4.01/1.78	YES
4.01/1.79	proof of /export/starexec/sandbox/benchmark/theBenchmark.xml
4.01/1.79	# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty
4.01/1.79	
4.01/1.79	
4.01/1.79	Termination w.r.t. Q of the given QTRS could be proven:
4.01/1.79	
4.01/1.79	(0) QTRS
4.01/1.79	(1) DependencyPairsProof [EQUIVALENT, 0 ms]
4.01/1.79	(2) QDP
4.01/1.79	(3) DependencyGraphProof [EQUIVALENT, 0 ms]
4.01/1.79	(4) AND
4.01/1.79	    (5) QDP
4.01/1.79	        (6) UsableRulesProof [EQUIVALENT, 1 ms]
4.01/1.79	        (7) QDP
4.01/1.79	        (8) QReductionProof [EQUIVALENT, 0 ms]
4.01/1.79	        (9) QDP
4.01/1.79	        (10) QDPSizeChangeProof [EQUIVALENT, 0 ms]
4.01/1.79	        (11) YES
4.01/1.79	    (12) QDP
4.01/1.79	        (13) UsableRulesProof [EQUIVALENT, 0 ms]
4.01/1.79	        (14) QDP
4.01/1.79	        (15) QReductionProof [EQUIVALENT, 0 ms]
4.01/1.79	        (16) QDP
4.01/1.79	        (17) QDPOrderProof [EQUIVALENT, 0 ms]
4.01/1.79	        (18) QDP
4.01/1.79	        (19) QDPOrderProof [EQUIVALENT, 0 ms]
4.01/1.79	        (20) QDP
4.01/1.79	        (21) PisEmptyProof [EQUIVALENT, 0 ms]
4.01/1.79	        (22) YES
4.01/1.79	
4.01/1.79	
4.01/1.79	----------------------------------------
4.01/1.79	
4.01/1.79	(0)
4.01/1.79	Obligation:
4.01/1.79	Q restricted rewrite system:
4.01/1.79	The TRS R consists of the following rules:
4.01/1.79	
4.01/1.79	   quot(0, s(y), s(z)) -> 0
4.01/1.79	   quot(s(x), s(y), z) -> quot(x, y, z)
4.01/1.79	   plus(0, y) -> y
4.01/1.79	   plus(s(x), y) -> s(plus(x, y))
4.01/1.79	   quot(x, 0, s(z)) -> s(quot(x, plus(z, s(0)), s(z)))
4.01/1.79	
4.01/1.79	The set Q consists of the following terms:
4.01/1.79	
4.01/1.79	   quot(0, s(x0), s(x1))
4.01/1.79	   quot(s(x0), s(x1), x2)
4.01/1.79	   plus(0, x0)
4.01/1.79	   plus(s(x0), x1)
4.01/1.79	   quot(x0, 0, s(x1))
4.01/1.79	
4.01/1.79	
4.01/1.79	----------------------------------------
4.01/1.79	
4.01/1.79	(1) DependencyPairsProof (EQUIVALENT)
4.01/1.79	Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.
4.01/1.79	----------------------------------------
4.01/1.79	
4.01/1.79	(2)
4.01/1.79	Obligation:
4.01/1.79	Q DP problem:
4.01/1.79	The TRS P consists of the following rules:
4.01/1.79	
4.01/1.79	   QUOT(s(x), s(y), z) -> QUOT(x, y, z)
4.01/1.79	   PLUS(s(x), y) -> PLUS(x, y)
4.01/1.79	   QUOT(x, 0, s(z)) -> QUOT(x, plus(z, s(0)), s(z))
4.01/1.79	   QUOT(x, 0, s(z)) -> PLUS(z, s(0))
4.01/1.79	
4.01/1.79	The TRS R consists of the following rules:
4.01/1.79	
4.01/1.79	   quot(0, s(y), s(z)) -> 0
4.01/1.79	   quot(s(x), s(y), z) -> quot(x, y, z)
4.01/1.79	   plus(0, y) -> y
4.01/1.79	   plus(s(x), y) -> s(plus(x, y))
4.01/1.79	   quot(x, 0, s(z)) -> s(quot(x, plus(z, s(0)), s(z)))
4.01/1.79	
4.01/1.79	The set Q consists of the following terms:
4.01/1.79	
4.01/1.79	   quot(0, s(x0), s(x1))
4.01/1.79	   quot(s(x0), s(x1), x2)
4.01/1.79	   plus(0, x0)
4.01/1.79	   plus(s(x0), x1)
4.01/1.79	   quot(x0, 0, s(x1))
4.01/1.79	
4.01/1.79	We have to consider all minimal (P,Q,R)-chains.
