20.77/6.19	YES
21.10/6.23	proof of /export/starexec/sandbox/benchmark/theBenchmark.xml
21.10/6.23	# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty
21.10/6.23	
21.10/6.23	
21.10/6.23	Termination w.r.t. Q of the given QTRS could be proven:
21.10/6.23	
21.10/6.23	(0) QTRS
21.10/6.23	(1) FlatCCProof [EQUIVALENT, 0 ms]
21.10/6.23	(2) QTRS
21.10/6.23	(3) RootLabelingProof [EQUIVALENT, 0 ms]
21.10/6.23	(4) QTRS
21.10/6.23	(5) QTRSRRRProof [EQUIVALENT, 42 ms]
21.10/6.23	(6) QTRS
21.10/6.23	(7) DependencyPairsProof [EQUIVALENT, 0 ms]
21.10/6.23	(8) QDP
21.10/6.23	(9) DependencyGraphProof [EQUIVALENT, 0 ms]
21.10/6.23	(10) AND
21.10/6.23	    (11) QDP
21.10/6.23	        (12) UsableRulesProof [EQUIVALENT, 2 ms]
21.10/6.23	        (13) QDP
21.10/6.23	        (14) QDPSizeChangeProof [EQUIVALENT, 0 ms]
21.10/6.23	        (15) YES
21.10/6.23	    (16) QDP
21.10/6.23	        (17) UsableRulesProof [EQUIVALENT, 1 ms]
21.10/6.23	        (18) QDP
21.10/6.23	        (19) QDPSizeChangeProof [EQUIVALENT, 0 ms]
21.10/6.23	        (20) YES
21.10/6.23	    (21) QDP
21.10/6.23	        (22) QDPOrderProof [EQUIVALENT, 105 ms]
21.10/6.23	        (23) QDP
21.10/6.23	        (24) PisEmptyProof [EQUIVALENT, 0 ms]
21.10/6.23	        (25) YES
21.10/6.23	
21.10/6.23	
21.10/6.23	----------------------------------------
21.10/6.23	
21.10/6.23	(0)
21.10/6.23	Obligation:
21.10/6.23	Q restricted rewrite system:
21.10/6.23	The TRS R consists of the following rules:
21.10/6.23	
21.10/6.23	   a(b(x1)) -> x1
21.10/6.23	   a(c(x1)) -> c(c(x1))
21.10/6.23	   b(c(x1)) -> a(b(a(b(x1))))
21.10/6.23	
21.10/6.23	Q is empty.
21.10/6.23	
21.10/6.23	----------------------------------------
21.10/6.23	
21.10/6.23	(1) FlatCCProof (EQUIVALENT)
21.10/6.23	We used flat context closure [ROOTLAB]
21.10/6.23	As Q is empty the flat context closure was sound AND complete.
21.10/6.23	
21.10/6.23	----------------------------------------
21.10/6.23	
21.10/6.23	(2)
21.10/6.23	Obligation:
21.10/6.23	Q restricted rewrite system:
21.10/6.23	The TRS R consists of the following rules:
21.10/6.23	
21.10/6.23	   a(a(b(x1))) -> a(x1)
21.10/6.23	   b(a(b(x1))) -> b(x1)
21.10/6.23	   c(a(b(x1))) -> c(x1)
21.10/6.23	   a(a(c(x1))) -> a(c(c(x1)))
21.10/6.23	   b(a(c(x1))) -> b(c(c(x1)))
21.10/6.23	   c(a(c(x1))) -> c(c(c(x1)))
21.10/6.23	   a(b(c(x1))) -> a(a(b(a(b(x1)))))
21.10/6.23	   b(b(c(x1))) -> b(a(b(a(b(x1)))))
21.10/6.23	   c(b(c(x1))) -> c(a(b(a(b(x1)))))
21.10/6.23	
21.10/6.23	Q is empty.
21.10/6.23	
21.10/6.23	----------------------------------------
21.10/6.23	
21.10/6.23	(3) RootLabelingProof (EQUIVALENT)
21.10/6.23	We used plain root labeling [ROOTLAB] with the following heuristic:
21.10/6.23	LabelAll: All function symbols get labeled
21.10/6.23	
21.10/6.23	As Q is empty the root labeling was sound AND complete.
