3.57/1.76	YES
3.57/1.77	proof of /export/starexec/sandbox/benchmark/theBenchmark.xml
3.57/1.77	# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty
3.57/1.77	
3.57/1.77	
3.57/1.77	Termination w.r.t. Q of the given QTRS could be proven:
3.57/1.77	
3.57/1.77	(0) QTRS
3.57/1.77	(1) DependencyPairsProof [EQUIVALENT, 0 ms]
3.57/1.77	(2) QDP
3.57/1.77	(3) DependencyGraphProof [EQUIVALENT, 2 ms]
3.57/1.77	(4) AND
3.57/1.77	    (5) QDP
3.57/1.77	        (6) UsableRulesProof [EQUIVALENT, 0 ms]
3.57/1.77	        (7) QDP
3.57/1.77	        (8) QReductionProof [EQUIVALENT, 0 ms]
3.57/1.77	        (9) QDP
3.57/1.77	        (10) QDPSizeChangeProof [EQUIVALENT, 0 ms]
3.57/1.77	        (11) YES
3.57/1.77	    (12) QDP
3.57/1.77	        (13) UsableRulesProof [EQUIVALENT, 0 ms]
3.57/1.77	        (14) QDP
3.57/1.77	        (15) QReductionProof [EQUIVALENT, 0 ms]
3.57/1.77	        (16) QDP
3.57/1.77	        (17) MRRProof [EQUIVALENT, 0 ms]
3.57/1.77	        (18) QDP
3.57/1.77	        (19) DependencyGraphProof [EQUIVALENT, 0 ms]
3.57/1.77	        (20) TRUE
3.57/1.77	
3.57/1.77	
3.57/1.77	----------------------------------------
3.57/1.77	
3.57/1.77	(0)
3.57/1.77	Obligation:
3.57/1.77	Q restricted rewrite system:
3.57/1.77	The TRS R consists of the following rules:
3.57/1.77	
3.57/1.77	   p(0) -> 0
3.57/1.77	   p(s(x)) -> x
3.57/1.77	   le(0, y) -> true
3.57/1.77	   le(s(x), 0) -> false
3.57/1.77	   le(s(x), s(y)) -> le(x, y)
3.57/1.77	   minus(x, 0) -> x
3.57/1.77	   minus(x, s(y)) -> if(le(x, s(y)), 0, p(minus(x, p(s(y)))))
3.57/1.77	   if(true, x, y) -> x
3.57/1.77	   if(false, x, y) -> y
3.57/1.77	
3.57/1.77	The set Q consists of the following terms:
3.57/1.77	
3.57/1.77	   p(0)
3.57/1.77	   p(s(x0))
3.57/1.77	   le(0, x0)
3.57/1.77	   le(s(x0), 0)
3.57/1.77	   le(s(x0), s(x1))
3.57/1.77	   minus(x0, 0)
3.57/1.77	   minus(x0, s(x1))
3.57/1.77	   if(true, x0, x1)
3.57/1.77	   if(false, x0, x1)
3.57/1.77	
3.57/1.77	
3.57/1.77	----------------------------------------
3.57/1.77	
3.57/1.77	(1) DependencyPairsProof (EQUIVALENT)
3.57/1.77	Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.
