3.71/1.91	YES
3.85/1.92	proof of /export/starexec/sandbox2/benchmark/theBenchmark.itrs
3.85/1.92	# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty
3.85/1.92	
3.85/1.92	
3.85/1.92	Termination of the given ITRS could be proven:
3.85/1.92	
3.85/1.92	(0) ITRS
3.85/1.92	(1) ITRStoIDPProof [EQUIVALENT, 0 ms]
3.85/1.92	(2) IDP
3.85/1.92	(3) UsableRulesProof [EQUIVALENT, 0 ms]
3.85/1.92	(4) IDP
3.85/1.92	(5) IDPNonInfProof [SOUND, 159 ms]
3.85/1.92	(6) IDP
3.85/1.92	(7) IDependencyGraphProof [EQUIVALENT, 0 ms]
3.85/1.92	(8) TRUE
3.85/1.92	
3.85/1.92	
3.85/1.92	----------------------------------------
3.85/1.92	
3.85/1.92	(0)
3.85/1.92	Obligation:
3.85/1.92	ITRS problem:
3.85/1.92	
3.85/1.92	The following function symbols are pre-defined:
3.85/1.92	<<<
3.85/1.92	 & 	~ 	Bwand: (Integer, Integer) -> Integer
3.85/1.92	 >= 	~ 	Ge: (Integer, Integer) -> Boolean
3.85/1.92	 | 	~ 	Bwor: (Integer, Integer) -> Integer
3.85/1.92	 / 	~ 	Div: (Integer, Integer) -> Integer
3.85/1.92	 != 	~ 	Neq: (Integer, Integer) -> Boolean
3.85/1.92	 && 	~ 	Land: (Boolean, Boolean) -> Boolean
3.85/1.92	 ! 	~ 	Lnot: (Boolean) -> Boolean
3.85/1.92	 = 	~ 	Eq: (Integer, Integer) -> Boolean
3.85/1.92	 <= 	~ 	Le: (Integer, Integer) -> Boolean
3.85/1.92	 ^ 	~ 	Bwxor: (Integer, Integer) -> Integer
3.85/1.92	 % 	~ 	Mod: (Integer, Integer) -> Integer
3.85/1.92	 + 	~ 	Add: (Integer, Integer) -> Integer
3.85/1.92	 > 	~ 	Gt: (Integer, Integer) -> Boolean
3.85/1.92	 -1 	~ 	UnaryMinus: (Integer) -> Integer
3.85/1.92	 < 	~ 	Lt: (Integer, Integer) -> Boolean
3.85/1.92	 || 	~ 	Lor: (Boolean, Boolean) -> Boolean
3.85/1.92	 - 	~ 	Sub: (Integer, Integer) -> Integer
3.85/1.92	 ~ 	~ 	Bwnot: (Integer) -> Integer
3.85/1.92	 * 	~ 	Mul: (Integer, Integer) -> Integer
3.85/1.92	>>>
3.85/1.92	
3.85/1.92	The TRS R consists of the following rules:
3.85/1.92	sum(x, y) -> sif(x >= y, x, y)
3.85/1.92	sif(TRUE, x, y) -> y + sum(x, y + 1)
3.85/1.92	sif(FALSE, x, y) -> 0
3.85/1.92	The set Q consists of the following terms:
3.85/1.92	sum(x0, x1)
3.85/1.92	sif(TRUE, x0, x1)
3.85/1.92	sif(FALSE, x0, x1)
3.85/1.92	
3.85/1.92	----------------------------------------
3.85/1.92	
3.85/1.92	(1) ITRStoIDPProof (EQUIVALENT)