21.10/6.23	
21.10/6.23	----------------------------------------
21.10/6.23	
21.10/6.23	(4)
21.10/6.23	Obligation:
21.10/6.23	Q restricted rewrite system:
21.10/6.23	The TRS R consists of the following rules:
21.10/6.23	
21.10/6.23	   a_{a_1}(a_{b_1}(b_{a_1}(x1))) -> a_{a_1}(x1)
21.10/6.23	   a_{a_1}(a_{b_1}(b_{b_1}(x1))) -> a_{b_1}(x1)
21.10/6.23	   a_{a_1}(a_{b_1}(b_{c_1}(x1))) -> a_{c_1}(x1)
21.10/6.23	   b_{a_1}(a_{b_1}(b_{a_1}(x1))) -> b_{a_1}(x1)
21.10/6.23	   b_{a_1}(a_{b_1}(b_{b_1}(x1))) -> b_{b_1}(x1)
21.10/6.23	   b_{a_1}(a_{b_1}(b_{c_1}(x1))) -> b_{c_1}(x1)
21.10/6.23	   c_{a_1}(a_{b_1}(b_{a_1}(x1))) -> c_{a_1}(x1)
21.10/6.23	   c_{a_1}(a_{b_1}(b_{b_1}(x1))) -> c_{b_1}(x1)
21.10/6.23	   c_{a_1}(a_{b_1}(b_{c_1}(x1))) -> c_{c_1}(x1)
21.10/6.23	   a_{a_1}(a_{c_1}(c_{a_1}(x1))) -> a_{c_1}(c_{c_1}(c_{a_1}(x1)))
21.10/6.23	   a_{a_1}(a_{c_1}(c_{b_1}(x1))) -> a_{c_1}(c_{c_1}(c_{b_1}(x1)))
21.10/6.23	   a_{a_1}(a_{c_1}(c_{c_1}(x1))) -> a_{c_1}(c_{c_1}(c_{c_1}(x1)))
21.10/6.23	   b_{a_1}(a_{c_1}(c_{a_1}(x1))) -> b_{c_1}(c_{c_1}(c_{a_1}(x1)))
21.10/6.23	   b_{a_1}(a_{c_1}(c_{b_1}(x1))) -> b_{c_1}(c_{c_1}(c_{b_1}(x1)))
21.10/6.23	   b_{a_1}(a_{c_1}(c_{c_1}(x1))) -> b_{c_1}(c_{c_1}(c_{c_1}(x1)))
21.10/6.23	   c_{a_1}(a_{c_1}(c_{a_1}(x1))) -> c_{c_1}(c_{c_1}(c_{a_1}(x1)))
21.10/6.23	   c_{a_1}(a_{c_1}(c_{b_1}(x1))) -> c_{c_1}(c_{c_1}(c_{b_1}(x1)))
21.10/6.23	   c_{a_1}(a_{c_1}(c_{c_1}(x1))) -> c_{c_1}(c_{c_1}(c_{c_1}(x1)))
21.10/6.23	   a_{b_1}(b_{c_1}(c_{a_1}(x1))) -> a_{a_1}(a_{b_1}(b_{a_1}(a_{b_1}(b_{a_1}(x1)))))
21.10/6.23	   a_{b_1}(b_{c_1}(c_{b_1}(x1))) -> a_{a_1}(a_{b_1}(b_{a_1}(a_{b_1}(b_{b_1}(x1)))))
21.10/6.23	   a_{b_1}(b_{c_1}(c_{c_1}(x1))) -> a_{a_1}(a_{b_1}(b_{a_1}(a_{b_1}(b_{c_1}(x1)))))
21.10/6.23	   b_{b_1}(b_{c_1}(c_{a_1}(x1))) -> b_{a_1}(a_{b_1}(b_{a_1}(a_{b_1}(b_{a_1}(x1)))))
21.10/6.23	   b_{b_1}(b_{c_1}(c_{b_1}(x1))) -> b_{a_1}(a_{b_1}(b_{a_1}(a_{b_1}(b_{b_1}(x1)))))
21.10/6.23	   b_{b_1}(b_{c_1}(c_{c_1}(x1))) -> b_{a_1}(a_{b_1}(b_{a_1}(a_{b_1}(b_{c_1}(x1)))))
21.10/6.23	   c_{b_1}(b_{c_1}(c_{a_1}(x1))) -> c_{a_1}(a_{b_1}(b_{a_1}(a_{b_1}(b_{a_1}(x1)))))
21.10/6.23	   c_{b_1}(b_{c_1}(c_{b_1}(x1))) -> c_{a_1}(a_{b_1}(b_{a_1}(a_{b_1}(b_{b_1}(x1)))))
21.10/6.23	   c_{b_1}(b_{c_1}(c_{c_1}(x1))) -> c_{a_1}(a_{b_1}(b_{a_1}(a_{b_1}(b_{c_1}(x1)))))
21.10/6.23	
21.10/6.23	Q is empty.