4.01/1.79	----------------------------------------
4.01/1.79	
4.01/1.79	(3) DependencyGraphProof (EQUIVALENT)
4.01/1.79	The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 2 SCCs with 1 less node.
4.01/1.79	----------------------------------------
4.01/1.79	
4.01/1.79	(4)
4.01/1.79	Complex Obligation (AND)
4.01/1.79	
4.01/1.79	----------------------------------------
4.01/1.79	
4.01/1.79	(5)
4.01/1.79	Obligation:
4.01/1.79	Q DP problem:
4.01/1.79	The TRS P consists of the following rules:
4.01/1.79	
4.01/1.79	   PLUS(s(x), y) -> PLUS(x, y)
4.01/1.79	
4.01/1.79	The TRS R consists of the following rules:
4.01/1.79	
4.01/1.79	   quot(0, s(y), s(z)) -> 0
4.01/1.79	   quot(s(x), s(y), z) -> quot(x, y, z)
4.01/1.79	   plus(0, y) -> y
4.01/1.79	   plus(s(x), y) -> s(plus(x, y))
4.01/1.79	   quot(x, 0, s(z)) -> s(quot(x, plus(z, s(0)), s(z)))
4.01/1.79	
4.01/1.79	The set Q consists of the following terms:
4.01/1.79	
4.01/1.79	   quot(0, s(x0), s(x1))
4.01/1.79	   quot(s(x0), s(x1), x2)
4.01/1.79	   plus(0, x0)
4.01/1.79	   plus(s(x0), x1)
4.01/1.79	   quot(x0, 0, s(x1))
4.01/1.79	
4.01/1.79	We have to consider all minimal (P,Q,R)-chains.
4.01/1.79	----------------------------------------
4.01/1.79	
4.01/1.79	(6) UsableRulesProof (EQUIVALENT)
4.01/1.79	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.
4.01/1.79	----------------------------------------
4.01/1.79	
4.01/1.79	(7)
4.01/1.79	Obligation:
4.01/1.79	Q DP problem:
4.01/1.79	The TRS P consists of the following rules:
4.01/1.79	
4.01/1.79	   PLUS(s(x), y) -> PLUS(x, y)
4.01/1.79	
4.01/1.79	R is empty.
4.01/1.79	The set Q consists of the following terms:
4.01/1.79	
4.01/1.79	   quot(0, s(x0), s(x1))
4.01/1.79	   quot(s(x0), s(x1), x2)
4.01/1.79	   plus(0, x0)
4.01/1.79	   plus(s(x0), x1)
4.01/1.79	   quot(x0, 0, s(x1))
4.01/1.79	
4.01/1.79	We have to consider all minimal (P,Q,R)-chains.
4.01/1.79	----------------------------------------
4.01/1.79	
4.01/1.79	(8) QReductionProof (EQUIVALENT)
4.01/1.79	We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].
4.01/1.79	
4.01/1.79	   quot(0, s(x0), s(x1))
4.01/1.79	   quot(s(x0), s(x1), x2)
4.01/1.79	   plus(0, x0)
4.01/1.79	   plus(s(x0), x1)
4.01/1.79	   quot(x0, 0, s(x1))
4.01/1.79	
4.01/1.79	
4.01/1.79	----------------------------------------
4.01/1.79	
4.01/1.79	(9)
4.01/1.79	Obligation:
4.01/1.79	Q DP problem:
4.01/1.79	The TRS P consists of the following rules:
4.01/1.79	
4.01/1.79	   PLUS(s(x), y) -> PLUS(x, y)
4.01/1.79	
4.01/1.79	R is empty.
4.01/1.79	Q is empty.
4.01/1.79	We have to consider all minimal (P,Q,R)-chains.
4.01/1.79	----------------------------------------
4.01/1.79	
4.01/1.79	(10) QDPSizeChangeProof (EQUIVALENT)
4.01/1.79	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. 
