16.30/5.08	YES
16.76/5.24	proof of /export/starexec/sandbox2/benchmark/theBenchmark.xml
16.76/5.24	# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty
16.76/5.24	
16.76/5.24	
16.76/5.24	Termination w.r.t. Q of the given QTRS could be proven:
16.76/5.24	
16.76/5.24	(0) QTRS
16.76/5.24	(1) QTRS Reverse [EQUIVALENT, 0 ms]
16.76/5.24	(2) QTRS
16.76/5.24	(3) RootLabelingProof [EQUIVALENT, 0 ms]
16.76/5.24	(4) QTRS
16.76/5.24	(5) QTRSRRRProof [EQUIVALENT, 3 ms]
16.76/5.24	(6) QTRS
16.76/5.24	(7) QTRSRRRProof [EQUIVALENT, 1 ms]
16.76/5.24	(8) QTRS
16.76/5.24	(9) DependencyPairsProof [EQUIVALENT, 2 ms]
16.76/5.24	(10) QDP
16.76/5.24	(11) QDPSizeChangeProof [EQUIVALENT, 0 ms]
16.76/5.24	(12) YES
16.76/5.24	
16.76/5.24	
16.76/5.24	----------------------------------------
16.76/5.24	
16.76/5.24	(0)
16.76/5.24	Obligation:
16.76/5.24	Q restricted rewrite system:
16.76/5.24	The TRS R consists of the following rules:
16.76/5.24	
16.76/5.24	   a(a(b(a(x1)))) -> a(b(b(a(x1))))
16.76/5.24	   b(a(b(b(x1)))) -> b(b(a(b(x1))))
16.76/5.24	   b(a(b(a(x1)))) -> b(b(a(a(x1))))
16.76/5.24	
16.76/5.24	Q is empty.
16.76/5.24	
16.76/5.24	----------------------------------------
16.76/5.24	
16.76/5.24	(1) QTRS Reverse (EQUIVALENT)
16.76/5.24	We applied the QTRS Reverse Processor [REVERSE].
16.76/5.24	----------------------------------------
16.76/5.24	
16.76/5.24	(2)
16.76/5.24	Obligation:
16.76/5.24	Q restricted rewrite system:
16.76/5.24	The TRS R consists of the following rules:
16.76/5.24	
16.76/5.24	   a(b(a(a(x1)))) -> a(b(b(a(x1))))
16.76/5.24	   b(b(a(b(x1)))) -> b(a(b(b(x1))))
16.76/5.24	   a(b(a(b(x1)))) -> a(a(b(b(x1))))
16.76/5.24	
16.76/5.24	Q is empty.
16.76/5.24	
16.76/5.24	----------------------------------------
16.76/5.24	
16.76/5.24	(3) RootLabelingProof (EQUIVALENT)
16.76/5.24	We used plain root labeling [ROOTLAB] with the following heuristic:
16.76/5.24	LabelAll: All function symbols get labeled
16.76/5.24	
16.76/5.24	As Q is empty the root labeling was sound AND complete.
16.76/5.24	
16.76/5.24	----------------------------------------
16.76/5.24	
16.76/5.24	(4)
16.76/5.24	Obligation:
16.76/5.24	Q restricted rewrite system:
16.76/5.24	The TRS R consists of the following rules:
16.76/5.24	
16.76/5.24	   a_{b_1}(b_{a_1}(a_{a_1}(a_{a_1}(x1)))) -> a_{b_1}(b_{b_1}(b_{a_1}(a_{a_1}(x1))))
16.76/5.24	   a_{b_1}(b_{a_1}(a_{a_1}(a_{b_1}(x1)))) -> a_{b_1}(b_{b_1}(b_{a_1}(a_{b_1}(x1))))
16.76/5.24	   b_{b_1}(b_{a_1}(a_{b_1}(b_{a_1}(x1)))) -> b_{a_1}(a_{b_1}(b_{b_1}(b_{a_1}(x1))))
16.76/5.24	   b_{b_1}(b_{a_1}(a_{b_1}(b_{b_1}(x1)))) -> b_{a_1}(a_{b_1}(b_{b_1}(b_{b_1}(x1))))
16.76/5.24	   a_{b_1}(b_{a_1}(a_{b_1}(b_{a_1}(x1)))) -> a_{a_1}(a_{b_1}(b_{b_1}(b_{a_1}(x1))))
16.76/5.24	   a_{b_1}(b_{a_1}(a_{b_1}(b_{b_1}(x1)))) -> a_{a_1}(a_{b_1}(b_{b_1}(b_{b_1}(x1))))
16.76/5.24	
16.76/5.24	Q is empty.