21.10/6.23	
21.10/6.23	----------------------------------------
21.10/6.23	
21.10/6.23	(5) QTRSRRRProof (EQUIVALENT)
21.10/6.23	Used ordering:
21.10/6.23	Polynomial interpretation [POLO]:
21.10/6.23	
21.10/6.23	   POL(a_{a_1}(x_1)) = x_1
21.10/6.23	   POL(a_{b_1}(x_1)) = x_1
21.10/6.23	   POL(a_{c_1}(x_1)) = x_1
21.10/6.23	   POL(b_{a_1}(x_1)) = x_1
21.10/6.23	   POL(b_{b_1}(x_1)) = 2 + x_1
21.10/6.23	   POL(b_{c_1}(x_1)) = x_1
21.10/6.23	   POL(c_{a_1}(x_1)) = 2 + x_1
21.10/6.23	   POL(c_{b_1}(x_1)) = 3 + x_1
21.10/6.23	   POL(c_{c_1}(x_1)) = x_1
21.10/6.23	With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:
21.10/6.23	
21.10/6.23	   a_{a_1}(a_{b_1}(b_{b_1}(x1))) -> a_{b_1}(x1)
21.10/6.23	   c_{a_1}(a_{b_1}(b_{b_1}(x1))) -> c_{b_1}(x1)
21.10/6.23	   c_{a_1}(a_{b_1}(b_{c_1}(x1))) -> c_{c_1}(x1)
21.10/6.23	   c_{a_1}(a_{c_1}(c_{a_1}(x1))) -> c_{c_1}(c_{c_1}(c_{a_1}(x1)))
21.10/6.23	   c_{a_1}(a_{c_1}(c_{b_1}(x1))) -> c_{c_1}(c_{c_1}(c_{b_1}(x1)))
21.10/6.23	   c_{a_1}(a_{c_1}(c_{c_1}(x1))) -> c_{c_1}(c_{c_1}(c_{c_1}(x1)))
21.10/6.23	   a_{b_1}(b_{c_1}(c_{a_1}(x1))) -> a_{a_1}(a_{b_1}(b_{a_1}(a_{b_1}(b_{a_1}(x1)))))
21.10/6.23	   a_{b_1}(b_{c_1}(c_{b_1}(x1))) -> a_{a_1}(a_{b_1}(b_{a_1}(a_{b_1}(b_{b_1}(x1)))))
21.10/6.23	   b_{b_1}(b_{c_1}(c_{a_1}(x1))) -> b_{a_1}(a_{b_1}(b_{a_1}(a_{b_1}(b_{a_1}(x1)))))
21.10/6.23	   b_{b_1}(b_{c_1}(c_{b_1}(x1))) -> b_{a_1}(a_{b_1}(b_{a_1}(a_{b_1}(b_{b_1}(x1)))))
21.10/6.23	   b_{b_1}(b_{c_1}(c_{c_1}(x1))) -> b_{a_1}(a_{b_1}(b_{a_1}(a_{b_1}(b_{c_1}(x1)))))
21.10/6.23	   c_{b_1}(b_{c_1}(c_{a_1}(x1))) -> c_{a_1}(a_{b_1}(b_{a_1}(a_{b_1}(b_{a_1}(x1)))))
21.10/6.23	   c_{b_1}(b_{c_1}(c_{b_1}(x1))) -> c_{a_1}(a_{b_1}(b_{a_1}(a_{b_1}(b_{b_1}(x1)))))
21.10/6.23	   c_{b_1}(b_{c_1}(c_{c_1}(x1))) -> c_{a_1}(a_{b_1}(b_{a_1}(a_{b_1}(b_{c_1}(x1)))))
21.10/6.23	
21.10/6.23	
21.10/6.23	
21.10/6.23	
21.10/6.23	----------------------------------------
21.10/6.23	
21.10/6.23	(6)
21.10/6.23	Obligation:
21.10/6.23	Q restricted rewrite system:
21.10/6.23	The TRS R consists of the following rules:
21.10/6.23	
21.10/6.23	   a_{a_1}(a_{b_1}(b_{a_1}(x1))) -> a_{a_1}(x1)
21.10/6.23	   a_{a_1}(a_{b_1}(b_{c_1}(x1))) -> a_{c_1}(x1)
21.10/6.23	   b_{a_1}(a_{b_1}(b_{a_1}(x1))) -> b_{a_1}(x1)
21.10/6.23	   b_{a_1}(a_{b_1}(b_{b_1}(x1))) -> b_{b_1}(x1)
21.10/6.23	   b_{a_1}(a_{b_1}(b_{c_1}(x1))) -> b_{c_1}(x1)
21.10/6.23	   c_{a_1}(a_{b_1}(b_{a_1}(x1))) -> c_{a_1}(x1)
