17.39/5.32	YES
17.39/5.33	proof of /export/starexec/sandbox/benchmark/theBenchmark.xml
17.39/5.33	# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty
17.39/5.33	
17.39/5.33	
17.39/5.33	Termination w.r.t. Q of the given QTRS could be proven:
17.39/5.33	
17.39/5.33	(0) QTRS
17.39/5.33	(1) DependencyPairsProof [EQUIVALENT, 4 ms]
17.39/5.33	(2) QDP
17.39/5.33	(3) DependencyGraphProof [EQUIVALENT, 2 ms]
17.39/5.33	(4) AND
17.39/5.33	    (5) QDP
17.39/5.33	        (6) UsableRulesProof [EQUIVALENT, 3 ms]
17.39/5.33	        (7) QDP
17.39/5.33	        (8) DependencyGraphProof [EQUIVALENT, 0 ms]
17.39/5.33	        (9) QDP
17.39/5.33	        (10) MRRProof [EQUIVALENT, 23 ms]
17.39/5.33	        (11) QDP
17.39/5.33	        (12) PisEmptyProof [EQUIVALENT, 0 ms]
17.39/5.33	        (13) YES
17.39/5.33	    (14) QDP
17.39/5.33	        (15) QDPOrderProof [EQUIVALENT, 10 ms]
17.39/5.33	        (16) QDP
17.39/5.33	        (17) PisEmptyProof [EQUIVALENT, 0 ms]
17.39/5.33	        (18) YES
17.39/5.33	
17.39/5.33	
17.39/5.33	----------------------------------------
17.39/5.33	
17.39/5.33	(0)
17.39/5.33	Obligation:
17.39/5.33	Q restricted rewrite system:
17.39/5.33	The TRS R consists of the following rules:
17.39/5.33	
17.39/5.33	   b(a(a(b(x1)))) -> b(b(b(b(x1))))
17.39/5.33	   b(b(b(a(x1)))) -> b(b(b(b(x1))))
17.39/5.33	   a(b(b(b(x1)))) -> b(a(a(a(x1))))
17.39/5.33	
17.39/5.33	Q is empty.
17.39/5.33	
17.39/5.33	----------------------------------------
17.39/5.33	
17.39/5.33	(1) DependencyPairsProof (EQUIVALENT)
17.39/5.33	Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.
17.39/5.33	----------------------------------------
17.39/5.33	
17.39/5.33	(2)
17.39/5.33	Obligation:
17.39/5.33	Q DP problem:
17.39/5.33	The TRS P consists of the following rules:
17.39/5.33	
17.39/5.33	   B(a(a(b(x1)))) -> B(b(b(b(x1))))
17.39/5.33	   B(a(a(b(x1)))) -> B(b(b(x1)))
17.39/5.33	   B(a(a(b(x1)))) -> B(b(x1))
17.39/5.33	   B(b(b(a(x1)))) -> B(b(b(b(x1))))
17.39/5.33	   B(b(b(a(x1)))) -> B(b(b(x1)))
17.39/5.33	   B(b(b(a(x1)))) -> B(b(x1))
17.39/5.33	   B(b(b(a(x1)))) -> B(x1)
17.39/5.33	   A(b(b(b(x1)))) -> B(a(a(a(x1))))
17.39/5.33	   A(b(b(b(x1)))) -> A(a(a(x1)))
17.39/5.33	   A(b(b(b(x1)))) -> A(a(x1))
17.39/5.33	   A(b(b(b(x1)))) -> A(x1)
17.39/5.33	
17.39/5.33	The TRS R consists of the following rules:
17.39/5.33	
17.39/5.33	   b(a(a(b(x1)))) -> b(b(b(b(x1))))
17.39/5.33	   b(b(b(a(x1)))) -> b(b(b(b(x1))))
17.39/5.33	   a(b(b(b(x1)))) -> b(a(a(a(x1))))
17.39/5.33	
17.39/5.33	Q is empty.
17.39/5.33	We have to consider all minimal (P,Q,R)-chains.
17.39/5.33	----------------------------------------
17.39/5.33	
17.39/5.33	(3) DependencyGraphProof (EQUIVALENT)
17.39/5.33	The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 2 SCCs with 1 less node.
