17.38/5.37	YES
19.93/6.05	proof of /export/starexec/sandbox2/benchmark/theBenchmark.xml
19.93/6.05	# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty
19.93/6.05	
19.93/6.05	
19.93/6.05	Termination w.r.t. Q of the given QTRS could be proven:
19.93/6.05	
19.93/6.05	(0) QTRS
19.93/6.05	(1) FlatCCProof [EQUIVALENT, 0 ms]
19.93/6.05	(2) QTRS
19.93/6.05	(3) RootLabelingProof [EQUIVALENT, 0 ms]
19.93/6.05	(4) QTRS
19.93/6.05	(5) QTRSRRRProof [EQUIVALENT, 32 ms]
19.93/6.05	(6) QTRS
19.93/6.05	(7) DependencyPairsProof [EQUIVALENT, 0 ms]
19.93/6.05	(8) QDP
19.93/6.05	(9) DependencyGraphProof [EQUIVALENT, 0 ms]
19.93/6.05	(10) QDP
19.93/6.05	(11) QDPOrderProof [EQUIVALENT, 114 ms]
19.93/6.05	(12) QDP
19.93/6.05	(13) DependencyGraphProof [EQUIVALENT, 0 ms]
19.93/6.05	(14) TRUE
19.93/6.05	
19.93/6.05	
19.93/6.05	----------------------------------------
19.93/6.05	
19.93/6.05	(0)
19.93/6.05	Obligation:
19.93/6.05	Q restricted rewrite system:
19.93/6.05	The TRS R consists of the following rules:
19.93/6.05	
19.93/6.05	   a(x1) -> x1
19.93/6.05	   a(a(x1)) -> a(b(a(c(a(x1)))))
19.93/6.05	   c(b(x1)) -> a(c(x1))
19.93/6.05	
19.93/6.05	Q is empty.
19.93/6.05	
19.93/6.05	----------------------------------------
19.93/6.05	
19.93/6.05	(1) FlatCCProof (EQUIVALENT)
19.93/6.05	We used flat context closure [ROOTLAB]
19.93/6.05	As Q is empty the flat context closure was sound AND complete.
19.93/6.05	
19.93/6.05	----------------------------------------
19.93/6.05	
19.93/6.05	(2)
19.93/6.05	Obligation:
19.93/6.05	Q restricted rewrite system:
19.93/6.05	The TRS R consists of the following rules:
19.93/6.05	
19.93/6.05	   a(a(x1)) -> a(b(a(c(a(x1)))))
19.93/6.05	   a(a(x1)) -> a(x1)
19.93/6.05	   b(a(x1)) -> b(x1)
19.93/6.05	   c(a(x1)) -> c(x1)
19.93/6.05	   a(c(b(x1))) -> a(a(c(x1)))
19.93/6.05	   b(c(b(x1))) -> b(a(c(x1)))
19.93/6.05	   c(c(b(x1))) -> c(a(c(x1)))
19.93/6.05	
19.93/6.05	Q is empty.
19.93/6.05	
19.93/6.05	----------------------------------------
19.93/6.05	
19.93/6.05	(3) RootLabelingProof (EQUIVALENT)
19.93/6.05	We used plain root labeling [ROOTLAB] with the following heuristic:
19.93/6.05	LabelAll: All function symbols get labeled
19.93/6.05	
19.93/6.05	As Q is empty the root labeling was sound AND complete.