3.57/1.77	----------------------------------------
3.57/1.77	
3.57/1.77	(2)
3.57/1.77	Obligation:
3.57/1.77	Q DP problem:
3.57/1.77	The TRS P consists of the following rules:
3.57/1.77	
3.57/1.77	   LE(s(x), s(y)) -> LE(x, y)
3.57/1.77	   MINUS(x, s(y)) -> IF(le(x, s(y)), 0, p(minus(x, p(s(y)))))
3.57/1.77	   MINUS(x, s(y)) -> LE(x, s(y))
3.57/1.77	   MINUS(x, s(y)) -> P(minus(x, p(s(y))))
3.57/1.77	   MINUS(x, s(y)) -> MINUS(x, p(s(y)))
3.57/1.77	   MINUS(x, s(y)) -> P(s(y))
3.57/1.77	
3.57/1.77	The TRS R consists of the following rules:
3.57/1.77	
3.57/1.77	   p(0) -> 0
3.57/1.77	   p(s(x)) -> x
3.57/1.77	   le(0, y) -> true
3.57/1.77	   le(s(x), 0) -> false
3.57/1.77	   le(s(x), s(y)) -> le(x, y)
3.57/1.77	   minus(x, 0) -> x
3.57/1.77	   minus(x, s(y)) -> if(le(x, s(y)), 0, p(minus(x, p(s(y)))))
3.57/1.77	   if(true, x, y) -> x
3.57/1.77	   if(false, x, y) -> y
3.57/1.77	
3.57/1.77	The set Q consists of the following terms:
3.57/1.77	
3.57/1.77	   p(0)
3.57/1.77	   p(s(x0))
3.57/1.77	   le(0, x0)
3.57/1.77	   le(s(x0), 0)
3.57/1.77	   le(s(x0), s(x1))
3.57/1.77	   minus(x0, 0)
3.57/1.77	   minus(x0, s(x1))
3.57/1.77	   if(true, x0, x1)
3.57/1.77	   if(false, x0, x1)
3.57/1.77	
3.57/1.77	We have to consider all minimal (P,Q,R)-chains.
3.57/1.77	----------------------------------------
3.57/1.77	
3.57/1.77	(3) DependencyGraphProof (EQUIVALENT)
3.57/1.77	The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 2 SCCs with 4 less nodes.
3.57/1.77	----------------------------------------
3.57/1.77	
3.57/1.77	(4)
3.57/1.77	Complex Obligation (AND)
3.57/1.77	
3.57/1.77	----------------------------------------
3.57/1.77	
3.57/1.77	(5)
3.57/1.77	Obligation:
3.57/1.77	Q DP problem:
3.57/1.77	The TRS P consists of the following rules:
3.57/1.77	
3.57/1.77	   LE(s(x), s(y)) -> LE(x, y)
3.57/1.77	
3.57/1.77	The TRS R consists of the following rules:
3.57/1.77	
3.57/1.77	   p(0) -> 0
3.57/1.77	   p(s(x)) -> x
3.57/1.77	   le(0, y) -> true
3.57/1.77	   le(s(x), 0) -> false
3.57/1.77	   le(s(x), s(y)) -> le(x, y)
3.57/1.77	   minus(x, 0) -> x
3.57/1.77	   minus(x, s(y)) -> if(le(x, s(y)), 0, p(minus(x, p(s(y)))))
3.57/1.77	   if(true, x, y) -> x
3.57/1.77	   if(false, x, y) -> y
3.57/1.77	
3.57/1.77	The set Q consists of the following terms:
3.57/1.77	
3.57/1.77	   p(0)
3.57/1.77	   p(s(x0))
3.57/1.77	   le(0, x0)
3.57/1.77	   le(s(x0), 0)
3.57/1.77	   le(s(x0), s(x1))
3.57/1.77	   minus(x0, 0)
3.57/1.77	   minus(x0, s(x1))
3.57/1.77	   if(true, x0, x1)
3.57/1.77	   if(false, x0, x1)
3.57/1.77	
3.57/1.77	We have to consider all minimal (P,Q,R)-chains.
3.57/1.77	----------------------------------------
3.57/1.77	
3.57/1.77	(6) UsableRulesProof (EQUIVALENT)
3.57/1.77	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.
3.57/1.77	----------------------------------------
3.57/1.77	
3.57/1.77	(7)
3.57/1.77	Obligation:
3.57/1.77	Q DP problem:
3.57/1.77	The TRS P consists of the following rules:
3.57/1.77	
3.57/1.77	   LE(s(x), s(y)) -> LE(x, y)
3.57/1.77	
3.57/1.77	R is empty.