21.10/6.23	   a_{a_1}(a_{c_1}(c_{a_1}(x1))) -> a_{c_1}(c_{c_1}(c_{a_1}(x1)))
21.10/6.23	   a_{a_1}(a_{c_1}(c_{b_1}(x1))) -> a_{c_1}(c_{c_1}(c_{b_1}(x1)))
21.10/6.23	   a_{a_1}(a_{c_1}(c_{c_1}(x1))) -> a_{c_1}(c_{c_1}(c_{c_1}(x1)))
21.10/6.23	   b_{a_1}(a_{c_1}(c_{a_1}(x1))) -> b_{c_1}(c_{c_1}(c_{a_1}(x1)))
21.10/6.23	   b_{a_1}(a_{c_1}(c_{b_1}(x1))) -> b_{c_1}(c_{c_1}(c_{b_1}(x1)))
21.10/6.23	   b_{a_1}(a_{c_1}(c_{c_1}(x1))) -> b_{c_1}(c_{c_1}(c_{c_1}(x1)))
21.10/6.23	   a_{b_1}(b_{c_1}(c_{c_1}(x1))) -> a_{a_1}(a_{b_1}(b_{a_1}(a_{b_1}(b_{c_1}(x1)))))
21.10/6.23	
21.10/6.23	Q is empty.
21.10/6.23	
21.10/6.23	----------------------------------------
21.10/6.23	
21.10/6.23	(7) DependencyPairsProof (EQUIVALENT)
21.10/6.23	Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.
21.10/6.23	----------------------------------------
21.10/6.23	
21.10/6.23	(8)
21.10/6.23	Obligation:
21.10/6.23	Q DP problem:
21.10/6.23	The TRS P consists of the following rules:
21.10/6.23	
21.10/6.23	   A_{A_1}(a_{b_1}(b_{a_1}(x1))) -> A_{A_1}(x1)
21.10/6.23	   C_{A_1}(a_{b_1}(b_{a_1}(x1))) -> C_{A_1}(x1)
21.10/6.23	   A_{B_1}(b_{c_1}(c_{c_1}(x1))) -> A_{A_1}(a_{b_1}(b_{a_1}(a_{b_1}(b_{c_1}(x1)))))
21.10/6.23	   A_{B_1}(b_{c_1}(c_{c_1}(x1))) -> A_{B_1}(b_{a_1}(a_{b_1}(b_{c_1}(x1))))
21.10/6.23	   A_{B_1}(b_{c_1}(c_{c_1}(x1))) -> B_{A_1}(a_{b_1}(b_{c_1}(x1)))
21.10/6.23	   A_{B_1}(b_{c_1}(c_{c_1}(x1))) -> A_{B_1}(b_{c_1}(x1))
21.10/6.23	
21.10/6.23	The TRS R consists of the following rules:
21.10/6.23	
21.10/6.23	   a_{a_1}(a_{b_1}(b_{a_1}(x1))) -> a_{a_1}(x1)
21.10/6.23	   a_{a_1}(a_{b_1}(b_{c_1}(x1))) -> a_{c_1}(x1)
21.10/6.23	   b_{a_1}(a_{b_1}(b_{a_1}(x1))) -> b_{a_1}(x1)
21.10/6.23	   b_{a_1}(a_{b_1}(b_{b_1}(x1))) -> b_{b_1}(x1)
21.10/6.23	   b_{a_1}(a_{b_1}(b_{c_1}(x1))) -> b_{c_1}(x1)
21.10/6.23	   c_{a_1}(a_{b_1}(b_{a_1}(x1))) -> c_{a_1}(x1)
21.10/6.23	   a_{a_1}(a_{c_1}(c_{a_1}(x1))) -> a_{c_1}(c_{c_1}(c_{a_1}(x1)))
21.10/6.23	   a_{a_1}(a_{c_1}(c_{b_1}(x1))) -> a_{c_1}(c_{c_1}(c_{b_1}(x1)))
21.10/6.23	   a_{a_1}(a_{c_1}(c_{c_1}(x1))) -> a_{c_1}(c_{c_1}(c_{c_1}(x1)))
21.10/6.23	   b_{a_1}(a_{c_1}(c_{a_1}(x1))) -> b_{c_1}(c_{c_1}(c_{a_1}(x1)))
21.10/6.23	   b_{a_1}(a_{c_1}(c_{b_1}(x1))) -> b_{c_1}(c_{c_1}(c_{b_1}(x1)))
21.10/6.23	   b_{a_1}(a_{c_1}(c_{c_1}(x1))) -> b_{c_1}(c_{c_1}(c_{c_1}(x1)))
21.10/6.23	   a_{b_1}(b_{c_1}(c_{c_1}(x1))) -> a_{a_1}(a_{b_1}(b_{a_1}(a_{b_1}(b_{c_1}(x1)))))
21.10/6.23	
21.10/6.23	Q is empty.
