3.72/1.78	YES
3.72/1.79	proof of /export/starexec/sandbox/benchmark/theBenchmark.xml
3.72/1.79	# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty
3.72/1.79	
3.72/1.79	
3.72/1.79	Termination w.r.t. Q of the given QTRS could be proven:
3.72/1.79	
3.72/1.79	(0) QTRS
3.72/1.79	(1) DependencyPairsProof [EQUIVALENT, 0 ms]
3.72/1.79	(2) QDP
3.72/1.79	(3) DependencyGraphProof [EQUIVALENT, 0 ms]
3.72/1.79	(4) AND
3.72/1.79	    (5) QDP
3.72/1.79	        (6) UsableRulesProof [EQUIVALENT, 0 ms]
3.72/1.79	        (7) QDP
3.72/1.79	        (8) QReductionProof [EQUIVALENT, 0 ms]
3.72/1.79	        (9) QDP
3.72/1.79	        (10) MRRProof [EQUIVALENT, 6 ms]
3.72/1.79	        (11) QDP
3.72/1.79	        (12) PisEmptyProof [EQUIVALENT, 0 ms]
3.72/1.79	        (13) YES
3.72/1.79	    (14) QDP
3.72/1.79	        (15) UsableRulesProof [EQUIVALENT, 0 ms]
3.72/1.79	        (16) QDP
3.72/1.79	        (17) QReductionProof [EQUIVALENT, 0 ms]
3.72/1.79	        (18) QDP
3.72/1.79	        (19) MRRProof [EQUIVALENT, 4 ms]
3.72/1.79	        (20) QDP
3.72/1.79	        (21) PisEmptyProof [EQUIVALENT, 0 ms]
3.72/1.79	        (22) YES
3.72/1.79	
3.72/1.79	
3.72/1.79	----------------------------------------
3.72/1.79	
3.72/1.79	(0)
3.72/1.79	Obligation:
3.72/1.79	Q restricted rewrite system:
3.72/1.79	The TRS R consists of the following rules:
3.72/1.79	
3.72/1.79	   f(c(s(x), y)) -> f(c(x, s(y)))
3.72/1.79	   g(c(x, s(y))) -> g(c(s(x), y))
3.72/1.79	
3.72/1.79	The set Q consists of the following terms:
3.72/1.79	
3.72/1.79	   f(c(s(x0), x1))
3.72/1.79	   g(c(x0, s(x1)))
3.72/1.79	
3.72/1.79	
3.72/1.79	----------------------------------------
3.72/1.79	
3.72/1.79	(1) DependencyPairsProof (EQUIVALENT)
3.72/1.79	Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.
3.72/1.79	----------------------------------------
3.72/1.79	
3.72/1.79	(2)
3.72/1.79	Obligation:
3.72/1.79	Q DP problem:
3.72/1.79	The TRS P consists of the following rules:
3.72/1.79	
3.72/1.79	   F(c(s(x), y)) -> F(c(x, s(y)))
3.72/1.79	   G(c(x, s(y))) -> G(c(s(x), y))
3.72/1.79	
3.72/1.79	The TRS R consists of the following rules:
3.72/1.79	
3.72/1.79	   f(c(s(x), y)) -> f(c(x, s(y)))
3.72/1.79	   g(c(x, s(y))) -> g(c(s(x), y))
3.72/1.79	
3.72/1.79	The set Q consists of the following terms:
3.72/1.79	
3.72/1.79	   f(c(s(x0), x1))
3.72/1.79	   g(c(x0, s(x1)))
3.72/1.79	
3.72/1.79	We have to consider all minimal (P,Q,R)-chains.
3.72/1.79	----------------------------------------
3.72/1.79	
3.72/1.79	(3) DependencyGraphProof (EQUIVALENT)
3.72/1.79	The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 2 SCCs.