17.39/5.33	----------------------------------------
17.39/5.33	
17.39/5.33	(4)
17.39/5.33	Complex Obligation (AND)
17.39/5.33	
17.39/5.33	----------------------------------------
17.39/5.33	
17.39/5.33	(5)
17.39/5.33	Obligation:
17.39/5.33	Q DP problem:
17.39/5.33	The TRS P consists of the following rules:
17.39/5.33	
17.39/5.33	   B(b(b(a(x1)))) -> B(b(b(b(x1))))
17.39/5.33	   B(b(b(a(x1)))) -> B(b(b(x1)))
17.39/5.33	   B(b(b(a(x1)))) -> B(b(x1))
17.39/5.33	   B(b(b(a(x1)))) -> B(x1)
17.39/5.33	   B(a(a(b(x1)))) -> B(b(b(b(x1))))
17.39/5.33	   B(a(a(b(x1)))) -> B(b(b(x1)))
17.39/5.33	   B(a(a(b(x1)))) -> B(b(x1))
17.39/5.33	
17.39/5.33	The TRS R consists of the following rules:
17.39/5.33	
17.39/5.33	   b(a(a(b(x1)))) -> b(b(b(b(x1))))
17.39/5.33	   b(b(b(a(x1)))) -> b(b(b(b(x1))))
17.39/5.33	   a(b(b(b(x1)))) -> b(a(a(a(x1))))
17.39/5.33	
17.39/5.33	Q is empty.
17.39/5.33	We have to consider all minimal (P,Q,R)-chains.
17.39/5.33	----------------------------------------
17.39/5.33	
17.39/5.33	(6) UsableRulesProof (EQUIVALENT)
17.39/5.33	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.
17.39/5.33	----------------------------------------
17.39/5.33	
17.39/5.33	(7)
17.39/5.33	Obligation:
17.39/5.33	Q DP problem:
17.39/5.33	The TRS P consists of the following rules:
17.39/5.33	
17.39/5.33	   B(b(b(a(x1)))) -> B(b(b(b(x1))))
17.39/5.33	   B(b(b(a(x1)))) -> B(b(b(x1)))
17.39/5.33	   B(b(b(a(x1)))) -> B(b(x1))
17.39/5.33	   B(b(b(a(x1)))) -> B(x1)
17.39/5.33	   B(a(a(b(x1)))) -> B(b(b(b(x1))))
17.39/5.33	   B(a(a(b(x1)))) -> B(b(b(x1)))
17.39/5.33	   B(a(a(b(x1)))) -> B(b(x1))
17.39/5.33	
17.39/5.33	The TRS R consists of the following rules:
17.39/5.33	
17.39/5.33	   b(b(b(a(x1)))) -> b(b(b(b(x1))))
17.39/5.33	   b(a(a(b(x1)))) -> b(b(b(b(x1))))
17.39/5.33	
17.39/5.33	Q is empty.
17.39/5.33	We have to consider all minimal (P,Q,R)-chains.
17.39/5.33	----------------------------------------
17.39/5.33	
17.39/5.33	(8) DependencyGraphProof (EQUIVALENT)
17.39/5.33	The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 2 less nodes.
17.39/5.33	----------------------------------------
17.39/5.33	
17.39/5.33	(9)
17.39/5.33	Obligation:
17.39/5.33	Q DP problem:
17.39/5.33	The TRS P consists of the following rules:
17.39/5.33	
17.39/5.33	   B(b(b(a(x1)))) -> B(b(x1))
17.39/5.33	   B(b(b(a(x1)))) -> B(b(b(x1)))
17.39/5.33	   B(b(b(a(x1)))) -> B(x1)
17.39/5.33	   B(a(a(b(x1)))) -> B(b(b(x1)))
17.39/5.33	   B(a(a(b(x1)))) -> B(b(x1))
17.39/5.33	
17.39/5.33	The TRS R consists of the following rules:
17.39/5.33	
17.39/5.33	   b(b(b(a(x1)))) -> b(b(b(b(x1))))
17.39/5.33	   b(a(a(b(x1)))) -> b(b(b(b(x1))))
17.39/5.33	
17.39/5.33	Q is empty.
17.39/5.33	We have to consider all minimal (P,Q,R)-chains.
17.39/5.33	----------------------------------------
17.39/5.33	
17.39/5.33	(10) MRRProof (EQUIVALENT)
17.39/5.33	By using the rule removal processor [LPAR04] with the following ordering, at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented.