21.10/6.23	We have to consider all minimal (P,Q,R)-chains.
21.10/6.23	----------------------------------------
21.10/6.23	
21.10/6.23	(9) DependencyGraphProof (EQUIVALENT)
21.10/6.23	The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 3 SCCs with 2 less nodes.
21.10/6.23	----------------------------------------
21.10/6.23	
21.10/6.23	(10)
21.10/6.23	Complex Obligation (AND)
21.10/6.23	
21.10/6.23	----------------------------------------
21.10/6.23	
21.10/6.23	(11)
21.10/6.23	Obligation:
21.10/6.23	Q DP problem:
21.10/6.23	The TRS P consists of the following rules:
21.10/6.23	
21.10/6.23	   C_{A_1}(a_{b_1}(b_{a_1}(x1))) -> C_{A_1}(x1)
21.10/6.23	
21.10/6.23	The TRS R consists of the following rules:
21.10/6.23	
21.10/6.23	   a_{a_1}(a_{b_1}(b_{a_1}(x1))) -> a_{a_1}(x1)
21.10/6.23	   a_{a_1}(a_{b_1}(b_{c_1}(x1))) -> a_{c_1}(x1)
21.10/6.23	   b_{a_1}(a_{b_1}(b_{a_1}(x1))) -> b_{a_1}(x1)
21.10/6.23	   b_{a_1}(a_{b_1}(b_{b_1}(x1))) -> b_{b_1}(x1)
21.10/6.23	   b_{a_1}(a_{b_1}(b_{c_1}(x1))) -> b_{c_1}(x1)
21.10/6.23	   c_{a_1}(a_{b_1}(b_{a_1}(x1))) -> c_{a_1}(x1)
21.10/6.23	   a_{a_1}(a_{c_1}(c_{a_1}(x1))) -> a_{c_1}(c_{c_1}(c_{a_1}(x1)))
21.10/6.23	   a_{a_1}(a_{c_1}(c_{b_1}(x1))) -> a_{c_1}(c_{c_1}(c_{b_1}(x1)))
21.10/6.23	   a_{a_1}(a_{c_1}(c_{c_1}(x1))) -> a_{c_1}(c_{c_1}(c_{c_1}(x1)))
21.10/6.23	   b_{a_1}(a_{c_1}(c_{a_1}(x1))) -> b_{c_1}(c_{c_1}(c_{a_1}(x1)))
21.10/6.23	   b_{a_1}(a_{c_1}(c_{b_1}(x1))) -> b_{c_1}(c_{c_1}(c_{b_1}(x1)))
21.10/6.23	   b_{a_1}(a_{c_1}(c_{c_1}(x1))) -> b_{c_1}(c_{c_1}(c_{c_1}(x1)))
21.10/6.23	   a_{b_1}(b_{c_1}(c_{c_1}(x1))) -> a_{a_1}(a_{b_1}(b_{a_1}(a_{b_1}(b_{c_1}(x1)))))
21.10/6.23	
21.10/6.23	Q is empty.
21.10/6.23	We have to consider all minimal (P,Q,R)-chains.
21.10/6.23	----------------------------------------
21.10/6.23	
21.10/6.23	(12) UsableRulesProof (EQUIVALENT)
21.10/6.23	We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R.
21.10/6.23	----------------------------------------
21.10/6.23	
21.10/6.23	(13)
21.10/6.23	Obligation:
21.10/6.23	Q DP problem:
21.10/6.23	The TRS P consists of the following rules:
21.10/6.23	
21.10/6.23	   C_{A_1}(a_{b_1}(b_{a_1}(x1))) -> C_{A_1}(x1)
21.10/6.23	
21.10/6.23	R is empty.
21.10/6.23	Q is empty.
21.10/6.23	We have to consider all minimal (P,Q,R)-chains.
21.10/6.23	----------------------------------------
21.10/6.23	
21.10/6.23	(14) QDPSizeChangeProof (EQUIVALENT)
21.10/6.23	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. 
