3.61/1.71	YES
3.61/1.72	proof of /export/starexec/sandbox/benchmark/theBenchmark.xml
3.61/1.72	# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty
3.61/1.72	
3.61/1.72	
3.61/1.72	Termination w.r.t. Q of the given QTRS could be proven:
3.61/1.72	
3.61/1.72	(0) QTRS
3.61/1.72	(1) DependencyPairsProof [EQUIVALENT, 19 ms]
3.61/1.72	(2) QDP
3.61/1.72	(3) DependencyGraphProof [EQUIVALENT, 0 ms]
3.61/1.72	(4) AND
3.61/1.72	    (5) QDP
3.61/1.72	        (6) UsableRulesProof [EQUIVALENT, 0 ms]
3.61/1.72	        (7) QDP
3.61/1.72	        (8) QReductionProof [EQUIVALENT, 0 ms]
3.61/1.72	        (9) QDP
3.61/1.72	        (10) QDPSizeChangeProof [EQUIVALENT, 0 ms]
3.61/1.72	        (11) YES
3.61/1.72	    (12) QDP
3.61/1.72	        (13) UsableRulesProof [EQUIVALENT, 0 ms]
3.61/1.72	        (14) QDP
3.61/1.72	        (15) QReductionProof [EQUIVALENT, 0 ms]
3.61/1.72	        (16) QDP
3.61/1.72	        (17) QDPSizeChangeProof [EQUIVALENT, 0 ms]
3.61/1.72	        (18) YES
3.61/1.72	
3.61/1.72	
3.61/1.72	----------------------------------------
3.61/1.72	
3.61/1.72	(0)
3.61/1.72	Obligation:
3.61/1.72	Q restricted rewrite system:
3.61/1.72	The TRS R consists of the following rules:
3.61/1.72	
3.61/1.72	   f(x, c(x), c(y)) -> f(y, y, f(y, x, y))
3.61/1.72	   f(s(x), y, z) -> f(x, s(c(y)), c(z))
3.61/1.72	   f(c(x), x, y) -> c(y)
3.61/1.72	   g(x, y) -> x
3.61/1.72	   g(x, y) -> y
3.61/1.72	
3.61/1.72	The set Q consists of the following terms:
3.61/1.72	
3.61/1.72	   f(x0, c(x0), c(x1))
3.61/1.72	   f(s(x0), x1, x2)
3.61/1.72	   f(c(x0), x0, x1)
3.61/1.72	   g(x0, x1)
3.61/1.72	
3.61/1.72	
3.61/1.72	----------------------------------------
3.61/1.72	
3.61/1.72	(1) DependencyPairsProof (EQUIVALENT)
3.61/1.72	Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.
3.61/1.72	----------------------------------------
3.61/1.72	
3.61/1.72	(2)
3.61/1.72	Obligation:
3.61/1.72	Q DP problem:
3.61/1.72	The TRS P consists of the following rules:
3.61/1.72	
3.61/1.72	   F(x, c(x), c(y)) -> F(y, y, f(y, x, y))
3.61/1.72	   F(x, c(x), c(y)) -> F(y, x, y)
3.61/1.72	   F(s(x), y, z) -> F(x, s(c(y)), c(z))
3.61/1.72	
3.61/1.72	The TRS R consists of the following rules:
3.61/1.72	
3.61/1.72	   f(x, c(x), c(y)) -> f(y, y, f(y, x, y))
3.61/1.72	   f(s(x), y, z) -> f(x, s(c(y)), c(z))
3.61/1.72	   f(c(x), x, y) -> c(y)
3.61/1.72	   g(x, y) -> x
3.61/1.72	   g(x, y) -> y
3.61/1.72	
3.61/1.72	The set Q consists of the following terms:
3.61/1.72	
3.61/1.72	   f(x0, c(x0), c(x1))
3.61/1.72	   f(s(x0), x1, x2)
3.61/1.72	   f(c(x0), x0, x1)
3.61/1.72	   g(x0, x1)
3.61/1.72	
3.61/1.72	We have to consider all minimal (P,Q,R)-chains.