4.01/1.79	
4.01/1.79	From the DPs we obtained the following set of size-change graphs:
4.01/1.79	*PLUS(s(x), y) -> PLUS(x, y)
4.01/1.79	The graph contains the following edges 1 > 1, 2 >= 2
4.01/1.79	
4.01/1.79	
4.01/1.79	----------------------------------------
4.01/1.79	
4.01/1.79	(11)
4.01/1.79	YES
4.01/1.79	
4.01/1.79	----------------------------------------
4.01/1.79	
4.01/1.79	(12)
4.01/1.79	Obligation:
4.01/1.79	Q DP problem:
4.01/1.79	The TRS P consists of the following rules:
4.01/1.79	
4.01/1.79	   QUOT(x, 0, s(z)) -> QUOT(x, plus(z, s(0)), s(z))
4.01/1.79	   QUOT(s(x), s(y), z) -> QUOT(x, y, z)
4.01/1.79	
4.01/1.79	The TRS R consists of the following rules:
4.01/1.79	
4.01/1.79	   quot(0, s(y), s(z)) -> 0
4.01/1.79	   quot(s(x), s(y), z) -> quot(x, y, z)
4.01/1.79	   plus(0, y) -> y
4.01/1.79	   plus(s(x), y) -> s(plus(x, y))
4.01/1.79	   quot(x, 0, s(z)) -> s(quot(x, plus(z, s(0)), s(z)))
4.01/1.79	
4.01/1.79	The set Q consists of the following terms:
4.01/1.79	
4.01/1.79	   quot(0, s(x0), s(x1))
4.01/1.79	   quot(s(x0), s(x1), x2)
4.01/1.79	   plus(0, x0)
4.01/1.79	   plus(s(x0), x1)
4.01/1.79	   quot(x0, 0, s(x1))
4.01/1.79	
4.01/1.79	We have to consider all minimal (P,Q,R)-chains.
4.01/1.79	----------------------------------------
4.01/1.79	
4.01/1.79	(13) UsableRulesProof (EQUIVALENT)
4.01/1.79	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.
4.01/1.79	----------------------------------------
4.01/1.79	
4.01/1.79	(14)
4.01/1.79	Obligation:
4.01/1.79	Q DP problem:
4.01/1.79	The TRS P consists of the following rules:
4.01/1.79	
4.01/1.79	   QUOT(x, 0, s(z)) -> QUOT(x, plus(z, s(0)), s(z))
4.01/1.79	   QUOT(s(x), s(y), z) -> QUOT(x, y, z)
4.01/1.79	
4.01/1.79	The TRS R consists of the following rules:
4.01/1.79	
4.01/1.79	   plus(0, y) -> y
4.01/1.79	   plus(s(x), y) -> s(plus(x, y))
4.01/1.79	
4.01/1.79	The set Q consists of the following terms:
4.01/1.79	
4.01/1.79	   quot(0, s(x0), s(x1))
4.01/1.79	   quot(s(x0), s(x1), x2)
4.01/1.79	   plus(0, x0)
4.01/1.79	   plus(s(x0), x1)
4.01/1.79	   quot(x0, 0, s(x1))
4.01/1.79	
4.01/1.79	We have to consider all minimal (P,Q,R)-chains.
4.01/1.79	----------------------------------------
4.01/1.79	
4.01/1.79	(15) QReductionProof (EQUIVALENT)
4.01/1.79	We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].
4.01/1.79	
4.01/1.79	   quot(0, s(x0), s(x1))
4.01/1.79	   quot(s(x0), s(x1), x2)
4.01/1.79	   quot(x0, 0, s(x1))
4.01/1.79	
4.01/1.79	
4.01/1.79	----------------------------------------
4.01/1.79	
4.01/1.79	(16)
4.01/1.79	Obligation:
4.01/1.79	Q DP problem:
4.01/1.79	The TRS P consists of the following rules:
4.01/1.79	
4.01/1.79	   QUOT(x, 0, s(z)) -> QUOT(x, plus(z, s(0)), s(z))
4.01/1.79	   QUOT(s(x), s(y), z) -> QUOT(x, y, z)
4.01/1.79	
4.01/1.79	The TRS R consists of the following rules:
4.01/1.79	
4.01/1.79	   plus(0, y) -> y
4.01/1.79	   plus(s(x), y) -> s(plus(x, y))
4.01/1.79	
4.01/1.79	The set Q consists of the following terms:
4.01/1.79	
4.01/1.79	   plus(0, x0)
4.01/1.79	   plus(s(x0), x1)
4.01/1.79	
4.01/1.79	We have to consider all minimal (P,Q,R)-chains.