3.72/1.79	----------------------------------------
3.72/1.79	
3.72/1.79	(4)
3.72/1.79	Complex Obligation (AND)
3.72/1.79	
3.72/1.79	----------------------------------------
3.72/1.79	
3.72/1.79	(5)
3.72/1.79	Obligation:
3.72/1.79	Q DP problem:
3.72/1.79	The TRS P consists of the following rules:
3.72/1.79	
3.72/1.79	   G(c(x, s(y))) -> G(c(s(x), y))
3.72/1.79	
3.72/1.79	The TRS R consists of the following rules:
3.72/1.79	
3.72/1.79	   f(c(s(x), y)) -> f(c(x, s(y)))
3.72/1.79	   g(c(x, s(y))) -> g(c(s(x), y))
3.72/1.79	
3.72/1.79	The set Q consists of the following terms:
3.72/1.79	
3.72/1.79	   f(c(s(x0), x1))
3.72/1.79	   g(c(x0, s(x1)))
3.72/1.79	
3.72/1.79	We have to consider all minimal (P,Q,R)-chains.
3.72/1.79	----------------------------------------
3.72/1.79	
3.72/1.79	(6) UsableRulesProof (EQUIVALENT)
3.72/1.79	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.
3.72/1.79	----------------------------------------
3.72/1.79	
3.72/1.79	(7)
3.72/1.79	Obligation:
3.72/1.79	Q DP problem:
3.72/1.79	The TRS P consists of the following rules:
3.72/1.79	
3.72/1.79	   G(c(x, s(y))) -> G(c(s(x), y))
3.72/1.79	
3.72/1.79	R is empty.
3.72/1.79	The set Q consists of the following terms:
3.72/1.79	
3.72/1.79	   f(c(s(x0), x1))
3.72/1.79	   g(c(x0, s(x1)))
3.72/1.79	
3.72/1.79	We have to consider all minimal (P,Q,R)-chains.
3.72/1.79	----------------------------------------
3.72/1.79	
3.72/1.79	(8) QReductionProof (EQUIVALENT)
3.72/1.79	We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].
3.72/1.79	
3.72/1.79	   f(c(s(x0), x1))
3.72/1.79	   g(c(x0, s(x1)))
3.72/1.79	
3.72/1.79	
3.72/1.79	----------------------------------------
3.72/1.79	
3.72/1.79	(9)
3.72/1.79	Obligation:
3.72/1.79	Q DP problem:
3.72/1.79	The TRS P consists of the following rules:
3.72/1.79	
3.72/1.79	   G(c(x, s(y))) -> G(c(s(x), y))
3.72/1.79	
3.72/1.79	R is empty.
3.72/1.79	Q is empty.
3.72/1.79	We have to consider all minimal (P,Q,R)-chains.
3.72/1.79	----------------------------------------
3.72/1.79	
3.72/1.79	(10) MRRProof (EQUIVALENT)
3.72/1.79	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.
3.72/1.79	
3.72/1.79	Strictly oriented dependency pairs:
3.72/1.79	
3.72/1.79	   G(c(x, s(y))) -> G(c(s(x), y))
3.72/1.79	
3.72/1.79	
3.72/1.79	Used ordering: Polynomial interpretation [POLO]:
3.72/1.79	
3.72/1.79	   POL(G(x_1)) = 2*x_1
3.72/1.79	   POL(c(x_1, x_2)) = x_1 + 2*x_2
3.72/1.79	   POL(s(x_1)) = 2 + x_1
3.72/1.79	
3.72/1.79	
3.72/1.79	----------------------------------------
3.72/1.79	
3.72/1.79	(11)
3.72/1.79	Obligation:
3.72/1.79	Q DP problem:
3.72/1.79	P is empty.
3.72/1.79	R is empty.
3.72/1.79	Q is empty.
3.72/1.79	We have to consider all minimal (P,Q,R)-chains.
3.72/1.79	----------------------------------------
3.72/1.79	
3.72/1.79	(12) PisEmptyProof (EQUIVALENT)
3.72/1.79	The TRS P is empty. Hence, there is no (P,Q,R) chain.
