3.55/2.08	YES
3.55/2.09	proof of /export/starexec/sandbox/benchmark/theBenchmark.xml
3.55/2.09	# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty
3.55/2.09	
3.55/2.09	
3.55/2.09	Termination w.r.t. Q of the given QTRS could be proven:
3.55/2.09	
3.55/2.09	(0) QTRS
3.55/2.09	(1) QTRSRRRProof [EQUIVALENT, 67 ms]
3.55/2.09	(2) QTRS
3.55/2.09	(3) DependencyPairsProof [EQUIVALENT, 15 ms]
3.55/2.09	(4) QDP
3.55/2.09	(5) DependencyGraphProof [EQUIVALENT, 0 ms]
3.55/2.09	(6) AND
3.55/2.09	    (7) QDP
3.55/2.09	        (8) UsableRulesProof [EQUIVALENT, 0 ms]
3.55/2.09	        (9) QDP
3.55/2.09	        (10) QReductionProof [EQUIVALENT, 0 ms]
3.55/2.09	        (11) QDP
3.55/2.09	        (12) MRRProof [EQUIVALENT, 0 ms]
3.55/2.09	        (13) QDP
3.55/2.09	        (14) PisEmptyProof [EQUIVALENT, 0 ms]
3.55/2.09	        (15) YES
3.55/2.09	    (16) QDP
3.55/2.09	        (17) UsableRulesProof [EQUIVALENT, 0 ms]
3.55/2.09	        (18) QDP
3.55/2.09	        (19) QReductionProof [EQUIVALENT, 0 ms]
3.55/2.09	        (20) QDP
3.55/2.09	        (21) MRRProof [EQUIVALENT, 0 ms]
3.55/2.09	        (22) QDP
3.55/2.09	        (23) PisEmptyProof [EQUIVALENT, 0 ms]
3.55/2.09	        (24) YES
3.55/2.09	
3.55/2.09	
3.55/2.09	----------------------------------------
3.55/2.09	
3.55/2.09	(0)
3.55/2.09	Obligation:
3.55/2.09	Q restricted rewrite system:
3.55/2.09	The TRS R consists of the following rules:
3.55/2.09	
3.55/2.09	   f(c(s(x), y)) -> f(c(x, s(y)))
3.55/2.09	   g(c(x, s(y))) -> g(c(s(x), y))
3.55/2.09	   g(s(f(x))) -> g(f(x))
3.55/2.09	
3.55/2.09	The set Q consists of the following terms:
3.55/2.09	
3.55/2.09	   f(c(s(x0), x1))
3.55/2.09	   g(c(x0, s(x1)))
3.55/2.09	   g(s(f(x0)))
3.55/2.09	
3.55/2.09	
3.55/2.09	----------------------------------------
3.55/2.09	
3.55/2.09	(1) QTRSRRRProof (EQUIVALENT)
3.55/2.09	Used ordering:
3.55/2.09	Polynomial interpretation [POLO]:
3.55/2.09	
3.55/2.09	   POL(c(x_1, x_2)) = 2*x_1 + 2*x_2
3.55/2.09	   POL(f(x_1)) = 2*x_1
3.55/2.09	   POL(g(x_1)) = 2*x_1
3.55/2.09	   POL(s(x_1)) = 1 + x_1
3.55/2.09	With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:
3.55/2.09	
3.55/2.09	   g(s(f(x))) -> g(f(x))
3.55/2.09	
3.55/2.09	
3.55/2.09	
3.55/2.09	
3.55/2.09	----------------------------------------
3.55/2.09	
3.55/2.09	(2)
3.55/2.09	Obligation:
3.55/2.09	Q restricted rewrite system:
3.55/2.09	The TRS R consists of the following rules:
3.55/2.09	
3.55/2.09	   f(c(s(x), y)) -> f(c(x, s(y)))
3.55/2.09	   g(c(x, s(y))) -> g(c(s(x), y))
3.55/2.09	
3.55/2.09	The set Q consists of the following terms:
3.55/2.09	
3.55/2.09	   f(c(s(x0), x1))
3.55/2.09	   g(c(x0, s(x1)))
3.55/2.09	   g(s(f(x0)))
3.55/2.09	
3.55/2.09	
3.55/2.09	----------------------------------------
3.55/2.09	
3.55/2.09	(3) DependencyPairsProof (EQUIVALENT)
3.55/2.09	Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.