19.93/6.05	
19.93/6.05	----------------------------------------
19.93/6.05	
19.93/6.05	(4)
19.93/6.05	Obligation:
19.93/6.05	Q restricted rewrite system:
19.93/6.05	The TRS R consists of the following rules:
19.93/6.05	
19.93/6.05	   a_{a_1}(a_{a_1}(x1)) -> a_{b_1}(b_{a_1}(a_{c_1}(c_{a_1}(a_{a_1}(x1)))))
19.93/6.05	   a_{a_1}(a_{b_1}(x1)) -> a_{b_1}(b_{a_1}(a_{c_1}(c_{a_1}(a_{b_1}(x1)))))
19.93/6.05	   a_{a_1}(a_{c_1}(x1)) -> a_{b_1}(b_{a_1}(a_{c_1}(c_{a_1}(a_{c_1}(x1)))))
19.93/6.05	   a_{a_1}(a_{a_1}(x1)) -> a_{a_1}(x1)
19.93/6.05	   a_{a_1}(a_{b_1}(x1)) -> a_{b_1}(x1)
19.93/6.05	   a_{a_1}(a_{c_1}(x1)) -> a_{c_1}(x1)
19.93/6.05	   b_{a_1}(a_{a_1}(x1)) -> b_{a_1}(x1)
19.93/6.05	   b_{a_1}(a_{b_1}(x1)) -> b_{b_1}(x1)
19.93/6.05	   b_{a_1}(a_{c_1}(x1)) -> b_{c_1}(x1)
19.93/6.05	   c_{a_1}(a_{a_1}(x1)) -> c_{a_1}(x1)
19.93/6.05	   c_{a_1}(a_{b_1}(x1)) -> c_{b_1}(x1)
19.93/6.05	   c_{a_1}(a_{c_1}(x1)) -> c_{c_1}(x1)
19.93/6.05	   a_{c_1}(c_{b_1}(b_{a_1}(x1))) -> a_{a_1}(a_{c_1}(c_{a_1}(x1)))
19.93/6.05	   a_{c_1}(c_{b_1}(b_{b_1}(x1))) -> a_{a_1}(a_{c_1}(c_{b_1}(x1)))
19.93/6.05	   a_{c_1}(c_{b_1}(b_{c_1}(x1))) -> a_{a_1}(a_{c_1}(c_{c_1}(x1)))
19.93/6.05	   b_{c_1}(c_{b_1}(b_{a_1}(x1))) -> b_{a_1}(a_{c_1}(c_{a_1}(x1)))
19.93/6.05	   b_{c_1}(c_{b_1}(b_{b_1}(x1))) -> b_{a_1}(a_{c_1}(c_{b_1}(x1)))
19.93/6.05	   b_{c_1}(c_{b_1}(b_{c_1}(x1))) -> b_{a_1}(a_{c_1}(c_{c_1}(x1)))
19.93/6.05	   c_{c_1}(c_{b_1}(b_{a_1}(x1))) -> c_{a_1}(a_{c_1}(c_{a_1}(x1)))
19.93/6.05	   c_{c_1}(c_{b_1}(b_{b_1}(x1))) -> c_{a_1}(a_{c_1}(c_{b_1}(x1)))
19.93/6.05	   c_{c_1}(c_{b_1}(b_{c_1}(x1))) -> c_{a_1}(a_{c_1}(c_{c_1}(x1)))
19.93/6.05	
19.93/6.05	Q is empty.
19.93/6.05	
19.93/6.05	----------------------------------------
19.93/6.05	
19.93/6.05	(5) QTRSRRRProof (EQUIVALENT)
19.93/6.05	Used ordering:
19.93/6.05	Polynomial interpretation [POLO]:
19.93/6.05	
19.93/6.05	   POL(a_{a_1}(x_1)) = 1 + x_1
19.93/6.05	   POL(a_{b_1}(x_1)) = 1 + x_1
19.93/6.05	   POL(a_{c_1}(x_1)) = x_1
19.93/6.05	   POL(b_{a_1}(x_1)) = x_1
19.93/6.05	   POL(b_{b_1}(x_1)) = 1 + x_1
19.93/6.05	   POL(b_{c_1}(x_1)) = x_1
19.93/6.05	   POL(c_{a_1}(x_1)) = x_1
19.93/6.05	   POL(c_{b_1}(x_1)) = 1 + x_1