17.39/5.33	
17.39/5.33	Strictly oriented dependency pairs:
17.39/5.33	
17.39/5.33	   B(b(b(a(x1)))) -> B(b(x1))
17.39/5.33	   B(b(b(a(x1)))) -> B(b(b(x1)))
17.39/5.33	   B(b(b(a(x1)))) -> B(x1)
17.39/5.33	   B(a(a(b(x1)))) -> B(b(b(x1)))
17.39/5.33	   B(a(a(b(x1)))) -> B(b(x1))
17.39/5.33	
17.39/5.33	Strictly oriented rules of the TRS R:
17.39/5.33	
17.39/5.33	   b(b(b(a(x1)))) -> b(b(b(b(x1))))
17.39/5.33	   b(a(a(b(x1)))) -> b(b(b(b(x1))))
17.39/5.33	
17.39/5.33	Used ordering: Polynomial interpretation [POLO]:
17.39/5.33	
17.39/5.33	   POL(B(x_1)) = 2*x_1
17.39/5.33	   POL(a(x_1)) = 2 + x_1
17.39/5.33	   POL(b(x_1)) = x_1
17.39/5.33	
17.39/5.33	
17.39/5.33	----------------------------------------
17.39/5.33	
17.39/5.33	(11)
17.39/5.33	Obligation:
17.39/5.33	Q DP problem:
17.39/5.33	P is empty.
17.39/5.33	R is empty.
17.39/5.33	Q is empty.
17.39/5.33	We have to consider all minimal (P,Q,R)-chains.
17.39/5.33	----------------------------------------
17.39/5.33	
17.39/5.33	(12) PisEmptyProof (EQUIVALENT)
17.39/5.33	The TRS P is empty. Hence, there is no (P,Q,R) chain.
17.39/5.33	----------------------------------------
17.39/5.33	
17.39/5.33	(13)
17.39/5.33	YES
17.39/5.33	
17.39/5.33	----------------------------------------
17.39/5.33	
17.39/5.33	(14)
17.39/5.33	Obligation:
17.39/5.33	Q DP problem:
17.39/5.33	The TRS P consists of the following rules:
17.39/5.33	
17.39/5.33	   A(b(b(b(x1)))) -> A(a(x1))
17.39/5.33	   A(b(b(b(x1)))) -> A(a(a(x1)))
17.39/5.33	   A(b(b(b(x1)))) -> A(x1)
17.39/5.33	
17.39/5.33	The TRS R consists of the following rules:
17.39/5.33	
17.39/5.33	   b(a(a(b(x1)))) -> b(b(b(b(x1))))
17.39/5.33	   b(b(b(a(x1)))) -> b(b(b(b(x1))))
17.39/5.33	   a(b(b(b(x1)))) -> b(a(a(a(x1))))
17.39/5.33	
17.39/5.33	Q is empty.
17.39/5.33	We have to consider all minimal (P,Q,R)-chains.
17.39/5.33	----------------------------------------
17.39/5.33	
17.39/5.33	(15) QDPOrderProof (EQUIVALENT)
17.39/5.33	We use the reduction pair processor [LPAR04,JAR06].
17.39/5.33	
17.39/5.33	
17.39/5.33	The following pairs can be oriented strictly and are deleted.
17.39/5.33	
17.39/5.33	   A(b(b(b(x1)))) -> A(a(x1))
17.39/5.33	   A(b(b(b(x1)))) -> A(a(a(x1)))
17.39/5.33	   A(b(b(b(x1)))) -> A(x1)
17.39/5.33	The remaining pairs can at least be oriented weakly.
17.39/5.33	Used ordering:  Polynomial interpretation [POLO]:
17.39/5.33	
17.39/5.33	   POL(A(x_1)) = x_1
17.39/5.33	   POL(a(x_1)) = 1 + x_1
17.39/5.33	   POL(b(x_1)) = 1 + x_1
17.39/5.33	
17.39/5.33	The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:
17.39/5.33	
17.39/5.33	   a(b(b(b(x1)))) -> b(a(a(a(x1))))
17.39/5.33	   b(b(b(a(x1)))) -> b(b(b(b(x1))))
17.39/5.33	   b(a(a(b(x1)))) -> b(b(b(b(x1))))
17.39/5.33	
17.39/5.33	
17.39/5.33	----------------------------------------
17.39/5.33	
17.39/5.33	(16)
17.39/5.33	Obligation:
17.39/5.33	Q DP problem:
17.39/5.33	P is empty.
17.39/5.33	The TRS R consists of the following rules:
17.39/5.33	
17.39/5.33	   b(a(a(b(x1)))) -> b(b(b(b(x1))))
17.39/5.33	   b(b(b(a(x1)))) -> b(b(b(b(x1))))
17.39/5.33	   a(b(b(b(x1)))) -> b(a(a(a(x1))))
17.39/5.33	
17.39/5.33	Q is empty.
17.39/5.33	We have to consider all minimal (P,Q,R)-chains.
17.39/5.33	----------------------------------------
17.39/5.33	
17.39/5.33	(17) PisEmptyProof (EQUIVALENT)
17.39/5.33	The TRS P is empty. Hence, there is no (P,Q,R) chain.
17.39/5.33	----------------------------------------
17.39/5.33	
17.39/5.33	(18)
17.39/5.33	YES
17.72/5.41	EOF