3.58/1.73	YES
3.58/1.73	proof of /export/starexec/sandbox/benchmark/theBenchmark.xml
3.58/1.73	# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty
3.58/1.73	
3.58/1.73	
3.58/1.73	Termination w.r.t. Q of the given QTRS could be proven:
3.58/1.73	
3.58/1.73	(0) QTRS
3.58/1.73	(1) DependencyPairsProof [EQUIVALENT, 0 ms]
3.58/1.73	(2) QDP
3.58/1.73	(3) DependencyGraphProof [EQUIVALENT, 0 ms]
3.58/1.73	(4) AND
3.58/1.73	    (5) QDP
3.58/1.73	        (6) UsableRulesProof [EQUIVALENT, 0 ms]
3.58/1.73	        (7) QDP
3.58/1.73	        (8) QReductionProof [EQUIVALENT, 0 ms]
3.58/1.73	        (9) QDP
3.58/1.73	        (10) QDPSizeChangeProof [EQUIVALENT, 0 ms]
3.58/1.73	        (11) YES
3.58/1.73	    (12) QDP
3.58/1.73	        (13) UsableRulesProof [EQUIVALENT, 0 ms]
3.58/1.73	        (14) QDP
3.58/1.73	        (15) QReductionProof [EQUIVALENT, 0 ms]
3.58/1.73	        (16) QDP
3.58/1.73	        (17) TransformationProof [EQUIVALENT, 0 ms]
3.58/1.73	        (18) QDP
3.58/1.73	        (19) DependencyGraphProof [EQUIVALENT, 0 ms]
3.58/1.73	        (20) TRUE
3.58/1.73	
3.58/1.73	
3.58/1.73	----------------------------------------
3.58/1.73	
3.58/1.73	(0)
3.58/1.73	Obligation:
3.58/1.73	Q restricted rewrite system:
3.58/1.73	The TRS R consists of the following rules:
3.58/1.73	
3.58/1.73	   f(g(x), s(0), y) -> f(g(s(0)), y, g(x))
3.58/1.73	   g(s(x)) -> s(g(x))
3.58/1.73	   g(0) -> 0
3.58/1.73	
3.58/1.73	The set Q consists of the following terms:
3.58/1.73	
3.58/1.73	   f(g(x0), s(0), x1)
3.58/1.73	   g(s(x0))
3.58/1.73	   g(0)
3.58/1.73	
3.58/1.73	
3.58/1.73	----------------------------------------
3.58/1.73	
3.58/1.73	(1) DependencyPairsProof (EQUIVALENT)
3.58/1.73	Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.
3.58/1.73	----------------------------------------
3.58/1.73	
3.58/1.73	(2)
3.58/1.73	Obligation:
3.58/1.73	Q DP problem:
3.58/1.73	The TRS P consists of the following rules:
3.58/1.73	
3.58/1.73	   F(g(x), s(0), y) -> F(g(s(0)), y, g(x))
3.58/1.73	   F(g(x), s(0), y) -> G(s(0))
3.58/1.73	   G(s(x)) -> G(x)
3.58/1.73	
3.58/1.73	The TRS R consists of the following rules:
3.58/1.73	
3.58/1.73	   f(g(x), s(0), y) -> f(g(s(0)), y, g(x))
3.58/1.73	   g(s(x)) -> s(g(x))
3.58/1.73	   g(0) -> 0
3.58/1.73	
3.58/1.73	The set Q consists of the following terms:
3.58/1.73	
3.58/1.73	   f(g(x0), s(0), x1)
3.58/1.73	   g(s(x0))
3.58/1.73	   g(0)
3.58/1.73	
3.58/1.73	We have to consider all minimal (P,Q,R)-chains.
3.58/1.73	----------------------------------------
3.58/1.73	
3.58/1.73	(3) DependencyGraphProof (EQUIVALENT)
3.58/1.73	The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 2 SCCs with 1 less node.