21.10/6.23	
21.10/6.23	From the DPs we obtained the following set of size-change graphs:
21.10/6.23	*C_{A_1}(a_{b_1}(b_{a_1}(x1))) -> C_{A_1}(x1)
21.10/6.23	The graph contains the following edges 1 > 1
21.10/6.23	
21.10/6.23	
21.10/6.23	----------------------------------------
21.10/6.23	
21.10/6.23	(15)
21.10/6.23	YES
21.10/6.23	
21.10/6.23	----------------------------------------
21.10/6.23	
21.10/6.23	(16)
21.10/6.23	Obligation:
21.10/6.23	Q DP problem:
21.10/6.23	The TRS P consists of the following rules:
21.10/6.23	
21.10/6.23	   A_{A_1}(a_{b_1}(b_{a_1}(x1))) -> A_{A_1}(x1)
21.10/6.23	
21.10/6.23	The TRS R consists of the following rules:
21.10/6.23	
21.10/6.23	   a_{a_1}(a_{b_1}(b_{a_1}(x1))) -> a_{a_1}(x1)
21.10/6.23	   a_{a_1}(a_{b_1}(b_{c_1}(x1))) -> a_{c_1}(x1)
21.10/6.23	   b_{a_1}(a_{b_1}(b_{a_1}(x1))) -> b_{a_1}(x1)
21.10/6.23	   b_{a_1}(a_{b_1}(b_{b_1}(x1))) -> b_{b_1}(x1)
21.10/6.23	   b_{a_1}(a_{b_1}(b_{c_1}(x1))) -> b_{c_1}(x1)
21.10/6.23	   c_{a_1}(a_{b_1}(b_{a_1}(x1))) -> c_{a_1}(x1)
21.10/6.23	   a_{a_1}(a_{c_1}(c_{a_1}(x1))) -> a_{c_1}(c_{c_1}(c_{a_1}(x1)))
21.10/6.23	   a_{a_1}(a_{c_1}(c_{b_1}(x1))) -> a_{c_1}(c_{c_1}(c_{b_1}(x1)))
21.10/6.23	   a_{a_1}(a_{c_1}(c_{c_1}(x1))) -> a_{c_1}(c_{c_1}(c_{c_1}(x1)))
21.10/6.23	   b_{a_1}(a_{c_1}(c_{a_1}(x1))) -> b_{c_1}(c_{c_1}(c_{a_1}(x1)))
21.10/6.23	   b_{a_1}(a_{c_1}(c_{b_1}(x1))) -> b_{c_1}(c_{c_1}(c_{b_1}(x1)))
21.10/6.23	   b_{a_1}(a_{c_1}(c_{c_1}(x1))) -> b_{c_1}(c_{c_1}(c_{c_1}(x1)))
21.10/6.23	   a_{b_1}(b_{c_1}(c_{c_1}(x1))) -> a_{a_1}(a_{b_1}(b_{a_1}(a_{b_1}(b_{c_1}(x1)))))
21.10/6.23	
21.10/6.23	Q is empty.
21.10/6.23	We have to consider all minimal (P,Q,R)-chains.
21.10/6.23	----------------------------------------
21.10/6.23	
21.10/6.23	(17) UsableRulesProof (EQUIVALENT)
21.10/6.23	We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R.
21.10/6.23	----------------------------------------
21.10/6.23	
21.10/6.23	(18)
21.10/6.23	Obligation:
21.10/6.23	Q DP problem:
21.10/6.23	The TRS P consists of the following rules:
21.10/6.23	
21.10/6.23	   A_{A_1}(a_{b_1}(b_{a_1}(x1))) -> A_{A_1}(x1)
21.10/6.23	
21.10/6.23	R is empty.
21.10/6.23	Q is empty.
21.10/6.23	We have to consider all minimal (P,Q,R)-chains.
21.10/6.23	----------------------------------------
21.10/6.23	
21.10/6.23	(19) QDPSizeChangeProof (EQUIVALENT)
21.10/6.23	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. 
