(x1))))
35.88/10.06	   b_{a_1}(a_{b_1}(x1)) -> b_{b_1}(x1)
35.88/10.06	   a_{b_1}(b_{b_1}(b_{b_1}(b_{b_1}(b_{a_1}(x1))))) -> a_{a_1}(a_{a_1}(x1))
35.88/10.06	
35.88/10.06	Q is empty.
35.88/10.06	We have to consider all minimal (P,Q,R)-chains.
35.88/10.06	----------------------------------------
35.88/10.06	
35.88/10.06	(13) QDPOrderProof (EQUIVALENT)
35.88/10.06	We use the reduction pair processor [LPAR04,JAR06].
35.88/10.06	
35.88/10.06	
35.88/10.06	The following pairs can be oriented strictly and are deleted.
35.88/10.06	
35.88/10.06	   A_{B_1}(b_{b_1}(b_{b_1}(b_{b_1}(b_{a_1}(x1))))) -> A_{A_1}(a_{a_1}(x1))
35.88/10.06	The remaining pairs can at least be oriented weakly.
35.88/10.06	Used ordering:  Polynomial interpretation [POLO]:
35.88/10.06	
35.88/10.06	   POL(A_{A_1}(x_1)) = 2 + 4*x_1
35.88/10.06	   POL(A_{B_1}(x_1)) = 2*x_1
35.88/10.06	   POL(a_{a_1}(x_1)) = 2 + 4*x_1
35.88/10.06	   POL(a_{b_1}(x_1)) = 2*x_1
35.88/10.06	   POL(b_{a_1}(x_1)) = 1 + x_1
35.88/10.06	   POL(b_{b_1}(x_1)) = 2*x_1
35.88/10.06	
35.88/10.06	The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:
35.88/10.06	
35.88/10.06	   a_{b_1}(b_{b_1}(b_{b_1}(b_{b_1}(b_{a_1}(x1))))) -> a_{a_1}(a_{a_1}(x1))
35.88/10.06	   a_{a_1}(a_{b_1}(x1)) -> a_{b_1}(b_{a_1}(a_{b_1}(b_{b_1}(x1))))
35.88/10.06	   b_{a_1}(a_{b_1}(x1)) -> b_{b_1}(x1)
35.88/10.06	
35.88/10.06	
35.88/10.06	----------------------------------------
35.88/10.06	
35.88/10.06	(14)
35.88/10.06	Obligation:
35.88/10.06	Q DP problem:
35.88/10.06	The TRS P consists of the following rules:
35.88/10.06	
35.88/10.06	   A_{A_1}(a_{b_1}(x1)) -> A_{B_1}(b_{a_1}(a_{b_1}(b_{b_1}(x1))))
35.88/10.06	
35.88/10.06	The TRS R consists of the following rules:
35.88/10.06	
35.88/10.06	   a_{a_1}(a_{b_1}(x1)) -> a_{b_1}(b_{a_1}(a_{b_1}(b_{b_1}(x1))))
35.88/10.06	   b_{a_1}(a_{b_1}(x1)) -> b_{b_1}(x1)
35.88/10.06	   a_{b_1}(b_{b_1}(b_{b_1}(b_{b_1}(b_{a_1}(x1))))) -> a_{a_1}(a_{a_1}(x1))
35.88/10.06	
35.88/10.06	Q is empty.
35.88/10.06	We have to consider all minimal (P,Q,R)-chains.
35.88/10.06	----------------------------------------
35.88/10.06	
35.88/10.06	(15) DependencyGraphProof (EQUIVALENT)
35.88/10.06	The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 1 less node.
35.88/10.06	----------------------------------------
35.88/10.06	
35.88/10.06	(16)
35.88/10.06	TRUE
36.17/10.16	EOF