3.58/1.73	----------------------------------------
3.58/1.73	
3.58/1.73	(4)
3.58/1.73	Complex Obligation (AND)
3.58/1.73	
3.58/1.73	----------------------------------------
3.58/1.73	
3.58/1.73	(5)
3.58/1.73	Obligation:
3.58/1.73	Q DP problem:
3.58/1.73	The TRS P consists of the following rules:
3.58/1.73	
3.58/1.73	   G(s(x)) -> G(x)
3.58/1.73	
3.58/1.73	The TRS R consists of the following rules:
3.58/1.73	
3.58/1.73	   f(g(x), s(0), y) -> f(g(s(0)), y, g(x))
3.58/1.73	   g(s(x)) -> s(g(x))
3.58/1.73	   g(0) -> 0
3.58/1.73	
3.58/1.73	The set Q consists of the following terms:
3.58/1.73	
3.58/1.73	   f(g(x0), s(0), x1)
3.58/1.73	   g(s(x0))
3.58/1.73	   g(0)
3.58/1.73	
3.58/1.73	We have to consider all minimal (P,Q,R)-chains.
3.58/1.73	----------------------------------------
3.58/1.73	
3.58/1.73	(6) UsableRulesProof (EQUIVALENT)
3.58/1.73	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.
3.58/1.73	----------------------------------------
3.58/1.73	
3.58/1.73	(7)
3.58/1.73	Obligation:
3.58/1.73	Q DP problem:
3.58/1.73	The TRS P consists of the following rules:
3.58/1.73	
3.58/1.73	   G(s(x)) -> G(x)
3.58/1.73	
3.58/1.73	R is empty.
3.58/1.73	The set Q consists of the following terms:
3.58/1.73	
3.58/1.73	   f(g(x0), s(0), x1)
3.58/1.73	   g(s(x0))
3.58/1.73	   g(0)
3.58/1.73	
3.58/1.73	We have to consider all minimal (P,Q,R)-chains.
3.58/1.73	----------------------------------------
3.58/1.73	
3.58/1.73	(8) QReductionProof (EQUIVALENT)
3.58/1.73	We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].
3.58/1.73	
3.58/1.73	   f(g(x0), s(0), x1)
3.58/1.73	   g(s(x0))
3.58/1.73	   g(0)
3.58/1.73	
3.58/1.73	
3.58/1.73	----------------------------------------
3.58/1.73	
3.58/1.73	(9)
3.58/1.73	Obligation:
3.58/1.73	Q DP problem:
3.58/1.73	The TRS P consists of the following rules:
3.58/1.73	
3.58/1.73	   G(s(x)) -> G(x)
3.58/1.73	
3.58/1.73	R is empty.
3.58/1.73	Q is empty.
3.58/1.73	We have to consider all minimal (P,Q,R)-chains.
3.58/1.73	----------------------------------------
3.58/1.73	
3.58/1.73	(10) QDPSizeChangeProof (EQUIVALENT)
3.58/1.73	By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem. 
3.58/1.73	
3.58/1.73	From the DPs we obtained the following set of size-change graphs:
3.58/1.73	*G(s(x)) -> G(x)
3.58/1.73	The graph contains the following edges 1 > 1
3.58/1.73	
3.58/1.73	
3.58/1.73	----------------------------------------
3.58/1.73	
3.58/1.73	(11)
3.58/1.73	YES
3.58/1.73	
3.58/1.73	----------------------------------------
3.58/1.73	
3.58/1.73	(12)
3.58/1.73	Obligation:
3.58/1.73	Q DP problem:
3.58/1.73	The TRS P consists of the following rules:
3.58/1.73	
3.58/1.73	   F(g(x), s(0), y) -> F(g(s(0)), y, g(x))
3.58/1.73	
3.58/1.73	The TRS R consists of the following rules:
3.58/1.73	
3.58/1.73	   f(g(x), s(0), y) -> f(g(s(0)), y, g(x))
3.58/1.73	   g(s(x)) -> s(g(x))
3.58/1.73	   g(0) -> 0
3.58/1.73	
3.58/1.73	The set Q consists of the following terms:
3.58/1.73	
3.58/1.73	   f(g(x0), s(0), x1)
3.58/1.73	   g(s(x0))
3.58/1.73	   g(0)
3.58/1.73	
3.58/1.73	We have to consider all minimal (P,Q,R)-chains.