21.10/6.23	
21.10/6.23	From the DPs we obtained the following set of size-change graphs:
21.10/6.23	*A_{A_1}(a_{b_1}(b_{a_1}(x1))) -> A_{A_1}(x1)
21.10/6.23	The graph contains the following edges 1 > 1
21.10/6.23	
21.10/6.23	
21.10/6.23	----------------------------------------
21.10/6.23	
21.10/6.23	(20)
21.10/6.23	YES
21.10/6.23	
21.10/6.23	----------------------------------------
21.10/6.23	
21.10/6.23	(21)
21.10/6.23	Obligation:
21.10/6.23	Q DP problem:
21.10/6.23	The TRS P consists of the following rules:
21.10/6.23	
21.10/6.23	   A_{B_1}(b_{c_1}(c_{c_1}(x1))) -> A_{B_1}(b_{c_1}(x1))
21.10/6.23	   A_{B_1}(b_{c_1}(c_{c_1}(x1))) -> A_{B_1}(b_{a_1}(a_{b_1}(b_{c_1}(x1))))
21.10/6.23	
21.10/6.23	The TRS R consists of the following rules:
21.10/6.23	
21.10/6.23	   a_{a_1}(a_{b_1}(b_{a_1}(x1))) -> a_{a_1}(x1)
21.10/6.23	   a_{a_1}(a_{b_1}(b_{c_1}(x1))) -> a_{c_1}(x1)
21.10/6.23	   b_{a_1}(a_{b_1}(b_{a_1}(x1))) -> b_{a_1}(x1)
21.10/6.23	   b_{a_1}(a_{b_1}(b_{b_1}(x1))) -> b_{b_1}(x1)
21.10/6.23	   b_{a_1}(a_{b_1}(b_{c_1}(x1))) -> b_{c_1}(x1)
21.10/6.23	   c_{a_1}(a_{b_1}(b_{a_1}(x1))) -> c_{a_1}(x1)
21.10/6.23	   a_{a_1}(a_{c_1}(c_{a_1}(x1))) -> a_{c_1}(c_{c_1}(c_{a_1}(x1)))
21.10/6.23	   a_{a_1}(a_{c_1}(c_{b_1}(x1))) -> a_{c_1}(c_{c_1}(c_{b_1}(x1)))
21.10/6.23	   a_{a_1}(a_{c_1}(c_{c_1}(x1))) -> a_{c_1}(c_{c_1}(c_{c_1}(x1)))
21.10/6.23	   b_{a_1}(a_{c_1}(c_{a_1}(x1))) -> b_{c_1}(c_{c_1}(c_{a_1}(x1)))
21.10/6.23	   b_{a_1}(a_{c_1}(c_{b_1}(x1))) -> b_{c_1}(c_{c_1}(c_{b_1}(x1)))
21.10/6.23	   b_{a_1}(a_{c_1}(c_{c_1}(x1))) -> b_{c_1}(c_{c_1}(c_{c_1}(x1)))
21.10/6.23	   a_{b_1}(b_{c_1}(c_{c_1}(x1))) -> a_{a_1}(a_{b_1}(b_{a_1}(a_{b_1}(b_{c_1}(x1)))))
21.10/6.23	
21.10/6.23	Q is empty.
21.10/6.23	We have to consider all minimal (P,Q,R)-chains.
21.10/6.23	----------------------------------------
21.10/6.23	
21.10/6.23	(22) QDPOrderProof (EQUIVALENT)
21.10/6.23	We use the reduction pair processor [LPAR04,JAR06].
21.10/6.23	
21.10/6.23	
21.10/6.23	The following pairs can be oriented strictly and are deleted.
21.10/6.23	
21.10/6.23	   A_{B_1}(b_{c_1}(c_{c_1}(x1))) -> A_{B_1}(b_{c_1}(x1))
21.10/6.23	   A_{B_1}(b_{c_1}(c_{c_1}(x1))) -> A_{B_1}(b_{a_1}(a_{b_1}(b_{c_1}(x1))))
21.10/6.23	The remaining pairs can at least be oriented weakly.