19.93/6.05	   POL(c_{c_1}(x_1)) = x_1
19.93/6.05	With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:
19.93/6.05	
19.93/6.05	   a_{a_1}(a_{a_1}(x1)) -> a_{a_1}(x1)
19.93/6.05	   a_{a_1}(a_{b_1}(x1)) -> a_{b_1}(x1)
19.93/6.05	   a_{a_1}(a_{c_1}(x1)) -> a_{c_1}(x1)
19.93/6.05	   b_{a_1}(a_{a_1}(x1)) -> b_{a_1}(x1)
19.93/6.05	   c_{a_1}(a_{a_1}(x1)) -> c_{a_1}(x1)
19.93/6.05	   b_{c_1}(c_{b_1}(b_{a_1}(x1))) -> b_{a_1}(a_{c_1}(c_{a_1}(x1)))
19.93/6.05	   b_{c_1}(c_{b_1}(b_{b_1}(x1))) -> b_{a_1}(a_{c_1}(c_{b_1}(x1)))
19.93/6.05	   b_{c_1}(c_{b_1}(b_{c_1}(x1))) -> b_{a_1}(a_{c_1}(c_{c_1}(x1)))
19.93/6.05	   c_{c_1}(c_{b_1}(b_{a_1}(x1))) -> c_{a_1}(a_{c_1}(c_{a_1}(x1)))
19.93/6.05	   c_{c_1}(c_{b_1}(b_{b_1}(x1))) -> c_{a_1}(a_{c_1}(c_{b_1}(x1)))
19.93/6.05	   c_{c_1}(c_{b_1}(b_{c_1}(x1))) -> c_{a_1}(a_{c_1}(c_{c_1}(x1)))
19.93/6.05	
19.93/6.05	
19.93/6.05	
19.93/6.05	
19.93/6.05	----------------------------------------
19.93/6.05	
19.93/6.05	(6)
19.93/6.05	Obligation:
19.93/6.05	Q restricted rewrite system:
19.93/6.05	The TRS R consists of the following rules:
19.93/6.05	
19.93/6.05	   a_{a_1}(a_{a_1}(x1)) -> a_{b_1}(b_{a_1}(a_{c_1}(c_{a_1}(a_{a_1}(x1)))))
19.93/6.05	   a_{a_1}(a_{b_1}(x1)) -> a_{b_1}(b_{a_1}(a_{c_1}(c_{a_1}(a_{b_1}(x1)))))
19.93/6.05	   a_{a_1}(a_{c_1}(x1)) -> a_{b_1}(b_{a_1}(a_{c_1}(c_{a_1}(a_{c_1}(x1)))))
19.93/6.05	   b_{a_1}(a_{b_1}(x1)) -> b_{b_1}(x1)
19.93/6.05	   b_{a_1}(a_{c_1}(x1)) -> b_{c_1}(x1)
19.93/6.05	   c_{a_1}(a_{b_1}(x1)) -> c_{b_1}(x1)
19.93/6.05	   c_{a_1}(a_{c_1}(x1)) -> c_{c_1}(x1)
19.93/6.05	   a_{c_1}(c_{b_1}(b_{a_1}(x1))) -> a_{a_1}(a_{c_1}(c_{a_1}(x1)))
19.93/6.05	   a_{c_1}(c_{b_1}(b_{b_1}(x1))) -> a_{a_1}(a_{c_1}(c_{b_1}(x1)))
19.93/6.05	   a_{c_1}(c_{b_1}(b_{c_1}(x1))) -> a_{a_1}(a_{c_1}(c_{c_1}(x1)))
19.93/6.05	
19.93/6.05	Q is empty.
19.93/6.05	
19.93/6.05	----------------------------------------
19.93/6.05	
19.93/6.05	(7) DependencyPairsProof (EQUIVALENT)
19.93/6.05	Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.
19.93/6.05	----------------------------------------
19.93/6.05	
19.93/6.05	(8)
19.93/6.05	Obligation:
19.93/6.05	Q DP problem:
19.93/6.05	The TRS P consists of the following rules:
19.93/6.05	
19.93/6.05	   A_{A_1}(a_{a_1}(x1)) -> B_{A_1}(a_{c_1}(c_{a_1}(a_{a_1}(x1))))
19.93/6.05	   A_{A_1}(a_{a_1}(x1)) -> A_{C_1}(c_{a_1}(a_{a_1}(x1)))
19.93/6.05	   A_{A_1}(a_{a_1}(x1)) -> C_{A_1}(a_{a_1}(x1))
19.93/6.05	   A_{A_1}(a_{b_1}(x1)) -> B_{A_1}(a_{c_1}(c_{a_1}(a_{b_1}(x1))))
19.93/6.05	   A_{A_1}(a_{b_1}(x1)) -> A_{C_1}(c_{a_1}(a_{b_1}(x1)))
19.93/6.05	   A_{A_1}(a_{b_1}(x1)) -> C_{A_1}(a_{b_1}(x1))
19.93/6.05	   A_{A_1}(a_{c_1}(x1)) -> B_{A_1}(a_{c_1}(c_{a_1}(a_{c_1}(x1))))
19.93/6.05	   A_{A_1}(a_{c_1}(x1)) -> A_{C_1}(c_{a_1}(a_{c_1}(x1)))
19.93/6.05	   A_{A_1}(a_{c_1}(x1)) -> C_{A_1}(a_{c_1}(x1))
19.93/6.05	   A_{C_1}(c_{b_1}(b_{a_1}(x1))) -> A_{A_1}(a_{c_1}(c_{a_1}(x1)))
19.93/6.05	   A_{C_1}(c_{b_1}(b_{a_1}(x1))) -> A_{C_1}(c_{a_1}(x1))
19.93/6.05	   A_{C_1}(c_{b_1}(b_{a_1}(x1))) -> C_{A_1}(x1)
19.93/6.05	   A_{C_1}(c_{b_1}(b_{b_1}(x1))) -> A_{A_1}(a_{c_1}(c_{b_1}(x1)))
19.93/6.05	   A_{C_1}(c_{b_1}(b_{b_1}(x1))) -> A_{C_1}(c_{b_1}(x1))
19.93/6.05	   A_{C_1}(c_{b_1}(b_{c_1}(x1))) -> A_{A_1}(a_{c_1}(c_{c_1}(x1)))
19.93/6.05	   A_{C_1}(c_{b_1}(b_{c_1}(x1))) -> A_{C_1}(c_{c_1}(x1))
19.93/6.05	
19.93/6.05	The TRS R consists of the following rules:
19.93/6.05	
19.93/6.05	   a_{a_1}(a_{a_1}(x1)) -> a_{b_1}(b_{a_1}(a_{c_1}(c_{a_1}(a_{a_1}(x1)))))
19.93/6.05	   a_{a_1}(a_{b_1}(x1)) -> a_{b_1}(b_{a_1}(a_{c_1}(c_{a_1}(a_{b_1}(x1)))))
19.93/6.05	   a_{a_1}(a_{c_1}(x1)) -> a_{b_1}(b_{a_1}(a_{c_1}(c_{a_1}(a_{c_1}(x1)))))
19.93/6.05	   b_{a_1}(a_{b_1}(x1)) -> b_{b_1}(x1)
19.93/6.05	   b_{a_1}(a_{c_1}(x1)) -> b_{c_1}(x1)
19.93/6.05	   c_{a_1}(a_{b_1}(x1)) -> c_{b_1}(x1)
19.93/6.05	   c_{a_1}(a_{c_1}(x1)) -> c_{c_1}(x1)
19.93/6.05	   a_{c_1}(c_{b_1}(b_{a_1}(x1))) -> a_{a_1}(a_{c_1}(c_{a_1}(x1)))
19.93/6.05	   a_{c_1}(c_{b_1}(b_{b_1}(x1))) -> a_{a_1}(a_{c_1}(c_{b_1}(x1)))
19.93/6.05	   a_{c_1}(c_{b_1}(b_{c_1}(x1))) -> a_{a_1}(a_{c_1}(c_{c_1}(x1)))
19.93/6.05	
19.93/6.05	Q is empty.
19.93/6.05	We have to consider all minimal (P,Q,R)-chains.
19.93/6.05	----------------------------------------
19.93/6.05	
19.93/6.05	(9) DependencyGraphProof (EQUIVALENT)
19.93/6.05	The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 8 less nodes.