3.57/1.77	The set Q consists of the following terms:
3.57/1.77	
3.57/1.77	   p(0)
3.57/1.77	   p(s(x0))
3.57/1.77	   le(0, x0)
3.57/1.77	   le(s(x0), 0)
3.57/1.77	   le(s(x0), s(x1))
3.57/1.77	   minus(x0, 0)
3.57/1.77	   minus(x0, s(x1))
3.57/1.77	   if(true, x0, x1)
3.57/1.77	   if(false, x0, x1)
3.57/1.77	
3.57/1.77	We have to consider all minimal (P,Q,R)-chains.
3.57/1.77	----------------------------------------
3.57/1.77	
3.57/1.77	(8) QReductionProof (EQUIVALENT)
3.57/1.77	We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].
3.57/1.77	
3.57/1.77	   p(0)
3.57/1.77	   p(s(x0))
3.57/1.77	   le(0, x0)
3.57/1.77	   le(s(x0), 0)
3.57/1.77	   le(s(x0), s(x1))
3.57/1.77	   minus(x0, 0)
3.57/1.77	   minus(x0, s(x1))
3.57/1.77	   if(true, x0, x1)
3.57/1.77	   if(false, x0, x1)
3.57/1.77	
3.57/1.77	
3.57/1.77	----------------------------------------
3.57/1.77	
3.57/1.77	(9)
3.57/1.77	Obligation:
3.57/1.77	Q DP problem:
3.57/1.77	The TRS P consists of the following rules:
3.57/1.77	
3.57/1.77	   LE(s(x), s(y)) -> LE(x, y)
3.57/1.77	
3.57/1.77	R is empty.
3.57/1.77	Q is empty.
3.57/1.77	We have to consider all minimal (P,Q,R)-chains.
3.57/1.77	----------------------------------------
3.57/1.77	
3.57/1.77	(10) QDPSizeChangeProof (EQUIVALENT)
3.57/1.77	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. 
3.57/1.77	
3.57/1.77	From the DPs we obtained the following set of size-change graphs:
3.57/1.77	*LE(s(x), s(y)) -> LE(x, y)
3.57/1.77	The graph contains the following edges 1 > 1, 2 > 2
3.57/1.77	
3.57/1.77	
3.57/1.77	----------------------------------------
3.57/1.77	
3.57/1.77	(11)
3.57/1.77	YES
3.57/1.77	
3.57/1.77	----------------------------------------
3.57/1.77	
3.57/1.77	(12)
3.57/1.77	Obligation:
3.57/1.77	Q DP problem:
3.57/1.77	The TRS P consists of the following rules:
3.57/1.77	
3.57/1.77	   MINUS(x, s(y)) -> MINUS(x, p(s(y)))
3.57/1.77	
3.57/1.77	The TRS R consists of the following rules:
3.57/1.77	
3.57/1.77	   p(0) -> 0
3.57/1.77	   p(s(x)) -> x
3.57/1.77	   le(0, y) -> true
3.57/1.77	   le(s(x), 0) -> false
3.57/1.77	   le(s(x), s(y)) -> le(x, y)
3.57/1.77	   minus(x, 0) -> x
3.57/1.77	   minus(x, s(y)) -> if(le(x, s(y)), 0, p(minus(x, p(s(y)))))
3.57/1.77	   if(true, x, y) -> x
3.57/1.77	   if(false, x, y) -> y
3.57/1.77	
3.57/1.77	The set Q consists of the following terms:
3.57/1.77	
3.57/1.77	   p(0)
3.57/1.77	   p(s(x0))
3.57/1.77	   le(0, x0)
3.57/1.77	   le(s(x0), 0)
3.57/1.77	   le(s(x0), s(x1))
3.57/1.77	   minus(x0, 0)
3.57/1.77	   minus(x0, s(x1))
3.57/1.77	   if(true, x0, x1)
3.57/1.77	   if(false, x0, x1)
3.57/1.77	
3.57/1.77	We have to consider all minimal (P,Q,R)-chains.
3.57/1.77	----------------------------------------
3.57/1.77	
3.57/1.77	(13) UsableRulesProof (EQUIVALENT)
3.57/1.77	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.
