16.09/4.98	YES
16.09/4.99	proof of /export/starexec/sandbox/benchmark/theBenchmark.xml
16.09/4.99	# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty
16.09/4.99	
16.09/4.99	
16.09/4.99	Termination w.r.t. Q of the given QTRS could be proven:
16.09/4.99	
16.09/4.99	(0) QTRS
16.09/4.99	(1) DependencyPairsProof [EQUIVALENT, 18 ms]
16.09/4.99	(2) QDP
16.09/4.99	(3) DependencyGraphProof [EQUIVALENT, 0 ms]
16.09/4.99	(4) AND
16.09/4.99	    (5) QDP
16.09/4.99	        (6) UsableRulesProof [EQUIVALENT, 0 ms]
16.09/4.99	        (7) QDP
16.09/4.99	        (8) MRRProof [EQUIVALENT, 42 ms]
16.09/4.99	        (9) QDP
16.09/4.99	        (10) DependencyGraphProof [EQUIVALENT, 0 ms]
16.09/4.99	        (11) TRUE
16.09/4.99	    (12) QDP
16.09/4.99	        (13) MNOCProof [EQUIVALENT, 0 ms]
16.09/4.99	        (14) QDP
16.09/4.99	        (15) UsableRulesProof [EQUIVALENT, 0 ms]
16.09/4.99	        (16) QDP
16.09/4.99	        (17) QReductionProof [EQUIVALENT, 0 ms]
16.09/4.99	        (18) QDP
16.09/4.99	        (19) QDPOrderProof [EQUIVALENT, 33 ms]
16.09/4.99	        (20) QDP
16.09/4.99	        (21) PisEmptyProof [EQUIVALENT, 0 ms]
16.09/4.99	        (22) YES
16.09/4.99	
16.09/4.99	
16.09/4.99	----------------------------------------
16.09/4.99	
16.09/4.99	(0)
16.09/4.99	Obligation:
16.09/4.99	Q restricted rewrite system:
16.09/4.99	The TRS R consists of the following rules:
16.09/4.99	
16.09/4.99	   b(a(a(x1))) -> a(b(c(x1)))
16.09/4.99	   c(a(x1)) -> a(c(x1))
16.09/4.99	   c(b(x1)) -> b(a(x1))
16.09/4.99	   L(a(a(x1))) -> L(a(b(c(x1))))
16.09/4.99	   c(R(x1)) -> b(a(R(x1)))
16.09/4.99	
16.09/4.99	Q is empty.
16.09/4.99	
16.09/4.99	----------------------------------------
16.09/4.99	
16.09/4.99	(1) DependencyPairsProof (EQUIVALENT)
16.09/4.99	Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.
16.09/4.99	----------------------------------------
16.09/4.99	
16.09/4.99	(2)
16.09/4.99	Obligation:
16.09/4.99	Q DP problem:
16.09/4.99	The TRS P consists of the following rules:
16.09/4.99	
16.09/4.99	   B(a(a(x1))) -> B(c(x1))
16.09/4.99	   B(a(a(x1))) -> C(x1)
16.09/4.99	   C(a(x1)) -> C(x1)
16.09/4.99	   C(b(x1)) -> B(a(x1))
16.09/4.99	   L^1(a(a(x1))) -> L^1(a(b(c(x1))))
16.09/4.99	   L^1(a(a(x1))) -> B(c(x1))
16.09/4.99	   L^1(a(a(x1))) -> C(x1)
16.09/4.99	   C(R(x1)) -> B(a(R(x1)))
16.09/4.99	
16.09/4.99	The TRS R consists of the following rules:
16.09/4.99	
16.09/4.99	   b(a(a(x1))) -> a(b(c(x1)))
16.09/4.99	   c(a(x1)) -> a(c(x1))
16.09/4.99	   c(b(x1)) -> b(a(x1))
16.09/4.99	   L(a(a(x1))) -> L(a(b(c(x1))))
16.09/4.99	   c(R(x1)) -> b(a(R(x1)))
16.09/4.99	
16.09/4.99	Q is empty.
16.09/4.99	We have to consider all minimal (P,Q,R)-chains.
16.09/4.99	----------------------------------------
16.09/4.99	
16.09/4.99	(3) DependencyGraphProof (EQUIVALENT)
16.09/4.99	The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 2 SCCs with 3 less nodes.