3.85/1.92	Added dependency pairs
3.85/1.92	----------------------------------------
3.85/1.92	
3.85/1.92	(2)
3.85/1.92	Obligation:
3.85/1.92	IDP problem:
3.85/1.92	The following function symbols are pre-defined:
3.85/1.92	<<<
3.85/1.92	 & 	~ 	Bwand: (Integer, Integer) -> Integer
3.85/1.92	 >= 	~ 	Ge: (Integer, Integer) -> Boolean
3.85/1.92	 | 	~ 	Bwor: (Integer, Integer) -> Integer
3.85/1.92	 / 	~ 	Div: (Integer, Integer) -> Integer
3.85/1.92	 != 	~ 	Neq: (Integer, Integer) -> Boolean
3.85/1.92	 && 	~ 	Land: (Boolean, Boolean) -> Boolean
3.85/1.92	 ! 	~ 	Lnot: (Boolean) -> Boolean
3.85/1.92	 = 	~ 	Eq: (Integer, Integer) -> Boolean
3.85/1.92	 <= 	~ 	Le: (Integer, Integer) -> Boolean
3.85/1.92	 ^ 	~ 	Bwxor: (Integer, Integer) -> Integer
3.85/1.92	 % 	~ 	Mod: (Integer, Integer) -> Integer
3.85/1.92	 + 	~ 	Add: (Integer, Integer) -> Integer
3.85/1.92	 > 	~ 	Gt: (Integer, Integer) -> Boolean
3.85/1.92	 -1 	~ 	UnaryMinus: (Integer) -> Integer
3.85/1.92	 < 	~ 	Lt: (Integer, Integer) -> Boolean
3.85/1.92	 || 	~ 	Lor: (Boolean, Boolean) -> Boolean
3.85/1.92	 - 	~ 	Sub: (Integer, Integer) -> Integer
3.85/1.92	 ~ 	~ 	Bwnot: (Integer) -> Integer
3.85/1.92	 * 	~ 	Mul: (Integer, Integer) -> Integer
3.85/1.92	>>>
3.85/1.92	
3.85/1.92	
3.85/1.92	The following domains are used:
3.85/1.92	   Integer
3.85/1.92	
3.85/1.92	The ITRS R consists of the following rules:
3.85/1.92	sum(x, y) -> sif(x >= y, x, y)
3.85/1.92	sif(TRUE, x, y) -> y + sum(x, y + 1)
3.85/1.92	sif(FALSE, x, y) -> 0
3.85/1.92	
3.85/1.92	The integer pair graph contains the following rules and edges:
3.85/1.92	(0): SUM(x[0], y[0]) -> SIF(x[0] >= y[0], x[0], y[0])
3.85/1.92	(1): SIF(TRUE, x[1], y[1]) -> SUM(x[1], y[1] + 1)
3.85/1.92	
3.85/1.92	   (0) -> (1), if (x[0] >= y[0]  & x[0] ->^* x[1] & y[0] ->^* y[1])
3.85/1.92	   (1) -> (0), if (x[1] ->^* x[0] & y[1] + 1 ->^* y[0])
3.85/1.92	
3.85/1.92	The set Q consists of the following terms:
3.85/1.92	sum(x0, x1)
3.85/1.92	sif(TRUE, x0, x1)
3.85/1.92	sif(FALSE, x0, x1)
3.85/1.92	
3.85/1.92	----------------------------------------
3.85/1.92	
3.85/1.92	(3) UsableRulesProof (EQUIVALENT)
3.85/1.92	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.
