3.55/1.69	YES
3.55/1.70	proof of /export/starexec/sandbox/benchmark/theBenchmark.xml
3.55/1.70	# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty
3.55/1.70	
3.55/1.70	
3.55/1.70	Termination w.r.t. Q of the given QTRS could be proven:
3.55/1.70	
3.55/1.70	(0) QTRS
3.55/1.70	(1) DependencyPairsProof [EQUIVALENT, 0 ms]
3.55/1.70	(2) QDP
3.55/1.70	(3) DependencyGraphProof [EQUIVALENT, 0 ms]
3.55/1.70	(4) AND
3.55/1.70	    (5) QDP
3.55/1.70	        (6) UsableRulesProof [EQUIVALENT, 0 ms]
3.55/1.70	        (7) QDP
3.55/1.70	        (8) QReductionProof [EQUIVALENT, 0 ms]
3.55/1.70	        (9) QDP
3.55/1.70	        (10) QDPSizeChangeProof [EQUIVALENT, 0 ms]
3.55/1.70	        (11) YES
3.55/1.70	    (12) QDP
3.55/1.70	        (13) UsableRulesProof [EQUIVALENT, 0 ms]
3.55/1.70	        (14) QDP
3.55/1.70	        (15) QReductionProof [EQUIVALENT, 0 ms]
3.55/1.70	        (16) QDP
3.55/1.70	        (17) QDPSizeChangeProof [EQUIVALENT, 0 ms]
3.55/1.70	        (18) YES
3.55/1.70	
3.55/1.70	
3.55/1.70	----------------------------------------
3.55/1.70	
3.55/1.70	(0)
3.55/1.70	Obligation:
3.55/1.70	Q restricted rewrite system:
3.55/1.70	The TRS R consists of the following rules:
3.55/1.70	
3.55/1.70	   f(x, c(y)) -> f(x, s(f(y, y)))
3.55/1.70	   f(s(x), s(y)) -> f(x, s(c(s(y))))
3.55/1.70	
3.55/1.70	The set Q consists of the following terms:
3.55/1.70	
3.55/1.70	   f(x0, c(x1))
3.55/1.70	   f(s(x0), s(x1))
3.55/1.70	
3.55/1.70	
3.55/1.70	----------------------------------------
3.55/1.70	
3.55/1.70	(1) DependencyPairsProof (EQUIVALENT)
3.55/1.70	Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.
3.55/1.70	----------------------------------------
3.55/1.70	
3.55/1.70	(2)
3.55/1.70	Obligation:
3.55/1.70	Q DP problem:
3.55/1.70	The TRS P consists of the following rules:
3.55/1.70	
3.55/1.70	   F(x, c(y)) -> F(x, s(f(y, y)))
3.55/1.70	   F(x, c(y)) -> F(y, y)
3.55/1.70	   F(s(x), s(y)) -> F(x, s(c(s(y))))
3.55/1.70	
3.55/1.70	The TRS R consists of the following rules:
3.55/1.70	
3.55/1.70	   f(x, c(y)) -> f(x, s(f(y, y)))
3.55/1.70	   f(s(x), s(y)) -> f(x, s(c(s(y))))
3.55/1.70	
3.55/1.70	The set Q consists of the following terms:
3.55/1.70	
3.55/1.70	   f(x0, c(x1))
3.55/1.70	   f(s(x0), s(x1))
3.55/1.70	
3.55/1.70	We have to consider all minimal (P,Q,R)-chains.
3.55/1.70	----------------------------------------
3.55/1.70	
3.55/1.70	(3) DependencyGraphProof (EQUIVALENT)
3.55/1.70	The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 2 SCCs with 1 less node.
3.55/1.70	----------------------------------------
3.55/1.70	
3.55/1.70	(4)
3.55/1.70	Complex Obligation (AND)
3.55/1.70	
3.55/1.70	----------------------------------------
3.55/1.70	
3.55/1.70	(5)
3.55/1.70	Obligation:
3.55/1.70	Q DP problem:
3.55/1.70	The TRS P consists of the following rules:
3.55/1.70	
3.55/1.70	   F(s(x), s(y)) -> F(x, s(c(s(y))))
3.55/1.70	
3.55/1.70	The TRS R consists of the following rules:
3.55/1.70	
3.55/1.70	   f(x, c(y)) -> f(x, s(f(y, y)))
3.55/1.70	   f(s(x), s(y)) -> f(x, s(c(s(y))))
3.55/1.70	
3.55/1.70	The set Q consists of the following terms:
3.55/1.70	
3.55/1.70	   f(x0, c(x1))
3.55/1.70	   f(s(x0), s(x1))
3.55/1.70	
3.55/1.70	We have to consider all minimal (P,Q,R)-chains.
3.55/1.70	----------------------------------------
3.55/1.70	
3.55/1.70	(6) UsableRulesProof (EQUIVALENT)
3.55/1.70	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.