3.57/1.77	----------------------------------------
3.57/1.77	
3.57/1.77	(14)
3.57/1.77	Obligation:
3.57/1.77	Q DP problem:
3.57/1.77	The TRS P consists of the following rules:
3.57/1.77	
3.57/1.77	   MINUS(x, s(y)) -> MINUS(x, p(s(y)))
3.57/1.77	
3.57/1.77	The TRS R consists of the following rules:
3.57/1.77	
3.57/1.77	   p(s(x)) -> x
3.57/1.77	
3.57/1.77	The set Q consists of the following terms:
3.57/1.77	
3.57/1.77	   p(0)
3.57/1.77	   p(s(x0))
3.57/1.77	   le(0, x0)
3.57/1.77	   le(s(x0), 0)
3.57/1.77	   le(s(x0), s(x1))
3.57/1.77	   minus(x0, 0)
3.57/1.77	   minus(x0, s(x1))
3.57/1.77	   if(true, x0, x1)
3.57/1.77	   if(false, x0, x1)
3.57/1.77	
3.57/1.77	We have to consider all minimal (P,Q,R)-chains.
3.57/1.77	----------------------------------------
3.57/1.77	
3.57/1.77	(15) QReductionProof (EQUIVALENT)
3.57/1.77	We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].
3.57/1.77	
3.57/1.77	   le(0, x0)
3.57/1.77	   le(s(x0), 0)
3.57/1.77	   le(s(x0), s(x1))
3.57/1.77	   minus(x0, 0)
3.57/1.77	   minus(x0, s(x1))
3.57/1.77	   if(true, x0, x1)
3.57/1.77	   if(false, x0, x1)
3.57/1.77	
3.57/1.77	
3.57/1.77	----------------------------------------
3.57/1.77	
3.57/1.77	(16)
3.57/1.77	Obligation:
3.57/1.77	Q DP problem:
3.57/1.77	The TRS P consists of the following rules:
3.57/1.77	
3.57/1.77	   MINUS(x, s(y)) -> MINUS(x, p(s(y)))
3.57/1.77	
3.57/1.77	The TRS R consists of the following rules:
3.57/1.77	
3.57/1.77	   p(s(x)) -> x
3.57/1.77	
3.57/1.77	The set Q consists of the following terms:
3.57/1.77	
3.57/1.77	   p(0)
3.57/1.77	   p(s(x0))
3.57/1.77	
3.57/1.77	We have to consider all minimal (P,Q,R)-chains.
3.57/1.77	----------------------------------------
3.57/1.77	
3.57/1.77	(17) MRRProof (EQUIVALENT)
3.57/1.77	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.
3.57/1.77	
3.57/1.77	
3.57/1.77	Strictly oriented rules of the TRS R:
3.57/1.77	
3.57/1.77	   p(s(x)) -> x
3.57/1.77	
3.57/1.77	Used ordering: Polynomial interpretation [POLO]:
3.57/1.77	
3.57/1.77	   POL(MINUS(x_1, x_2)) = x_1 + 2*x_2
3.57/1.77	   POL(p(x_1)) = x_1
3.57/1.77	   POL(s(x_1)) = 2 + x_1
3.57/1.77	
3.57/1.77	
3.57/1.77	----------------------------------------
3.57/1.77	
3.57/1.77	(18)
3.57/1.77	Obligation:
3.57/1.77	Q DP problem:
3.57/1.77	The TRS P consists of the following rules:
3.57/1.77	
3.57/1.77	   MINUS(x, s(y)) -> MINUS(x, p(s(y)))
3.57/1.77	
3.57/1.77	R is empty.
3.57/1.77	The set Q consists of the following terms:
3.57/1.77	
3.57/1.77	   p(0)
3.57/1.77	   p(s(x0))
3.57/1.77	
3.57/1.77	We have to consider all minimal (P,Q,R)-chains.
3.57/1.77	----------------------------------------
3.57/1.77	
3.57/1.77	(19) DependencyGraphProof (EQUIVALENT)
3.57/1.77	The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 1 less node.
3.57/1.77	----------------------------------------
3.57/1.77	
3.57/1.77	(20)
3.57/1.77	TRUE
3.69/1.81	EOF