3.85/1.92	----------------------------------------
3.85/1.92	
3.85/1.92	(4)
3.85/1.92	Obligation:
3.85/1.92	IDP problem:
3.85/1.92	The following function symbols are pre-defined:
3.85/1.92	<<<
3.85/1.92	 & 	~ 	Bwand: (Integer, Integer) -> Integer
3.85/1.92	 >= 	~ 	Ge: (Integer, Integer) -> Boolean
3.85/1.92	 | 	~ 	Bwor: (Integer, Integer) -> Integer
3.85/1.92	 / 	~ 	Div: (Integer, Integer) -> Integer
3.85/1.92	 != 	~ 	Neq: (Integer, Integer) -> Boolean
3.85/1.92	 && 	~ 	Land: (Boolean, Boolean) -> Boolean
3.85/1.92	 ! 	~ 	Lnot: (Boolean) -> Boolean
3.85/1.92	 = 	~ 	Eq: (Integer, Integer) -> Boolean
3.85/1.92	 <= 	~ 	Le: (Integer, Integer) -> Boolean
3.85/1.92	 ^ 	~ 	Bwxor: (Integer, Integer) -> Integer
3.85/1.92	 % 	~ 	Mod: (Integer, Integer) -> Integer
3.85/1.92	 + 	~ 	Add: (Integer, Integer) -> Integer
3.85/1.92	 > 	~ 	Gt: (Integer, Integer) -> Boolean
3.85/1.92	 -1 	~ 	UnaryMinus: (Integer) -> Integer
3.85/1.92	 < 	~ 	Lt: (Integer, Integer) -> Boolean
3.85/1.92	 || 	~ 	Lor: (Boolean, Boolean) -> Boolean
3.85/1.92	 - 	~ 	Sub: (Integer, Integer) -> Integer
3.85/1.92	 ~ 	~ 	Bwnot: (Integer) -> Integer
3.85/1.92	 * 	~ 	Mul: (Integer, Integer) -> Integer
3.85/1.92	>>>
3.85/1.92	
3.85/1.92	
3.85/1.92	The following domains are used:
3.85/1.92	   Integer
3.85/1.92	
3.85/1.92	R is empty.
3.85/1.92	
3.85/1.92	The integer pair graph contains the following rules and edges:
3.85/1.92	(0): SUM(x[0], y[0]) -> SIF(x[0] >= y[0], x[0], y[0])
3.85/1.92	(1): SIF(TRUE, x[1], y[1]) -> SUM(x[1], y[1] + 1)
3.85/1.92	
3.85/1.92	   (0) -> (1), if (x[0] >= y[0]  & x[0] ->^* x[1] & y[0] ->^* y[1])
3.85/1.92	   (1) -> (0), if (x[1] ->^* x[0] & y[1] + 1 ->^* y[0])
3.85/1.92	
3.85/1.92	The set Q consists of the following terms:
3.85/1.92	sum(x0, x1)
3.85/1.92	sif(TRUE, x0, x1)
3.85/1.92	sif(FALSE, x0, x1)
3.85/1.92	
3.85/1.92	----------------------------------------
3.85/1.92	
3.85/1.92	(5) IDPNonInfProof (SOUND)
3.85/1.92	Used the following options for this NonInfProof:
3.85/1.92	
3.85/1.92	IDPGPoloSolver:
3.85/1.92	Range: [(-1,2)]
3.85/1.92	IsNat: false
3.85/1.92	Interpretation Shape Heuristic: aprove.DPFramework.IDPProblem.Processors.nonInf.poly.IdpDefaultShapeHeuristic@701c2ff8
3.85/1.92	Constraint Generator: NonInfConstraintGenerator:
3.85/1.92	PathGenerator: MetricPathGenerator:
3.85/1.92	Max Left Steps: 1
3.85/1.92	Max Right Steps: 1
3.85/1.92	
3.85/1.92	
3.85/1.92	
3.85/1.92	The constraints were generated the following way:
3.85/1.92	
3.85/1.92	The DP Problem is simplified using the Induction Calculus [NONINF] with the following steps:
3.85/1.92	
3.85/1.92	Note that final constraints are written in bold face.