3.55/1.70	----------------------------------------
3.55/1.70	
3.55/1.70	(7)
3.55/1.70	Obligation:
3.55/1.70	Q DP problem:
3.55/1.70	The TRS P consists of the following rules:
3.55/1.70	
3.55/1.70	   F(s(x), s(y)) -> F(x, s(c(s(y))))
3.55/1.70	
3.55/1.70	R is empty.
3.55/1.70	The set Q consists of the following terms:
3.55/1.70	
3.55/1.70	   f(x0, c(x1))
3.55/1.70	   f(s(x0), s(x1))
3.55/1.70	
3.55/1.70	We have to consider all minimal (P,Q,R)-chains.
3.55/1.70	----------------------------------------
3.55/1.70	
3.55/1.70	(8) QReductionProof (EQUIVALENT)
3.55/1.70	We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].
3.55/1.70	
3.55/1.70	   f(x0, c(x1))
3.55/1.70	   f(s(x0), s(x1))
3.55/1.70	
3.55/1.70	
3.55/1.70	----------------------------------------
3.55/1.70	
3.55/1.70	(9)
3.55/1.70	Obligation:
3.55/1.70	Q DP problem:
3.55/1.70	The TRS P consists of the following rules:
3.55/1.70	
3.55/1.70	   F(s(x), s(y)) -> F(x, s(c(s(y))))
3.55/1.70	
3.55/1.70	R is empty.
3.55/1.70	Q is empty.
3.55/1.70	We have to consider all minimal (P,Q,R)-chains.
3.55/1.70	----------------------------------------
3.55/1.70	
3.55/1.70	(10) QDPSizeChangeProof (EQUIVALENT)
3.55/1.70	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. 
3.55/1.70	
3.55/1.70	From the DPs we obtained the following set of size-change graphs:
3.55/1.70	*F(s(x), s(y)) -> F(x, s(c(s(y))))
3.55/1.70	The graph contains the following edges 1 > 1
3.55/1.70	
3.55/1.70	
3.55/1.70	----------------------------------------
3.55/1.70	
3.55/1.70	(11)
3.55/1.70	YES
3.55/1.70	
3.55/1.70	----------------------------------------
3.55/1.70	
3.55/1.70	(12)
3.55/1.70	Obligation:
3.55/1.70	Q DP problem:
3.55/1.70	The TRS P consists of the following rules:
3.55/1.70	
3.55/1.70	   F(x, c(y)) -> F(y, y)
3.55/1.70	
3.55/1.70	The TRS R consists of the following rules:
3.55/1.70	
3.55/1.70	   f(x, c(y)) -> f(x, s(f(y, y)))
3.55/1.70	   f(s(x), s(y)) -> f(x, s(c(s(y))))
3.55/1.70	
3.55/1.70	The set Q consists of the following terms:
3.55/1.70	
3.55/1.70	   f(x0, c(x1))
3.55/1.70	   f(s(x0), s(x1))
3.55/1.70	
3.55/1.70	We have to consider all minimal (P,Q,R)-chains.
3.55/1.70	----------------------------------------
3.55/1.70	
3.55/1.70	(13) UsableRulesProof (EQUIVALENT)
3.55/1.70	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.
3.55/1.70	----------------------------------------
3.55/1.70	
3.55/1.70	(14)
3.55/1.70	Obligation:
3.55/1.70	Q DP problem:
3.55/1.70	The TRS P consists of the following rules:
3.55/1.70	
3.55/1.70	   F(x, c(y)) -> F(y, y)
3.55/1.70	
3.55/1.70	R is empty.
3.55/1.70	The set Q consists of the following terms:
3.55/1.70	
3.55/1.70	   f(x0, c(x1))
3.55/1.70	   f(s(x0), s(x1))
3.55/1.70	
3.55/1.70	We have to consider all minimal (P,Q,R)-chains.
3.55/1.70	----------------------------------------
3.55/1.70	
3.55/1.70	(15) QReductionProof (EQUIVALENT)
3.55/1.70	We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].
3.55/1.70	
3.55/1.70	   f(x0, c(x1))
3.55/1.70	   f(s(x0), s(x1))
3.55/1.70	
3.55/1.70	
3.55/1.70	----------------------------------------
3.55/1.70	
3.55/1.70	(16)
3.55/1.70	Obligation:
3.55/1.70	Q DP problem:
3.55/1.70	The TRS P consists of the following rules:
3.55/1.70	
3.55/1.70	   F(x, c(y)) -> F(y, y)
3.55/1.70	
3.55/1.70	R is empty.
3.55/1.70	Q is empty.
3.55/1.70	We have to consider all minimal (P,Q,R)-chains.
3.55/1.70	----------------------------------------
3.55/1.70	
3.55/1.70	(17) QDPSizeChangeProof (EQUIVALENT)
3.55/1.70	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. 
3.55/1.70	
3.55/1.70	From the DPs we obtained the following set of size-change graphs:
3.55/1.70	*F(x, c(y)) -> F(y, y)
3.55/1.70	The graph contains the following edges 2 > 1, 2 > 2
3.55/1.70	
3.55/1.70	
3.55/1.70	----------------------------------------
3.55/1.70	
3.55/1.70	(18)
3.55/1.70	YES
3.55/1.72	EOF