3.55/2.09	----------------------------------------
3.55/2.09	
3.55/2.09	(4)
3.55/2.09	Obligation:
3.55/2.09	Q DP problem:
3.55/2.09	The TRS P consists of the following rules:
3.55/2.09	
3.55/2.09	   F(c(s(x), y)) -> F(c(x, s(y)))
3.55/2.09	   G(c(x, s(y))) -> G(c(s(x), y))
3.55/2.09	
3.55/2.09	The TRS R consists of the following rules:
3.55/2.09	
3.55/2.09	   f(c(s(x), y)) -> f(c(x, s(y)))
3.55/2.09	   g(c(x, s(y))) -> g(c(s(x), y))
3.55/2.09	
3.55/2.09	The set Q consists of the following terms:
3.55/2.09	
3.55/2.09	   f(c(s(x0), x1))
3.55/2.09	   g(c(x0, s(x1)))
3.55/2.09	   g(s(f(x0)))
3.55/2.09	
3.55/2.09	We have to consider all minimal (P,Q,R)-chains.
3.55/2.09	----------------------------------------
3.55/2.09	
3.55/2.09	(5) DependencyGraphProof (EQUIVALENT)
3.55/2.09	The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 2 SCCs.
3.55/2.09	----------------------------------------
3.55/2.09	
3.55/2.09	(6)
3.55/2.09	Complex Obligation (AND)
3.55/2.09	
3.55/2.09	----------------------------------------
3.55/2.09	
3.55/2.09	(7)
3.55/2.09	Obligation:
3.55/2.09	Q DP problem:
3.55/2.09	The TRS P consists of the following rules:
3.55/2.09	
3.55/2.09	   G(c(x, s(y))) -> G(c(s(x), y))
3.55/2.09	
3.55/2.09	The TRS R consists of the following rules:
3.55/2.09	
3.55/2.09	   f(c(s(x), y)) -> f(c(x, s(y)))
3.55/2.09	   g(c(x, s(y))) -> g(c(s(x), y))
3.55/2.09	
3.55/2.09	The set Q consists of the following terms:
3.55/2.09	
3.55/2.09	   f(c(s(x0), x1))
3.55/2.09	   g(c(x0, s(x1)))
3.55/2.09	   g(s(f(x0)))
3.55/2.09	
3.55/2.09	We have to consider all minimal (P,Q,R)-chains.
3.55/2.09	----------------------------------------
3.55/2.09	
3.55/2.09	(8) UsableRulesProof (EQUIVALENT)
3.55/2.09	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.
3.55/2.09	----------------------------------------
3.55/2.09	
3.55/2.09	(9)
3.55/2.09	Obligation:
3.55/2.09	Q DP problem:
3.55/2.09	The TRS P consists of the following rules:
3.55/2.09	
3.55/2.09	   G(c(x, s(y))) -> G(c(s(x), y))
3.55/2.09	
3.55/2.09	R is empty.
3.55/2.09	The set Q consists of the following terms:
3.55/2.09	
3.55/2.09	   f(c(s(x0), x1))
3.55/2.09	   g(c(x0, s(x1)))
3.55/2.09	   g(s(f(x0)))
3.55/2.09	
3.55/2.09	We have to consider all minimal (P,Q,R)-chains.
3.55/2.09	----------------------------------------
3.55/2.09	
3.55/2.09	(10) QReductionProof (EQUIVALENT)
3.55/2.09	We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].
3.55/2.09	
3.55/2.09	   f(c(s(x0), x1))
3.55/2.09	   g(c(x0, s(x1)))
3.55/2.09	   g(s(f(x0)))
3.55/2.09	
3.55/2.09	
3.55/2.09	----------------------------------------
3.55/2.09	
3.55/2.09	(11)
3.55/2.09	Obligation:
3.55/2.09	Q DP problem:
3.55/2.09	The TRS P consists of the following rules:
3.55/2.09	
3.55/2.09	   G(c(x, s(y))) -> G(c(s(x), y))
3.55/2.09	
3.55/2.09	R is empty.
3.55/2.09	Q is empty.
3.55/2.09	We have to consider all minimal (P,Q,R)-chains.
3.55/2.09	----------------------------------------
3.55/2.09	
3.55/2.09	(12) MRRProof (EQUIVALENT)
3.55/2.09	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.
3.55/2.09	
3.55/2.09	Strictly oriented dependency pairs:
3.55/2.09	
3.55/2.09	   G(c(x, s(y))) -> G(c(s(x), y))
3.55/2.09	
3.55/2.09	
3.55/2.09	Used ordering: Polynomial interpretation [POLO]:
3.55/2.09	
3.55/2.09	   POL(G(x_1)) = 2*x_1
3.55/2.09	   POL(c(x_1, x_2)) = x_1 + 2*x_2
3.55/2.09	   POL(s(x_1)) = 2 + x_1
3.55/2.09	
3.55/2.09	
3.55/2.09	----------------------------------------
3.55/2.09	
3.55/2.09	(13)
3.55/2.09	Obligation:
3.55/2.09	Q DP problem:
3.55/2.09	P is empty.
3.55/2.09	R is empty.