3.85/1.92	
3.85/1.92	
3.85/1.92	
3.85/1.92	For Pair SUM(x, y) -> SIF(>=(x, y), x, y) the following chains were created:
3.85/1.92	*We consider the chain SUM(x[0], y[0]) -> SIF(>=(x[0], y[0]), x[0], y[0]), SIF(TRUE, x[1], y[1]) -> SUM(x[1], +(y[1], 1)) which results in the following constraint:
3.85/1.92	
3.85/1.92	(1)    (>=(x[0], y[0])=TRUE & x[0]=x[1] & y[0]=y[1]  ==>  SUM(x[0], y[0])_>=_NonInfC & SUM(x[0], y[0])_>=_SIF(>=(x[0], y[0]), x[0], y[0]) & (U^Increasing(SIF(>=(x[0], y[0]), x[0], y[0])), >=))
3.85/1.92	
3.85/1.92	
3.85/1.92	
3.85/1.92	We simplified constraint (1) using rule (IV) which results in the following new constraint:
3.85/1.92	
3.85/1.92	(2)    (>=(x[0], y[0])=TRUE  ==>  SUM(x[0], y[0])_>=_NonInfC & SUM(x[0], y[0])_>=_SIF(>=(x[0], y[0]), x[0], y[0]) & (U^Increasing(SIF(>=(x[0], y[0]), x[0], y[0])), >=))
3.85/1.92	
3.85/1.92	
3.85/1.92	
3.85/1.92	We simplified constraint (2) using rule (POLY_CONSTRAINTS) which results in the following new constraint:
3.85/1.92	
3.85/1.92	(3)    (x[0] + [-1]y[0] >= 0  ==>  (U^Increasing(SIF(>=(x[0], y[0]), x[0], y[0])), >=) & [(2)bni_11 + (-1)Bound*bni_11] + [(-1)bni_11]y[0] + [bni_11]x[0] >= 0 & [(-1)bso_12] >= 0)
3.85/1.92	
3.85/1.92	
3.85/1.92	
3.85/1.92	We simplified constraint (3) using rule (IDP_POLY_SIMPLIFY) which results in the following new constraint:
3.85/1.92	
3.85/1.92	(4)    (x[0] + [-1]y[0] >= 0  ==>  (U^Increasing(SIF(>=(x[0], y[0]), x[0], y[0])), >=) & [(2)bni_11 + (-1)Bound*bni_11] + [(-1)bni_11]y[0] + [bni_11]x[0] >= 0 & [(-1)bso_12] >= 0)
3.85/1.92	
3.85/1.92	
3.85/1.92	
3.85/1.92	We simplified constraint (4) using rule (POLY_REMOVE_MIN_MAX) which results in the following new constraint:
3.85/1.92	
3.85/1.92	(5)    (x[0] + [-1]y[0] >= 0  ==>  (U^Increasing(SIF(>=(x[0], y[0]), x[0], y[0])), >=) & [(2)bni_11 + (-1)Bound*bni_11] + [(-1)bni_11]y[0] + [bni_11]x[0] >= 0 & [(-1)bso_12] >= 0)
3.85/1.92	
3.85/1.92	
3.85/1.92	
3.85/1.92	We simplified constraint (5) using rule (IDP_SMT_SPLIT) which results in the following new constraint:
3.85/1.92	
3.85/1.92	(6)    (x[0] >= 0  ==>  (U^Increasing(SIF(>=(x[0], y[0]), x[0], y[0])), >=) & [(2)bni_11 + (-1)Bound*bni_11] + [bni_11]x[0] >= 0 & [(-1)bso_12] >= 0)
3.85/1.92	
3.85/1.92	
3.85/1.92	
3.85/1.92	We simplified constraint (6) using rule (IDP_SMT_SPLIT) which results in the following new constraints:
3.85/1.92	
3.85/1.92	(7)    (x[0] >= 0 & y[0] >= 0  ==>  (U^Increasing(SIF(>=(x[0], y[0]), x[0], y[0])), >=) & [(2)bni_11 + (-1)Bound*bni_11] + [bni_11]x[0] >= 0 & [(-1)bso_12] >= 0)
3.85/1.92	
3.85/1.92	(8)    (x[0] >= 0 & y[0] >= 0  ==>  (U^Increasing(SIF(>=(x[0], y[0]), x[0], y[0])), >=) & [(2)bni_11 + (-1)Bound*bni_11] + [bni_11]x[0] >= 0 & [(-1)bso_12] >= 0)
3.85/1.92	
3.85/1.92	
3.85/1.92	
3.85/1.92	
3.85/1.92	
3.85/1.92	
3.85/1.92	
3.85/1.92	
3.85/1.92	For Pair SIF(TRUE, x, y) -> SUM(x, +(y, 1)) the following chains were created:
3.85/1.92	*We consider the chain SUM(x[0], y[0]) -> SIF(>=(x[0], y[0]), x[0], y[0]), SIF(TRUE, x[1], y[1]) -> SUM(x[1], +(y[1], 1)), SUM(x[0], y[0]) -> SIF(>=(x[0], y[0]), x[0], y[0]) which results in the following constraint:
3.85/1.92	
3.85/1.92	(1)    (>=(x[0], y[0])=TRUE & x[0]=x[1] & y[0]=y[1] & x[1]=x[0]1 & +(y[1], 1)=y[0]1  ==>  SIF(TRUE, x[1], y[1])_>=_NonInfC & SIF(TRUE, x[1], y[1])_>=_SUM(x[1], +(y[1], 1)) & (U^Increasing(SUM(x[1], +(y[1], 1))), >=))
3.85/1.92	
3.85/1.92	
3.85/1.92	
3.85/1.92	We simplified constraint (1) using rules (III), (IV) which results in the following new constraint:
3.85/1.92	
3.85/1.92	(2)    (>=(x[0], y[0])=TRUE  ==>  SIF(TRUE, x[0], y[0])_>=_NonInfC & SIF(TRUE, x[0], y[0])_>=_SUM(x[0], +(y[0], 1)) & (U^Increasing(SUM(x[1], +(y[1], 1))), >=))
3.85/1.92	
3.85/1.92	
3.85/1.92	
3.85/1.92	We simplified constraint (2) using rule (POLY_CONSTRAINTS) which results in the following new constraint:
3.85/1.92	
3.85/1.92	(3)    (x[0] + [-1]y[0] >= 0  ==>  (U^Increasing(SUM(x[1], +(y[1], 1))), >=) & [(2)bni_13 + (-1)Bound*bni_13] + [(-1)bni_13]y[0] + [bni_13]x[0] >= 0 & [1 + (-1)bso_14] >= 0)
3.85/1.92	
3.85/1.92	
3.85/1.92	
3.85/1.92	We simplified constraint (3) using rule (IDP_POLY_SIMPLIFY) which results in the following new constraint:
3.85/1.92	
3.85/1.92	(4)    (x[0] + [-1]y[0] >= 0  ==>  (U^Increasing(SUM(x[1], +(y[1], 1))), >=) & [(2)bni_13 + (-1)Bound*bni_13] + [(-1)bni_13]y[0] + [bni_13]x[0] >= 0 & [1 + (-1)bso_14] >= 0)
3.85/1.92	
3.85/1.92	
3.85/1.92	
3.85/1.92	We simplified constraint (4) using rule (POLY_REMOVE_MIN_MAX) which results in the following new constraint:
3.85/1.92	
3.85/1.92	(5)    (x[0] + [-1]y[0] >= 0  ==>  (U^Increasing(SUM(x[1], +(y[1], 1))), >=) & [(2)bni_13 + (-1)Bound*bni_13] + [(-1)bni_13]y[0] + [bni_13]x[0] >= 0 & [1 + (-1)bso_14] >= 0)
3.85/1.92	
3.85/1.92	
3.85/1.92	
3.85/1.92	We simplified constraint (5) using rule (IDP_SMT_SPLIT) which results in the following new constraint:
3.85/1.92	
3.85/1.92	(6)    (x[0] >= 0  ==>  (U^Increasing(SUM(x[1], +(y[1], 1))), >=) & [(2)bni_13 + (-1)Bound*bni_13] + [bni_13]x[0] >= 0 & [1 + (-1)bso_14] >= 0)
3.85/1.92	
3.85/1.92	
3.85/1.92	
3.85/1.92	We simplified constraint (6) using rule (IDP_SMT_SPLIT) which results in the following new constraints:
3.85/1.92	
3.85/1.92	(7)    (x[0] >= 0 & y[0] >= 0  ==>  (U^Increasing(SUM(x[1], +(y[1], 1))), >=) & [(2)bni_13 + (-1)Bound*bni_13] + [bni_13]x[0] >= 0 & [1 + (-1)bso_14] >= 0)
3.85/1.92	
3.85/1.92	(8)    (x[0] >= 0 & y[0] >= 0  ==>  (U^Increasing(SUM(x[1], +(y[1], 1))), >=) & [(2)bni_13 + (-1)Bound*bni_13] + [bni_13]x[0] >= 0 & [1 + (-1)bso_14] >= 0)
3.85/1.92	
3.85/1.92	
3.85/1.92	
3.85/1.92	
3.85/1.92	
3.85/1.92	
3.85/1.92	
3.85/1.92	
3.85/1.92	To summarize, we get the following constraints P__>=_ for the following pairs.