3.61/1.72	----------------------------------------
3.61/1.72	
3.61/1.72	(3) DependencyGraphProof (EQUIVALENT)
3.61/1.72	The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 2 SCCs with 1 less node.
3.61/1.72	----------------------------------------
3.61/1.72	
3.61/1.72	(4)
3.61/1.72	Complex Obligation (AND)
3.61/1.72	
3.61/1.72	----------------------------------------
3.61/1.72	
3.61/1.72	(5)
3.61/1.72	Obligation:
3.61/1.72	Q DP problem:
3.61/1.72	The TRS P consists of the following rules:
3.61/1.72	
3.61/1.72	   F(s(x), y, z) -> F(x, s(c(y)), c(z))
3.61/1.72	
3.61/1.72	The TRS R consists of the following rules:
3.61/1.72	
3.61/1.72	   f(x, c(x), c(y)) -> f(y, y, f(y, x, y))
3.61/1.72	   f(s(x), y, z) -> f(x, s(c(y)), c(z))
3.61/1.72	   f(c(x), x, y) -> c(y)
3.61/1.72	   g(x, y) -> x
3.61/1.72	   g(x, y) -> y
3.61/1.72	
3.61/1.72	The set Q consists of the following terms:
3.61/1.72	
3.61/1.72	   f(x0, c(x0), c(x1))
3.61/1.72	   f(s(x0), x1, x2)
3.61/1.72	   f(c(x0), x0, x1)
3.61/1.72	   g(x0, x1)
3.61/1.72	
3.61/1.72	We have to consider all minimal (P,Q,R)-chains.
3.61/1.72	----------------------------------------
3.61/1.72	
3.61/1.72	(6) UsableRulesProof (EQUIVALENT)
3.61/1.72	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.
3.61/1.72	----------------------------------------
3.61/1.72	
3.61/1.72	(7)
3.61/1.72	Obligation:
3.61/1.72	Q DP problem:
3.61/1.72	The TRS P consists of the following rules:
3.61/1.72	
3.61/1.72	   F(s(x), y, z) -> F(x, s(c(y)), c(z))
3.61/1.72	
3.61/1.72	R is empty.
3.61/1.72	The set Q consists of the following terms:
3.61/1.72	
3.61/1.72	   f(x0, c(x0), c(x1))
3.61/1.72	   f(s(x0), x1, x2)
3.61/1.72	   f(c(x0), x0, x1)
3.61/1.72	   g(x0, x1)
3.61/1.72	
3.61/1.72	We have to consider all minimal (P,Q,R)-chains.
3.61/1.72	----------------------------------------
3.61/1.72	
3.61/1.72	(8) QReductionProof (EQUIVALENT)
3.61/1.72	We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].
3.61/1.72	
3.61/1.72	   f(x0, c(x0), c(x1))
3.61/1.72	   f(s(x0), x1, x2)
3.61/1.72	   f(c(x0), x0, x1)
3.61/1.72	   g(x0, x1)
3.61/1.72	
3.61/1.72	
3.61/1.72	----------------------------------------
3.61/1.72	
3.61/1.72	(9)
3.61/1.72	Obligation:
3.61/1.72	Q DP problem:
3.61/1.72	The TRS P consists of the following rules:
3.61/1.72	
3.61/1.72	   F(s(x), y, z) -> F(x, s(c(y)), c(z))
3.61/1.72	
3.61/1.72	R is empty.
3.61/1.72	Q is empty.
3.61/1.72	We have to consider all minimal (P,Q,R)-chains.
3.61/1.72	----------------------------------------
3.61/1.72	
3.61/1.72	(10) QDPSizeChangeProof (EQUIVALENT)
3.61/1.72	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. 