21.10/6.23	Used ordering:  Polynomial Order [NEGPOLO,POLO] with Interpretation:
21.10/6.23	
21.10/6.23	POL( A_{B_1}_1(x_1) ) = max{0, 2x_1 - 1}
21.10/6.23	POL( b_{a_1}_1(x_1) ) = x_1
21.10/6.23	POL( a_{b_1}_1(x_1) ) = x_1
21.10/6.23	POL( b_{c_1}_1(x_1) ) = x_1 + 1
21.10/6.23	POL( c_{c_1}_1(x_1) ) = 2x_1 + 1
21.10/6.23	POL( a_{a_1}_1(x_1) ) = 2x_1
21.10/6.23	POL( b_{b_1}_1(x_1) ) = 0
21.10/6.23	POL( a_{c_1}_1(x_1) ) = 2x_1 + 2
21.10/6.23	POL( c_{a_1}_1(x_1) ) = 2
21.10/6.23	POL( c_{b_1}_1(x_1) ) = 0
21.10/6.23	
21.10/6.23	The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:
21.10/6.23	
21.10/6.23	   a_{b_1}(b_{c_1}(c_{c_1}(x1))) -> a_{a_1}(a_{b_1}(b_{a_1}(a_{b_1}(b_{c_1}(x1)))))
21.10/6.23	   b_{a_1}(a_{b_1}(b_{a_1}(x1))) -> b_{a_1}(x1)
21.10/6.23	   b_{a_1}(a_{b_1}(b_{b_1}(x1))) -> b_{b_1}(x1)
21.10/6.23	   b_{a_1}(a_{b_1}(b_{c_1}(x1))) -> b_{c_1}(x1)
21.10/6.23	   b_{a_1}(a_{c_1}(c_{a_1}(x1))) -> b_{c_1}(c_{c_1}(c_{a_1}(x1)))
21.10/6.23	   b_{a_1}(a_{c_1}(c_{b_1}(x1))) -> b_{c_1}(c_{c_1}(c_{b_1}(x1)))
21.10/6.23	   b_{a_1}(a_{c_1}(c_{c_1}(x1))) -> b_{c_1}(c_{c_1}(c_{c_1}(x1)))
21.10/6.23	   a_{a_1}(a_{b_1}(b_{a_1}(x1))) -> a_{a_1}(x1)
21.10/6.23	   a_{a_1}(a_{b_1}(b_{c_1}(x1))) -> a_{c_1}(x1)
21.10/6.23	   a_{a_1}(a_{c_1}(c_{a_1}(x1))) -> a_{c_1}(c_{c_1}(c_{a_1}(x1)))
21.10/6.23	   a_{a_1}(a_{c_1}(c_{b_1}(x1))) -> a_{c_1}(c_{c_1}(c_{b_1}(x1)))
21.10/6.23	   a_{a_1}(a_{c_1}(c_{c_1}(x1))) -> a_{c_1}(c_{c_1}(c_{c_1}(x1)))
21.10/6.23	   c_{a_1}(a_{b_1}(b_{a_1}(x1))) -> c_{a_1}(x1)
21.10/6.23	
21.10/6.23	
21.10/6.23	----------------------------------------
21.10/6.23	
21.10/6.23	(23)
21.10/6.23	Obligation:
21.10/6.23	Q DP problem:
21.10/6.23	P is empty.
21.10/6.23	The TRS R consists of the following rules:
21.10/6.23	
21.10/6.23	   a_{a_1}(a_{b_1}(b_{a_1}(x1))) -> a_{a_1}(x1)
21.10/6.23	   a_{a_1}(a_{b_1}(b_{c_1}(x1))) -> a_{c_1}(x1)
21.10/6.23	   b_{a_1}(a_{b_1}(b_{a_1}(x1))) -> b_{a_1}(x1)
21.10/6.23	   b_{a_1}(a_{b_1}(b_{b_1}(x1))) -> b_{b_1}(x1)
21.10/6.23	   b_{a_1}(a_{b_1}(b_{c_1}(x1))) -> b_{c_1}(x1)
21.10/6.23	   c_{a_1}(a_{b_1}(b_{a_1}(x1))) -> c_{a_1}(x1)
21.10/6.23	   a_{a_1}(a_{c_1}(c_{a_1}(x1))) -> a_{c_1}(c_{c_1}(c_{a_1}(x1)))
21.10/6.23	   a_{a_1}(a_{c_1}(c_{b_1}(x1))) -> a_{c_1}(c_{c_1}(c_{b_1}(x1)))
21.10/6.23	   a_{a_1}(a_{c_1}(c_{c_1}(x1))) -> a_{c_1}(c_{c_1}(c_{c_1}(x1)))
21.10/6.23	   b_{a_1}(a_{c_1}(c_{a_1}(x1))) -> b_{c_1}(c_{c_1}(c_{a_1}(x1)))
21.10/6.23	   b_{a_1}(a_{c_1}(c_{b_1}(x1))) -> b_{c_1}(c_{c_1}(c_{b_1}(x1)))
21.10/6.23	   b_{a_1}(a_{c_1}(c_{c_1}(x1))) -> b_{c_1}(c_{c_1}(c_{c_1}(x1)))
21.10/6.23	   a_{b_1}(b_{c_1}(c_{c_1}(x1))) -> a_{a_1}(a_{b_1}(b_{a_1}(a_{b_1}(b_{c_1}(x1)))))
21.10/6.23	
21.10/6.23	Q is empty.
21.10/6.23	We have to consider all minimal (P,Q,R)-chains.
21.10/6.23	----------------------------------------
21.10/6.23	
21.10/6.23	(24) PisEmptyProof (EQUIVALENT)
21.10/6.23	The TRS P is empty. Hence, there is no (P,Q,R) chain.
21.10/6.23	----------------------------------------
21.10/6.23	
21.10/6.23	(25)
21.10/6.23	YES
21.33/6.30	EOF