3.85/1.92	
3.85/1.92	*SUM(x, y) -> SIF(>=(x, y), x, y)
3.85/1.92	
3.85/1.92	*(x[0] >= 0 & y[0] >= 0  ==>  (U^Increasing(SIF(>=(x[0], y[0]), x[0], y[0])), >=) & [(2)bni_11 + (-1)Bound*bni_11] + [bni_11]x[0] >= 0 & [(-1)bso_12] >= 0)
3.85/1.92	
3.85/1.92	
3.85/1.92	*(x[0] >= 0 & y[0] >= 0  ==>  (U^Increasing(SIF(>=(x[0], y[0]), x[0], y[0])), >=) & [(2)bni_11 + (-1)Bound*bni_11] + [bni_11]x[0] >= 0 & [(-1)bso_12] >= 0)
3.85/1.92	
3.85/1.92	
3.85/1.92	
3.85/1.92	
3.85/1.92	*SIF(TRUE, x, y) -> SUM(x, +(y, 1))
3.85/1.92	
3.85/1.92	*(x[0] >= 0 & y[0] >= 0  ==>  (U^Increasing(SUM(x[1], +(y[1], 1))), >=) & [(2)bni_13 + (-1)Bound*bni_13] + [bni_13]x[0] >= 0 & [1 + (-1)bso_14] >= 0)
3.85/1.92	
3.85/1.92	
3.85/1.92	*(x[0] >= 0 & y[0] >= 0  ==>  (U^Increasing(SUM(x[1], +(y[1], 1))), >=) & [(2)bni_13 + (-1)Bound*bni_13] + [bni_13]x[0] >= 0 & [1 + (-1)bso_14] >= 0)
3.85/1.92	
3.85/1.92	
3.85/1.92	
3.85/1.92	
3.85/1.92	
3.85/1.92	
3.85/1.92	
3.85/1.92	
3.85/1.92	The constraints for P_> respective P_bound are constructed from P__>=_ where we just replace every occurence of "t _>=_ s" in P__>=_ by  "t > s" respective "t _>=_ c". Here c stands for the fresh constant used for P_bound. 