3.61/1.72	
3.61/1.72	From the DPs we obtained the following set of size-change graphs:
3.61/1.72	*F(s(x), y, z) -> F(x, s(c(y)), c(z))
3.61/1.72	The graph contains the following edges 1 > 1
3.61/1.72	
3.61/1.72	
3.61/1.72	----------------------------------------
3.61/1.72	
3.61/1.72	(11)
3.61/1.72	YES
3.61/1.72	
3.61/1.72	----------------------------------------
3.61/1.72	
3.61/1.72	(12)
3.61/1.72	Obligation:
3.61/1.72	Q DP problem:
3.61/1.72	The TRS P consists of the following rules:
3.61/1.72	
3.61/1.72	   F(x, c(x), c(y)) -> F(y, x, y)
3.61/1.72	
3.61/1.72	The TRS R consists of the following rules:
3.61/1.72	
3.61/1.72	   f(x, c(x), c(y)) -> f(y, y, f(y, x, y))
3.61/1.72	   f(s(x), y, z) -> f(x, s(c(y)), c(z))
3.61/1.72	   f(c(x), x, y) -> c(y)
3.61/1.72	   g(x, y) -> x
3.61/1.72	   g(x, y) -> y
3.61/1.72	
3.61/1.72	The set Q consists of the following terms:
3.61/1.72	
3.61/1.72	   f(x0, c(x0), c(x1))
3.61/1.72	   f(s(x0), x1, x2)
3.61/1.72	   f(c(x0), x0, x1)
3.61/1.72	   g(x0, x1)
3.61/1.72	
3.61/1.72	We have to consider all minimal (P,Q,R)-chains.
3.61/1.72	----------------------------------------
3.61/1.72	
3.61/1.72	(13) UsableRulesProof (EQUIVALENT)
3.61/1.72	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.
3.61/1.72	----------------------------------------
3.61/1.72	
3.61/1.72	(14)
3.61/1.72	Obligation:
3.61/1.72	Q DP problem:
3.61/1.72	The TRS P consists of the following rules:
3.61/1.72	
3.61/1.72	   F(x, c(x), c(y)) -> F(y, x, y)
3.61/1.72	
3.61/1.72	R is empty.
3.61/1.72	The set Q consists of the following terms:
3.61/1.72	
3.61/1.72	   f(x0, c(x0), c(x1))
3.61/1.72	   f(s(x0), x1, x2)
3.61/1.72	   f(c(x0), x0, x1)
3.61/1.72	   g(x0, x1)
3.61/1.72	
3.61/1.72	We have to consider all minimal (P,Q,R)-chains.
3.61/1.72	----------------------------------------
3.61/1.72	
3.61/1.72	(15) QReductionProof (EQUIVALENT)
3.61/1.72	We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].
3.61/1.72	
3.61/1.72	   f(x0, c(x0), c(x1))
3.61/1.72	   f(s(x0), x1, x2)
3.61/1.72	   f(c(x0), x0, x1)
3.61/1.72	   g(x0, x1)
3.61/1.72	
3.61/1.72	
3.61/1.72	----------------------------------------
3.61/1.72	
3.61/1.72	(16)
3.61/1.72	Obligation:
3.61/1.72	Q DP problem:
3.61/1.72	The TRS P consists of the following rules:
3.61/1.72	
3.61/1.72	   F(x, c(x), c(y)) -> F(y, x, y)
3.61/1.72	
3.61/1.72	R is empty.
3.61/1.72	Q is empty.
3.61/1.72	We have to consider all minimal (P,Q,R)-chains.
3.61/1.72	----------------------------------------
3.61/1.72	
3.61/1.72	(17) QDPSizeChangeProof (EQUIVALENT)
3.61/1.72	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. 
3.61/1.72	
3.61/1.72	From the DPs we obtained the following set of size-change graphs:
3.61/1.72	*F(x, c(x), c(y)) -> F(y, x, y)
3.61/1.72	The graph contains the following edges 3 > 1, 1 >= 2, 2 > 2, 3 > 3
3.61/1.72	
3.61/1.72	
3.61/1.72	----------------------------------------
3.61/1.72	
3.61/1.72	(18)
3.61/1.72	YES
3.68/1.75	EOF