16.09/4.99	----------------------------------------
16.09/4.99	
16.09/4.99	(4)
16.09/4.99	Complex Obligation (AND)
16.09/4.99	
16.09/4.99	----------------------------------------
16.09/4.99	
16.09/4.99	(5)
16.09/4.99	Obligation:
16.09/4.99	Q DP problem:
16.09/4.99	The TRS P consists of the following rules:
16.09/4.99	
16.09/4.99	   B(a(a(x1))) -> C(x1)
16.09/4.99	   C(a(x1)) -> C(x1)
16.09/4.99	   C(b(x1)) -> B(a(x1))
16.09/4.99	   B(a(a(x1))) -> B(c(x1))
16.09/4.99	
16.09/4.99	The TRS R consists of the following rules:
16.09/4.99	
16.09/4.99	   b(a(a(x1))) -> a(b(c(x1)))
16.09/4.99	   c(a(x1)) -> a(c(x1))
16.09/4.99	   c(b(x1)) -> b(a(x1))
16.09/4.99	   L(a(a(x1))) -> L(a(b(c(x1))))
16.09/4.99	   c(R(x1)) -> b(a(R(x1)))
16.09/4.99	
16.09/4.99	Q is empty.
16.09/4.99	We have to consider all minimal (P,Q,R)-chains.
16.09/4.99	----------------------------------------
16.09/4.99	
16.09/4.99	(6) UsableRulesProof (EQUIVALENT)
16.09/4.99	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.
16.09/4.99	----------------------------------------
16.09/4.99	
16.09/4.99	(7)
16.09/4.99	Obligation:
16.09/4.99	Q DP problem:
16.09/4.99	The TRS P consists of the following rules:
16.09/4.99	
16.09/4.99	   B(a(a(x1))) -> C(x1)
16.09/4.99	   C(a(x1)) -> C(x1)
16.09/4.99	   C(b(x1)) -> B(a(x1))
16.09/4.99	   B(a(a(x1))) -> B(c(x1))
16.09/4.99	
16.09/4.99	The TRS R consists of the following rules:
16.09/4.99	
16.09/4.99	   c(a(x1)) -> a(c(x1))
16.09/4.99	   c(b(x1)) -> b(a(x1))
16.09/4.99	   c(R(x1)) -> b(a(R(x1)))
16.09/4.99	   b(a(a(x1))) -> a(b(c(x1)))
16.09/4.99	
16.09/4.99	Q is empty.
16.09/4.99	We have to consider all minimal (P,Q,R)-chains.
16.09/4.99	----------------------------------------
16.09/4.99	
16.09/4.99	(8) MRRProof (EQUIVALENT)
16.09/4.99	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.
16.09/4.99	
16.09/4.99	Strictly oriented dependency pairs:
16.09/4.99	
16.09/4.99	   B(a(a(x1))) -> C(x1)
16.09/4.99	   C(a(x1)) -> C(x1)
16.09/4.99	   B(a(a(x1))) -> B(c(x1))
16.09/4.99	
16.09/4.99	
16.09/4.99	Used ordering: Polynomial interpretation [POLO]:
16.09/4.99	
16.09/4.99	   POL(B(x_1)) = x_1
16.09/4.99	   POL(C(x_1)) = 2 + 2*x_1
16.09/4.99	   POL(R(x_1)) = x_1
16.09/4.99	   POL(a(x_1)) = 2 + 2*x_1
16.09/4.99	   POL(b(x_1)) = x_1
16.09/4.99	   POL(c(x_1)) = 2 + 2*x_1
16.09/4.99	
16.09/4.99	
16.09/4.99	----------------------------------------
16.09/4.99	
16.09/4.99	(9)
16.09/4.99	Obligation:
16.09/4.99	Q DP problem:
16.09/4.99	The TRS P consists of the following rules:
16.09/4.99	
16.09/4.99	   C(b(x1)) -> B(a(x1))
16.09/4.99	
16.09/4.99	The TRS R consists of the following rules:
16.09/4.99	
16.09/4.99	   c(a(x1)) -> a(c(x1))
16.09/4.99	   c(b(x1)) -> b(a(x1))
16.09/4.99	   c(R(x1)) -> b(a(R(x1)))
16.09/4.99	   b(a(a(x1))) -> a(b(c(x1)))
16.09/4.99	
16.09/4.99	Q is empty.