3.85/1.92	
3.85/1.92	Using the following integer polynomial ordering the  resulting constraints can be solved 
3.85/1.92	
3.85/1.92	Polynomial interpretation over integers[POLO]:
3.85/1.92	
3.85/1.92	   POL(TRUE) = [2]
3.85/1.92	   POL(FALSE) = 0
3.85/1.92	   POL(SUM(x_1, x_2)) = [2] + [-1]x_2 + x_1
3.85/1.92	   POL(SIF(x_1, x_2, x_3)) = [2] + [-1]x_3 + x_2
3.85/1.92	   POL(>=(x_1, x_2)) = 0
3.85/1.92	   POL(+(x_1, x_2)) = x_1 + x_2
3.85/1.92	   POL(1) = [1]
3.85/1.92	
3.85/1.92	
3.85/1.92	The following pairs are in P_>:
3.85/1.92	
3.85/1.92	
3.85/1.92	   SIF(TRUE, x[1], y[1]) -> SUM(x[1], +(y[1], 1))
3.85/1.92	
3.85/1.92	
3.85/1.92	The following pairs are in P_bound:
3.85/1.92	
3.85/1.92	
3.85/1.92	   SUM(x[0], y[0]) -> SIF(>=(x[0], y[0]), x[0], y[0])
3.85/1.92	   SIF(TRUE, x[1], y[1]) -> SUM(x[1], +(y[1], 1))
3.85/1.92	
3.85/1.92	
3.85/1.92	The following pairs are in P_>=:
3.85/1.92	
3.85/1.92	
3.85/1.92	   SUM(x[0], y[0]) -> SIF(>=(x[0], y[0]), x[0], y[0])
3.85/1.92	
3.85/1.92	
3.85/1.92	There are no usable rules.
3.85/1.92	----------------------------------------
3.85/1.92	
3.85/1.92	(6)
3.85/1.92	Obligation:
3.85/1.92	IDP problem:
3.85/1.92	The following function symbols are pre-defined:
3.85/1.92	<<<
3.85/1.92	 & 	~ 	Bwand: (Integer, Integer) -> Integer
3.85/1.92	 >= 	~ 	Ge: (Integer, Integer) -> Boolean
3.85/1.92	 | 	~ 	Bwor: (Integer, Integer) -> Integer
3.85/1.92	 / 	~ 	Div: (Integer, Integer) -> Integer
3.85/1.92	 != 	~ 	Neq: (Integer, Integer) -> Boolean
3.85/1.92	 && 	~ 	Land: (Boolean, Boolean) -> Boolean
3.85/1.92	 ! 	~ 	Lnot: (Boolean) -> Boolean
3.85/1.92	 = 	~ 	Eq: (Integer, Integer) -> Boolean
3.85/1.92	 <= 	~ 	Le: (Integer, Integer) -> Boolean
3.85/1.92	 ^ 	~ 	Bwxor: (Integer, Integer) -> Integer
3.85/1.92	 % 	~ 	Mod: (Integer, Integer) -> Integer
3.85/1.92	 + 	~ 	Add: (Integer, Integer) -> Integer
3.85/1.92	 > 	~ 	Gt: (Integer, Integer) -> Boolean
3.85/1.93	 -1 	~ 	UnaryMinus: (Integer) -> Integer
3.85/1.93	 < 	~ 	Lt: (Integer, Integer) -> Boolean
3.85/1.93	 || 	~ 	Lor: (Boolean, Boolean) -> Boolean
3.85/1.93	 - 	~ 	Sub: (Integer, Integer) -> Integer
3.85/1.93	 ~ 	~ 	Bwnot: (Integer) -> Integer
3.85/1.93	 * 	~ 	Mul: (Integer, Integer) -> Integer
3.85/1.93	>>>
3.85/1.93	
3.85/1.93	
3.85/1.93	The following domains are used:
3.85/1.93	   Integer
3.85/1.93	
3.85/1.93	R is empty.
3.85/1.93	
3.85/1.93	The integer pair graph contains the following rules and edges:
3.85/1.93	(0): SUM(x[0], y[0]) -> SIF(x[0] >= y[0], x[0], y[0])
3.85/1.93	
3.85/1.93	
3.85/1.93	The set Q consists of the following terms:
3.85/1.93	sum(x0, x1)
3.85/1.93	sif(TRUE, x0, x1)
3.85/1.93	sif(FALSE, x0, x1)
3.85/1.93	
3.85/1.93	----------------------------------------
3.85/1.93	
3.85/1.93	(7) IDependencyGraphProof (EQUIVALENT)
3.85/1.93	The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 1 less node.
3.85/1.93	----------------------------------------
3.85/1.93	
3.85/1.93	(8)
3.85/1.93	TRUE
3.85/1.95	EOF