3.72/1.79	----------------------------------------
3.72/1.79	
3.72/1.79	(13)
3.72/1.79	YES
3.72/1.79	
3.72/1.79	----------------------------------------
3.72/1.79	
3.72/1.79	(14)
3.72/1.79	Obligation:
3.72/1.79	Q DP problem:
3.72/1.79	The TRS P consists of the following rules:
3.72/1.79	
3.72/1.79	   F(c(s(x), y)) -> F(c(x, s(y)))
3.72/1.79	
3.72/1.79	The TRS R consists of the following rules:
3.72/1.79	
3.72/1.79	   f(c(s(x), y)) -> f(c(x, s(y)))
3.72/1.79	   g(c(x, s(y))) -> g(c(s(x), y))
3.72/1.79	
3.72/1.79	The set Q consists of the following terms:
3.72/1.79	
3.72/1.79	   f(c(s(x0), x1))
3.72/1.79	   g(c(x0, s(x1)))
3.72/1.79	
3.72/1.79	We have to consider all minimal (P,Q,R)-chains.
3.72/1.79	----------------------------------------
3.72/1.79	
3.72/1.79	(15) UsableRulesProof (EQUIVALENT)
3.72/1.79	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.
3.72/1.79	----------------------------------------
3.72/1.79	
3.72/1.79	(16)
3.72/1.79	Obligation:
3.72/1.79	Q DP problem:
3.72/1.79	The TRS P consists of the following rules:
3.72/1.79	
3.72/1.79	   F(c(s(x), y)) -> F(c(x, s(y)))
3.72/1.79	
3.72/1.79	R is empty.
3.72/1.79	The set Q consists of the following terms:
3.72/1.79	
3.72/1.79	   f(c(s(x0), x1))
3.72/1.79	   g(c(x0, s(x1)))
3.72/1.79	
3.72/1.79	We have to consider all minimal (P,Q,R)-chains.
3.72/1.79	----------------------------------------
3.72/1.79	
3.72/1.79	(17) QReductionProof (EQUIVALENT)
3.72/1.79	We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].
3.72/1.79	
3.72/1.79	   f(c(s(x0), x1))
3.72/1.79	   g(c(x0, s(x1)))
3.72/1.79	
3.72/1.79	
3.72/1.79	----------------------------------------
3.72/1.79	
3.72/1.79	(18)
3.72/1.79	Obligation:
3.72/1.79	Q DP problem:
3.72/1.79	The TRS P consists of the following rules:
3.72/1.79	
3.72/1.79	   F(c(s(x), y)) -> F(c(x, s(y)))
3.72/1.79	
3.72/1.79	R is empty.
3.72/1.79	Q is empty.
3.72/1.79	We have to consider all minimal (P,Q,R)-chains.
3.72/1.79	----------------------------------------
3.72/1.79	
3.72/1.79	(19) MRRProof (EQUIVALENT)
3.72/1.79	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.
3.72/1.79	
3.72/1.79	Strictly oriented dependency pairs:
3.72/1.79	
3.72/1.79	   F(c(s(x), y)) -> F(c(x, s(y)))
3.72/1.79	
3.72/1.79	
3.72/1.79	Used ordering: Polynomial interpretation [POLO]:
3.72/1.79	
3.72/1.79	   POL(F(x_1)) = 2*x_1
3.72/1.79	   POL(c(x_1, x_2)) = 2*x_1 + x_2
3.72/1.79	   POL(s(x_1)) = 2 + x_1
3.72/1.79	
3.72/1.79	
3.72/1.79	----------------------------------------
3.72/1.79	
3.72/1.79	(20)
3.72/1.79	Obligation:
3.72/1.79	Q DP problem:
3.72/1.79	P is empty.
3.72/1.79	R is empty.
3.72/1.79	Q is empty.
3.72/1.79	We have to consider all minimal (P,Q,R)-chains.
3.72/1.79	----------------------------------------
3.72/1.79	
3.72/1.79	(21) PisEmptyProof (EQUIVALENT)
3.72/1.79	The TRS P is empty. Hence, there is no (P,Q,R) chain.
3.72/1.79	----------------------------------------
3.72/1.79	
3.72/1.79	(22)
3.72/1.79	YES
3.80/1.81	EOF