16.09/4.99	We have to consider all minimal (P,Q,R)-chains.
16.09/4.99	----------------------------------------
16.09/4.99	
16.09/4.99	(10) DependencyGraphProof (EQUIVALENT)
16.09/4.99	The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 1 less node.
16.09/4.99	----------------------------------------
16.09/4.99	
16.09/4.99	(11)
16.09/4.99	TRUE
16.09/4.99	
16.09/4.99	----------------------------------------
16.09/4.99	
16.09/4.99	(12)
16.09/4.99	Obligation:
16.09/4.99	Q DP problem:
16.09/4.99	The TRS P consists of the following rules:
16.09/4.99	
16.09/4.99	   L^1(a(a(x1))) -> L^1(a(b(c(x1))))
16.09/4.99	
16.09/4.99	The TRS R consists of the following rules:
16.09/4.99	
16.09/4.99	   b(a(a(x1))) -> a(b(c(x1)))
16.09/4.99	   c(a(x1)) -> a(c(x1))
16.09/4.99	   c(b(x1)) -> b(a(x1))
16.09/4.99	   L(a(a(x1))) -> L(a(b(c(x1))))
16.09/4.99	   c(R(x1)) -> b(a(R(x1)))
16.09/4.99	
16.09/4.99	Q is empty.
16.09/4.99	We have to consider all minimal (P,Q,R)-chains.
16.09/4.99	----------------------------------------
16.09/4.99	
16.09/4.99	(13) MNOCProof (EQUIVALENT)
16.09/4.99	We use the modular non-overlap check [LPAR04] to enlarge Q to all left-hand sides of R.
16.09/4.99	----------------------------------------
16.09/4.99	
16.09/4.99	(14)
16.09/4.99	Obligation:
16.09/4.99	Q DP problem:
16.09/4.99	The TRS P consists of the following rules:
16.09/4.99	
16.09/4.99	   L^1(a(a(x1))) -> L^1(a(b(c(x1))))
16.09/4.99	
16.09/4.99	The TRS R consists of the following rules:
16.09/4.99	
16.09/4.99	   b(a(a(x1))) -> a(b(c(x1)))
16.09/4.99	   c(a(x1)) -> a(c(x1))
16.09/4.99	   c(b(x1)) -> b(a(x1))
16.09/4.99	   L(a(a(x1))) -> L(a(b(c(x1))))
16.09/4.99	   c(R(x1)) -> b(a(R(x1)))
16.09/4.99	
16.09/4.99	The set Q consists of the following terms:
16.09/4.99	
16.09/4.99	   b(a(a(x0)))
16.09/4.99	   c(a(x0))
16.09/4.99	   c(b(x0))
16.09/4.99	   L(a(a(x0)))
16.09/4.99	   c(R(x0))
16.09/4.99	
16.09/4.99	We have to consider all minimal (P,Q,R)-chains.
16.09/4.99	----------------------------------------
16.09/4.99	
16.09/4.99	(15) UsableRulesProof (EQUIVALENT)
16.09/4.99	As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.
16.09/4.99	----------------------------------------
16.09/4.99	
16.09/4.99	(16)
16.09/4.99	Obligation:
16.09/4.99	Q DP problem:
16.09/4.99	The TRS P consists of the following rules:
16.09/4.99	
16.09/4.99	   L^1(a(a(x1))) -> L^1(a(b(c(x1))))
16.09/4.99	
16.09/4.99	The TRS R consists of the following rules:
16.09/4.99	
16.09/4.99	   c(a(x1)) -> a(c(x1))
16.09/4.99	   c(b(x1)) -> b(a(x1))
16.09/4.99	   c(R(x1)) -> b(a(R(x1)))
16.09/4.99	   b(a(a(x1))) -> a(b(c(x1)))
16.09/4.99	
16.09/4.99	The set Q consists of the following terms:
16.09/4.99	
16.09/4.99	   b(a(a(x0)))
16.09/4.99	   c(a(x0))
16.09/4.99	   c(b(x0))
16.09/4.99	   L(a(a(x0)))
16.09/4.99	   c(R(x0))
16.09/4.99	
16.09/4.99	We have to consider all minimal (P,Q,R)-chains.