19.93/6.05	----------------------------------------
19.93/6.05	
19.93/6.05	(10)
19.93/6.05	Obligation:
19.93/6.05	Q DP problem:
19.93/6.05	The TRS P consists of the following rules:
19.93/6.05	
19.93/6.05	   A_{A_1}(a_{a_1}(x1)) -> A_{C_1}(c_{a_1}(a_{a_1}(x1)))
19.93/6.05	   A_{C_1}(c_{b_1}(b_{a_1}(x1))) -> A_{A_1}(a_{c_1}(c_{a_1}(x1)))
19.93/6.05	   A_{A_1}(a_{b_1}(x1)) -> A_{C_1}(c_{a_1}(a_{b_1}(x1)))
19.93/6.05	   A_{C_1}(c_{b_1}(b_{a_1}(x1))) -> A_{C_1}(c_{a_1}(x1))
19.93/6.05	   A_{C_1}(c_{b_1}(b_{b_1}(x1))) -> A_{A_1}(a_{c_1}(c_{b_1}(x1)))
19.93/6.05	   A_{A_1}(a_{c_1}(x1)) -> A_{C_1}(c_{a_1}(a_{c_1}(x1)))
19.93/6.05	   A_{C_1}(c_{b_1}(b_{b_1}(x1))) -> A_{C_1}(c_{b_1}(x1))
19.93/6.05	   A_{C_1}(c_{b_1}(b_{c_1}(x1))) -> A_{A_1}(a_{c_1}(c_{c_1}(x1)))
19.93/6.05	
19.93/6.05	The TRS R consists of the following rules:
19.93/6.05	
19.93/6.05	   a_{a_1}(a_{a_1}(x1)) -> a_{b_1}(b_{a_1}(a_{c_1}(c_{a_1}(a_{a_1}(x1)))))
19.93/6.05	   a_{a_1}(a_{b_1}(x1)) -> a_{b_1}(b_{a_1}(a_{c_1}(c_{a_1}(a_{b_1}(x1)))))
19.93/6.05	   a_{a_1}(a_{c_1}(x1)) -> a_{b_1}(b_{a_1}(a_{c_1}(c_{a_1}(a_{c_1}(x1)))))
19.93/6.05	   b_{a_1}(a_{b_1}(x1)) -> b_{b_1}(x1)
19.93/6.05	   b_{a_1}(a_{c_1}(x1)) -> b_{c_1}(x1)
19.93/6.05	   c_{a_1}(a_{b_1}(x1)) -> c_{b_1}(x1)
19.93/6.05	   c_{a_1}(a_{c_1}(x1)) -> c_{c_1}(x1)
19.93/6.05	   a_{c_1}(c_{b_1}(b_{a_1}(x1))) -> a_{a_1}(a_{c_1}(c_{a_1}(x1)))
19.93/6.05	   a_{c_1}(c_{b_1}(b_{b_1}(x1))) -> a_{a_1}(a_{c_1}(c_{b_1}(x1)))
19.93/6.05	   a_{c_1}(c_{b_1}(b_{c_1}(x1))) -> a_{a_1}(a_{c_1}(c_{c_1}(x1)))
19.93/6.05	
19.93/6.05	Q is empty.
19.93/6.05	We have to consider all minimal (P,Q,R)-chains.
19.93/6.05	----------------------------------------
19.93/6.05	
19.93/6.05	(11) QDPOrderProof (EQUIVALENT)
19.93/6.05	We use the reduction pair processor [LPAR04,JAR06].
19.93/6.05	
19.93/6.05	
19.93/6.05	The following pairs can be oriented strictly and are deleted.
19.93/6.05	
19.93/6.05	   A_{A_1}(a_{a_1}(x1)) -> A_{C_1}(c_{a_1}(a_{a_1}(x1)))
19.93/6.05	   A_{A_1}(a_{b_1}(x1)) -> A_{C_1}(c_{a_1}(a_{b_1}(x1)))
19.93/6.05	   A_{C_1}(c_{b_1}(b_{a_1}(x1))) -> A_{C_1}(c_{a_1}(x1))
19.93/6.05	   A_{A_1}(a_{c_1}(x1)) -> A_{C_1}(c_{a_1}(a_{c_1}(x1)))
19.93/6.05	   A_{C_1}(c_{b_1}(b_{b_1}(x1))) -> A_{C_1}(c_{b_1}(x1))
19.93/6.05	The remaining pairs can at least be oriented weakly.