16.76/5.24	
16.76/5.24	----------------------------------------
16.76/5.24	
16.76/5.24	(5) QTRSRRRProof (EQUIVALENT)
16.76/5.24	Used ordering:
16.76/5.24	Polynomial interpretation [POLO]:
16.76/5.24	
16.76/5.24	   POL(a_{a_1}(x_1)) = 1 + x_1
16.76/5.24	   POL(a_{b_1}(x_1)) = x_1
16.76/5.24	   POL(b_{a_1}(x_1)) = 1 + x_1
16.76/5.24	   POL(b_{b_1}(x_1)) = x_1
16.76/5.24	With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:
16.76/5.24	
16.76/5.24	   a_{b_1}(b_{a_1}(a_{a_1}(a_{a_1}(x1)))) -> a_{b_1}(b_{b_1}(b_{a_1}(a_{a_1}(x1))))
16.76/5.24	   a_{b_1}(b_{a_1}(a_{a_1}(a_{b_1}(x1)))) -> a_{b_1}(b_{b_1}(b_{a_1}(a_{b_1}(x1))))
16.76/5.24	
16.76/5.24	
16.76/5.24	
16.76/5.24	
16.76/5.24	----------------------------------------
16.76/5.24	
16.76/5.24	(6)
16.76/5.24	Obligation:
16.76/5.24	Q restricted rewrite system:
16.76/5.24	The TRS R consists of the following rules:
16.76/5.24	
16.76/5.24	   b_{b_1}(b_{a_1}(a_{b_1}(b_{a_1}(x1)))) -> b_{a_1}(a_{b_1}(b_{b_1}(b_{a_1}(x1))))
16.76/5.24	   b_{b_1}(b_{a_1}(a_{b_1}(b_{b_1}(x1)))) -> b_{a_1}(a_{b_1}(b_{b_1}(b_{b_1}(x1))))
16.76/5.24	   a_{b_1}(b_{a_1}(a_{b_1}(b_{a_1}(x1)))) -> a_{a_1}(a_{b_1}(b_{b_1}(b_{a_1}(x1))))
16.76/5.24	   a_{b_1}(b_{a_1}(a_{b_1}(b_{b_1}(x1)))) -> a_{a_1}(a_{b_1}(b_{b_1}(b_{b_1}(x1))))
16.76/5.24	
16.76/5.24	Q is empty.
16.76/5.24	
16.76/5.24	----------------------------------------
16.76/5.24	
16.76/5.24	(7) QTRSRRRProof (EQUIVALENT)
16.76/5.24	Used ordering:
16.76/5.24	Polynomial interpretation [POLO]:
16.76/5.24	
16.76/5.24	   POL(a_{a_1}(x_1)) = x_1
16.76/5.24	   POL(a_{b_1}(x_1)) = 1 + x_1
16.76/5.24	   POL(b_{a_1}(x_1)) = x_1
16.76/5.24	   POL(b_{b_1}(x_1)) = x_1
16.76/5.24	With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:
16.76/5.24	
16.76/5.24	   a_{b_1}(b_{a_1}(a_{b_1}(b_{a_1}(x1)))) -> a_{a_1}(a_{b_1}(b_{b_1}(b_{a_1}(x1))))
16.76/5.24	   a_{b_1}(b_{a_1}(a_{b_1}(b_{b_1}(x1)))) -> a_{a_1}(a_{b_1}(b_{b_1}(b_{b_1}(x1))))
16.76/5.24	
16.76/5.24	
16.76/5.24	
16.76/5.24	
16.76/5.24	----------------------------------------
16.76/5.24	
16.76/5.24	(8)
16.76/5.24	Obligation:
16.76/5.24	Q restricted rewrite system:
16.76/5.24	The TRS R consists of the following rules:
16.76/5.24	
16.76/5.24	   b_{b_1}(b_{a_1}(a_{b_1}(b_{a_1}(x1)))) -> b_{a_1}(a_{b_1}(b_{b_1}(b_{a_1}(x1))))
16.76/5.24	   b_{b_1}(b_{a_1}(a_{b_1}(b_{b_1}(x1)))) -> b_{a_1}(a_{b_1}(b_{b_1}(b_{b_1}(x1))))
16.76/5.24	
16.76/5.24	Q is empty.
16.76/5.24	
16.76/5.24	----------------------------------------
16.76/5.24	
16.76/5.24	(9) DependencyPairsProof (EQUIVALENT)
16.76/5.24	Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.
16.76/5.24	----------------------------------------
16.76/5.24	
16.76/5.24	(10)
16.76/5.24	Obligation:
16.76/5.24	Q DP problem:
16.76/5.24	The TRS P consists of the following rules:
16.76/5.24	
16.76/5.24	   B_{B_1}(b_{a_1}(a_{b_1}(b_{a_1}(x1)))) -> B_{B_1}(b_{a_1}(x1))
16.76/5.24	   B_{B_1}(b_{a_1}(a_{b_1}(b_{b_1}(x1)))) -> B_{B_1}(b_{b_1}(x1))
16.76/5.24	
16.76/5.24	The TRS R consists of the following rules:
16.76/5.24	
16.76/5.24	   b_{b_1}(b_{a_1}(a_{b_1}(b_{a_1}(x1)))) -> b_{a_1}(a_{b_1}(b_{b_1}(b_{a_1}(x1))))
16.76/5.24	   b_{b_1}(b_{a_1}(a_{b_1}(b_{b_1}(x1)))) -> b_{a_1}(a_{b_1}(b_{b_1}(b_{b_1}(x1))))
16.76/5.24	
16.76/5.24	Q is empty.
16.76/5.24	We have to consider all minimal (P,Q,R)-chains.
16.76/5.24	----------------------------------------
16.76/5.24	
16.76/5.24	(11) QDPSizeChangeProof (EQUIVALENT)
16.76/5.24	By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem. 
16.76/5.24	
16.76/5.24	From the DPs we obtained the following set of size-change graphs:
16.76/5.24	*B_{B_1}(b_{a_1}(a_{b_1}(b_{a_1}(x1)))) -> B_{B_1}(b_{a_1}(x1))
16.76/5.24	The graph contains the following edges 1 > 1
16.76/5.24	
16.76/5.24	
16.76/5.24	*B_{B_1}(b_{a_1}(a_{b_1}(b_{b_1}(x1)))) -> B_{B_1}(b_{b_1}(x1))
16.76/5.24	The graph contains the following edges 1 > 1
16.76/5.24	
16.76/5.24	
16.76/5.24	----------------------------------------
16.76/5.24	
16.76/5.24	(12)
16.76/5.24	YES
16.96/5.95	EOF