3.58/1.73	----------------------------------------
3.58/1.73	
3.58/1.73	(13) UsableRulesProof (EQUIVALENT)
3.58/1.73	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.
3.58/1.73	----------------------------------------
3.58/1.73	
3.58/1.73	(14)
3.58/1.73	Obligation:
3.58/1.73	Q DP problem:
3.58/1.73	The TRS P consists of the following rules:
3.58/1.73	
3.58/1.73	   F(g(x), s(0), y) -> F(g(s(0)), y, g(x))
3.58/1.73	
3.58/1.73	The TRS R consists of the following rules:
3.58/1.73	
3.58/1.73	   g(s(x)) -> s(g(x))
3.58/1.73	   g(0) -> 0
3.58/1.73	
3.58/1.73	The set Q consists of the following terms:
3.58/1.73	
3.58/1.73	   f(g(x0), s(0), x1)
3.58/1.73	   g(s(x0))
3.58/1.73	   g(0)
3.58/1.73	
3.58/1.73	We have to consider all minimal (P,Q,R)-chains.
3.58/1.73	----------------------------------------
3.58/1.73	
3.58/1.73	(15) QReductionProof (EQUIVALENT)
3.58/1.73	We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].
3.58/1.73	
3.58/1.73	   f(g(x0), s(0), x1)
3.58/1.73	
3.58/1.73	
3.58/1.73	----------------------------------------
3.58/1.73	
3.58/1.73	(16)
3.58/1.73	Obligation:
3.58/1.74	Q DP problem:
3.58/1.74	The TRS P consists of the following rules:
3.58/1.74	
3.58/1.74	   F(g(x), s(0), y) -> F(g(s(0)), y, g(x))
3.58/1.74	
3.58/1.74	The TRS R consists of the following rules:
3.58/1.74	
3.58/1.74	   g(s(x)) -> s(g(x))
3.58/1.74	   g(0) -> 0
3.58/1.74	
3.58/1.74	The set Q consists of the following terms:
3.58/1.74	
3.58/1.74	   g(s(x0))
3.58/1.74	   g(0)
3.58/1.74	
3.58/1.74	We have to consider all minimal (P,Q,R)-chains.
3.58/1.74	----------------------------------------
3.58/1.74	
3.58/1.74	(17) TransformationProof (EQUIVALENT)
3.58/1.74	By rewriting [LPAR04] the rule F(g(x), s(0), y) -> F(g(s(0)), y, g(x)) at position [0] we obtained the following new rules [LPAR04]:
3.58/1.74	
3.58/1.74	   (F(g(x), s(0), y) -> F(s(g(0)), y, g(x)),F(g(x), s(0), y) -> F(s(g(0)), y, g(x)))
3.58/1.74	
3.58/1.74	
3.58/1.74	----------------------------------------
3.58/1.74	
3.58/1.74	(18)
3.58/1.74	Obligation:
3.58/1.74	Q DP problem:
3.58/1.74	The TRS P consists of the following rules:
3.58/1.74	
3.58/1.74	   F(g(x), s(0), y) -> F(s(g(0)), y, g(x))
3.58/1.74	
3.58/1.74	The TRS R consists of the following rules:
3.58/1.74	
3.58/1.74	   g(s(x)) -> s(g(x))
3.58/1.74	   g(0) -> 0
3.58/1.74	
3.58/1.74	The set Q consists of the following terms:
3.58/1.74	
3.58/1.74	   g(s(x0))
3.58/1.74	   g(0)
3.58/1.74	
3.58/1.74	We have to consider all minimal (P,Q,R)-chains.
3.58/1.74	----------------------------------------
3.58/1.74	
3.58/1.74	(19) DependencyGraphProof (EQUIVALENT)
3.58/1.74	The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 1 less node.
3.58/1.74	----------------------------------------
3.58/1.74	
3.58/1.74	(20)
3.58/1.74	TRUE
3.58/1.75	EOF