3.55/2.09	Q is empty.
3.55/2.09	We have to consider all minimal (P,Q,R)-chains.
3.55/2.09	----------------------------------------
3.55/2.09	
3.55/2.09	(14) PisEmptyProof (EQUIVALENT)
3.55/2.09	The TRS P is empty. Hence, there is no (P,Q,R) chain.
3.55/2.09	----------------------------------------
3.55/2.09	
3.55/2.09	(15)
3.55/2.09	YES
3.55/2.09	
3.55/2.09	----------------------------------------
3.55/2.09	
3.55/2.09	(16)
3.55/2.09	Obligation:
3.55/2.09	Q DP problem:
3.55/2.09	The TRS P consists of the following rules:
3.55/2.09	
3.55/2.09	   F(c(s(x), y)) -> F(c(x, s(y)))
3.55/2.09	
3.55/2.09	The TRS R consists of the following rules:
3.55/2.09	
3.55/2.09	   f(c(s(x), y)) -> f(c(x, s(y)))
3.55/2.09	   g(c(x, s(y))) -> g(c(s(x), y))
3.55/2.09	
3.55/2.09	The set Q consists of the following terms:
3.55/2.09	
3.55/2.09	   f(c(s(x0), x1))
3.55/2.09	   g(c(x0, s(x1)))
3.55/2.09	   g(s(f(x0)))
3.55/2.09	
3.55/2.09	We have to consider all minimal (P,Q,R)-chains.
3.55/2.09	----------------------------------------
3.55/2.09	
3.55/2.09	(17) UsableRulesProof (EQUIVALENT)
3.55/2.09	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.
3.55/2.09	----------------------------------------
3.55/2.09	
3.55/2.09	(18)
3.55/2.09	Obligation:
3.55/2.09	Q DP problem:
3.55/2.09	The TRS P consists of the following rules:
3.55/2.09	
3.55/2.09	   F(c(s(x), y)) -> F(c(x, s(y)))
3.55/2.09	
3.55/2.09	R is empty.
3.55/2.09	The set Q consists of the following terms:
3.55/2.09	
3.55/2.09	   f(c(s(x0), x1))
3.55/2.09	   g(c(x0, s(x1)))
3.55/2.09	   g(s(f(x0)))
3.55/2.09	
3.55/2.09	We have to consider all minimal (P,Q,R)-chains.
3.55/2.09	----------------------------------------
3.55/2.09	
3.55/2.09	(19) QReductionProof (EQUIVALENT)
3.55/2.09	We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].
3.55/2.09	
3.55/2.09	   f(c(s(x0), x1))
3.55/2.09	   g(c(x0, s(x1)))
3.55/2.09	   g(s(f(x0)))
3.55/2.09	
3.55/2.09	
3.55/2.09	----------------------------------------
3.55/2.09	
3.55/2.09	(20)
3.55/2.09	Obligation:
3.55/2.09	Q DP problem:
3.55/2.09	The TRS P consists of the following rules:
3.55/2.09	
3.55/2.09	   F(c(s(x), y)) -> F(c(x, s(y)))
3.55/2.09	
3.55/2.09	R is empty.
3.55/2.09	Q is empty.
3.55/2.09	We have to consider all minimal (P,Q,R)-chains.
3.55/2.09	----------------------------------------
3.55/2.09	
3.55/2.09	(21) MRRProof (EQUIVALENT)
3.55/2.09	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.
3.55/2.09	
3.55/2.09	Strictly oriented dependency pairs:
3.55/2.09	
3.55/2.09	   F(c(s(x), y)) -> F(c(x, s(y)))
3.55/2.09	
3.55/2.09	
3.55/2.09	Used ordering: Knuth-Bendix order [KBO] with precedence:s_1 > c_2 > F_1
3.55/2.09	
3.55/2.09	and weight map:
3.55/2.09	
3.55/2.09	   F_1=1
3.55/2.09	   s_1=1
3.55/2.09	   c_2=0
3.55/2.09	
3.55/2.09	The variable weight is 1
3.55/2.09	
3.55/2.09	----------------------------------------
3.55/2.09	
3.55/2.09	(22)
3.55/2.09	Obligation:
3.55/2.09	Q DP problem:
3.55/2.09	P is empty.
3.55/2.09	R is empty.
3.55/2.09	Q is empty.
3.55/2.09	We have to consider all minimal (P,Q,R)-chains.
3.55/2.09	----------------------------------------
3.55/2.09	
3.55/2.09	(23) PisEmptyProof (EQUIVALENT)
3.55/2.09	The TRS P is empty. Hence, there is no (P,Q,R) chain.
3.55/2.09	----------------------------------------
3.55/2.09	
3.55/2.09	(24)
3.55/2.09	YES
3.74/2.10	EOF