4.01/1.79	----------------------------------------
4.01/1.79	
4.01/1.79	(17) QDPOrderProof (EQUIVALENT)
4.01/1.79	We use the reduction pair processor [LPAR04,JAR06].
4.01/1.79	
4.01/1.79	
4.01/1.79	The following pairs can be oriented strictly and are deleted.
4.01/1.79	
4.01/1.79	   QUOT(s(x), s(y), z) -> QUOT(x, y, z)
4.01/1.79	The remaining pairs can at least be oriented weakly.
4.01/1.79	Used ordering:  Combined order from the following AFS and order.
4.01/1.79	QUOT(x1, x2, x3)  =  x1
4.01/1.79	
4.01/1.79	s(x1)  =  s(x1)
4.01/1.79	
4.01/1.79	
4.01/1.79	Knuth-Bendix order [KBO] with precedence:trivial
4.01/1.79	
4.01/1.79	and weight map:
4.01/1.79	
4.01/1.79	   s_1=1
4.01/1.79	   dummyConstant=1
4.01/1.79	
4.01/1.79	The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:
4.01/1.79	none
4.01/1.79	
4.01/1.79	
4.01/1.79	----------------------------------------
4.01/1.79	
4.01/1.79	(18)
4.01/1.79	Obligation:
4.01/1.79	Q DP problem:
4.01/1.79	The TRS P consists of the following rules:
4.01/1.79	
4.01/1.79	   QUOT(x, 0, s(z)) -> QUOT(x, plus(z, s(0)), s(z))
4.01/1.79	
4.01/1.79	The TRS R consists of the following rules:
4.01/1.79	
4.01/1.79	   plus(0, y) -> y
4.01/1.79	   plus(s(x), y) -> s(plus(x, y))
4.01/1.79	
4.01/1.79	The set Q consists of the following terms:
4.01/1.79	
4.01/1.79	   plus(0, x0)
4.01/1.79	   plus(s(x0), x1)
4.01/1.79	
4.01/1.79	We have to consider all minimal (P,Q,R)-chains.
4.01/1.79	----------------------------------------
4.01/1.79	
4.01/1.79	(19) QDPOrderProof (EQUIVALENT)
4.01/1.79	We use the reduction pair processor [LPAR04,JAR06].
4.01/1.79	
4.01/1.79	
4.01/1.79	The following pairs can be oriented strictly and are deleted.
4.01/1.79	
4.01/1.79	   QUOT(x, 0, s(z)) -> QUOT(x, plus(z, s(0)), s(z))
4.01/1.79	The remaining pairs can at least be oriented weakly.
4.01/1.79	Used ordering:  Combined order from the following AFS and order.
4.01/1.79	QUOT(x1, x2, x3)  =  x2
4.01/1.79	
4.01/1.79	0  =  0
4.01/1.79	
4.01/1.79	plus(x1, x2)  =  x2
4.01/1.79	
4.01/1.79	s(x1)  =  s
4.01/1.79	
4.01/1.79	
4.01/1.79	Knuth-Bendix order [KBO] with precedence:trivial
4.01/1.79	
4.01/1.79	and weight map:
4.01/1.79	
4.01/1.79	   s=1
4.01/1.79	   0=2
4.01/1.79	
4.01/1.79	The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:
4.01/1.79	
4.01/1.79	   plus(0, y) -> y
4.01/1.79	   plus(s(x), y) -> s(plus(x, y))
4.01/1.79	
4.01/1.79	
4.01/1.79	----------------------------------------
4.01/1.79	
4.01/1.79	(20)
4.01/1.79	Obligation:
4.01/1.79	Q DP problem:
4.01/1.79	P is empty.
4.01/1.79	The TRS R consists of the following rules:
4.01/1.79	
4.01/1.79	   plus(0, y) -> y
4.01/1.79	   plus(s(x), y) -> s(plus(x, y))
4.01/1.79	
4.01/1.79	The set Q consists of the following terms:
4.01/1.79	
4.01/1.79	   plus(0, x0)
4.01/1.79	   plus(s(x0), x1)
4.01/1.79	
4.01/1.79	We have to consider all minimal (P,Q,R)-chains.
4.01/1.79	----------------------------------------
4.01/1.79	
4.01/1.79	(21) PisEmptyProof (EQUIVALENT)
4.01/1.79	The TRS P is empty. Hence, there is no (P,Q,R) chain.
4.01/1.79	----------------------------------------
4.01/1.79	
4.01/1.79	(22)
4.01/1.79	YES
4.08/1.82	EOF