19.93/6.05	Used ordering:  Polynomial Order [NEGPOLO,POLO] with Interpretation:
19.93/6.05	
19.93/6.05	POL( A_{A_1}_1(x_1) ) = 2x_1 + 2
19.93/6.05	POL( A_{C_1}_1(x_1) ) = max{0, 2x_1 - 2}
19.93/6.05	POL( c_{a_1}_1(x_1) ) = x_1
19.93/6.05	POL( a_{a_1}_1(x_1) ) = 2x_1 + 2
19.93/6.05	POL( a_{b_1}_1(x_1) ) = x_1
19.93/6.05	POL( b_{a_1}_1(x_1) ) = 2x_1 + 2
19.93/6.05	POL( a_{c_1}_1(x_1) ) = x_1
19.93/6.05	POL( c_{b_1}_1(x_1) ) = x_1
19.93/6.05	POL( c_{c_1}_1(x_1) ) = 0
19.93/6.05	POL( b_{b_1}_1(x_1) ) = 2x_1 + 2
19.93/6.05	POL( b_{c_1}_1(x_1) ) = 2
19.93/6.05	
19.93/6.05	The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:
19.93/6.05	
19.93/6.05	   a_{a_1}(a_{a_1}(x1)) -> a_{b_1}(b_{a_1}(a_{c_1}(c_{a_1}(a_{a_1}(x1)))))
19.93/6.05	   a_{a_1}(a_{b_1}(x1)) -> a_{b_1}(b_{a_1}(a_{c_1}(c_{a_1}(a_{b_1}(x1)))))
19.93/6.05	   a_{a_1}(a_{c_1}(x1)) -> a_{b_1}(b_{a_1}(a_{c_1}(c_{a_1}(a_{c_1}(x1)))))
19.93/6.05	   c_{a_1}(a_{b_1}(x1)) -> c_{b_1}(x1)
19.93/6.05	   c_{a_1}(a_{c_1}(x1)) -> c_{c_1}(x1)
19.93/6.05	   a_{c_1}(c_{b_1}(b_{a_1}(x1))) -> a_{a_1}(a_{c_1}(c_{a_1}(x1)))
19.93/6.05	   a_{c_1}(c_{b_1}(b_{b_1}(x1))) -> a_{a_1}(a_{c_1}(c_{b_1}(x1)))
19.93/6.05	   a_{c_1}(c_{b_1}(b_{c_1}(x1))) -> a_{a_1}(a_{c_1}(c_{c_1}(x1)))
19.93/6.05	   b_{a_1}(a_{b_1}(x1)) -> b_{b_1}(x1)
19.93/6.05	   b_{a_1}(a_{c_1}(x1)) -> b_{c_1}(x1)
19.93/6.05	
19.93/6.05	
19.93/6.05	----------------------------------------
19.93/6.05	
19.93/6.05	(12)
19.93/6.05	Obligation:
19.93/6.05	Q DP problem:
19.93/6.05	The TRS P consists of the following rules:
19.93/6.05	
19.93/6.05	   A_{C_1}(c_{b_1}(b_{a_1}(x1))) -> A_{A_1}(a_{c_1}(c_{a_1}(x1)))
19.93/6.05	   A_{C_1}(c_{b_1}(b_{b_1}(x1))) -> A_{A_1}(a_{c_1}(c_{b_1}(x1)))
19.93/6.05	   A_{C_1}(c_{b_1}(b_{c_1}(x1))) -> A_{A_1}(a_{c_1}(c_{c_1}(x1)))
19.93/6.05	
19.93/6.05	The TRS R consists of the following rules:
19.93/6.05	
19.93/6.05	   a_{a_1}(a_{a_1}(x1)) -> a_{b_1}(b_{a_1}(a_{c_1}(c_{a_1}(a_{a_1}(x1)))))
19.93/6.05	   a_{a_1}(a_{b_1}(x1)) -> a_{b_1}(b_{a_1}(a_{c_1}(c_{a_1}(a_{b_1}(x1)))))
19.93/6.05	   a_{a_1}(a_{c_1}(x1)) -> a_{b_1}(b_{a_1}(a_{c_1}(c_{a_1}(a_{c_1}(x1)))))
19.93/6.05	   b_{a_1}(a_{b_1}(x1)) -> b_{b_1}(x1)
19.93/6.05	   b_{a_1}(a_{c_1}(x1)) -> b_{c_1}(x1)
19.93/6.05	   c_{a_1}(a_{b_1}(x1)) -> c_{b_1}(x1)
19.93/6.05	   c_{a_1}(a_{c_1}(x1)) -> c_{c_1}(x1)
19.93/6.05	   a_{c_1}(c_{b_1}(b_{a_1}(x1))) -> a_{a_1}(a_{c_1}(c_{a_1}(x1)))
19.93/6.05	   a_{c_1}(c_{b_1}(b_{b_1}(x1))) -> a_{a_1}(a_{c_1}(c_{b_1}(x1)))
19.93/6.05	   a_{c_1}(c_{b_1}(b_{c_1}(x1))) -> a_{a_1}(a_{c_1}(c_{c_1}(x1)))
19.93/6.05	
19.93/6.05	Q is empty.
19.93/6.05	We have to consider all minimal (P,Q,R)-chains.
19.93/6.05	----------------------------------------
19.93/6.05	
19.93/6.05	(13) DependencyGraphProof (EQUIVALENT)
19.93/6.05	The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 3 less nodes.
19.93/6.05	----------------------------------------
19.93/6.05	
19.93/6.05	(14)
19.93/6.05	TRUE
20.47/6.17	EOF