16.09/4.99	----------------------------------------
16.09/4.99	
16.09/4.99	(17) QReductionProof (EQUIVALENT)
16.09/4.99	We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].
16.09/4.99	
16.09/4.99	   L(a(a(x0)))
16.09/4.99	
16.09/4.99	
16.09/4.99	----------------------------------------
16.09/4.99	
16.09/4.99	(18)
16.09/4.99	Obligation:
16.09/4.99	Q DP problem:
16.09/4.99	The TRS P consists of the following rules:
16.09/4.99	
16.09/4.99	   L^1(a(a(x1))) -> L^1(a(b(c(x1))))
16.09/4.99	
16.09/4.99	The TRS R consists of the following rules:
16.09/4.99	
16.09/4.99	   c(a(x1)) -> a(c(x1))
16.09/4.99	   c(b(x1)) -> b(a(x1))
16.09/4.99	   c(R(x1)) -> b(a(R(x1)))
16.09/4.99	   b(a(a(x1))) -> a(b(c(x1)))
16.09/4.99	
16.09/4.99	The set Q consists of the following terms:
16.09/4.99	
16.09/4.99	   b(a(a(x0)))
16.09/4.99	   c(a(x0))
16.09/4.99	   c(b(x0))
16.09/4.99	   c(R(x0))
16.09/4.99	
16.09/4.99	We have to consider all minimal (P,Q,R)-chains.
16.09/4.99	----------------------------------------
16.09/4.99	
16.09/4.99	(19) QDPOrderProof (EQUIVALENT)
16.09/4.99	We use the reduction pair processor [LPAR04,JAR06].
16.09/4.99	
16.09/4.99	
16.09/4.99	The following pairs can be oriented strictly and are deleted.
16.09/4.99	
16.09/4.99	   L^1(a(a(x1))) -> L^1(a(b(c(x1))))
16.09/4.99	The remaining pairs can at least be oriented weakly.
16.09/4.99	Used ordering:  Polynomial Order [NEGPOLO,POLO] with Interpretation:
16.09/4.99	
16.09/4.99	POL( L^1_1(x_1) ) = 2x_1 + 2
16.09/4.99	POL( a_1(x_1) ) = x_1 + 1
16.09/4.99	POL( b_1(x_1) ) = max{0, x_1 - 1}
16.09/4.99	POL( c_1(x_1) ) = x_1 + 1
16.09/4.99	POL( R_1(x_1) ) = 0
16.09/4.99	
16.09/4.99	The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:
16.09/4.99	
16.09/4.99	   c(a(x1)) -> a(c(x1))
16.09/4.99	   c(b(x1)) -> b(a(x1))
16.09/4.99	   c(R(x1)) -> b(a(R(x1)))
16.09/4.99	   b(a(a(x1))) -> a(b(c(x1)))
16.09/4.99	
16.09/4.99	
16.09/4.99	----------------------------------------
16.09/4.99	
16.09/4.99	(20)
16.09/4.99	Obligation:
16.09/4.99	Q DP problem:
16.09/4.99	P is empty.
16.09/4.99	The TRS R consists of the following rules:
16.09/4.99	
16.09/4.99	   c(a(x1)) -> a(c(x1))
16.09/4.99	   c(b(x1)) -> b(a(x1))
16.09/4.99	   c(R(x1)) -> b(a(R(x1)))
16.09/4.99	   b(a(a(x1))) -> a(b(c(x1)))
16.09/4.99	
16.09/4.99	The set Q consists of the following terms:
16.09/4.99	
16.09/4.99	   b(a(a(x0)))
16.09/4.99	   c(a(x0))
16.09/4.99	   c(b(x0))
16.09/4.99	   c(R(x0))
16.09/4.99	
16.09/4.99	We have to consider all minimal (P,Q,R)-chains.
16.09/4.99	----------------------------------------
16.09/4.99	
16.09/4.99	(21) PisEmptyProof (EQUIVALENT)
16.09/4.99	The TRS P is empty. Hence, there is no (P,Q,R) chain.
16.09/4.99	----------------------------------------
16.09/4.99	
16.09/4.99	(22)
16.09/4.99	YES
16.61/6.65	EOF