4.06/1.93	YES
4.06/1.95	proof of /export/starexec/sandbox/benchmark/theBenchmark.itrs
4.06/1.95	# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty
4.06/1.95	
4.06/1.95	
4.06/1.95	Termination of the given ITRS could be proven:
4.06/1.95	
4.06/1.95	(0) ITRS
4.06/1.95	(1) ITRStoIDPProof [EQUIVALENT, 0 ms]
4.06/1.95	(2) IDP
4.06/1.95	(3) UsableRulesProof [EQUIVALENT, 0 ms]
4.06/1.95	(4) IDP
4.06/1.95	(5) IDPNonInfProof [SOUND, 220 ms]
4.06/1.95	(6) IDP
4.06/1.95	(7) IDependencyGraphProof [EQUIVALENT, 0 ms]
4.06/1.95	(8) TRUE
4.06/1.95	
4.06/1.95	
4.06/1.95	----------------------------------------
4.06/1.95	
4.06/1.95	(0)
4.06/1.95	Obligation:
4.06/1.95	ITRS problem:
4.06/1.95	
4.06/1.95	The following function symbols are pre-defined:
4.06/1.95	<<<
4.06/1.95	 & 	~ 	Bwand: (Integer, Integer) -> Integer
4.06/1.95	 >= 	~ 	Ge: (Integer, Integer) -> Boolean
4.06/1.95	 / 	~ 	Div: (Integer, Integer) -> Integer
4.06/1.95	 | 	~ 	Bwor: (Integer, Integer) -> Integer
4.06/1.95	 != 	~ 	Neq: (Integer, Integer) -> Boolean
4.06/1.95	 && 	~ 	Land: (Boolean, Boolean) -> Boolean
4.06/1.95	 ! 	~ 	Lnot: (Boolean) -> Boolean
4.06/1.95	 = 	~ 	Eq: (Integer, Integer) -> Boolean
4.06/1.95	 <= 	~ 	Le: (Integer, Integer) -> Boolean
4.06/1.95	 ^ 	~ 	Bwxor: (Integer, Integer) -> Integer
4.06/1.95	 % 	~ 	Mod: (Integer, Integer) -> Integer
4.06/1.95	 > 	~ 	Gt: (Integer, Integer) -> Boolean
4.06/1.95	 + 	~ 	Add: (Integer, Integer) -> Integer
4.06/1.95	 -1 	~ 	UnaryMinus: (Integer) -> Integer
4.06/1.95	 < 	~ 	Lt: (Integer, Integer) -> Boolean
4.06/1.95	 || 	~ 	Lor: (Boolean, Boolean) -> Boolean
4.06/1.95	 - 	~ 	Sub: (Integer, Integer) -> Integer
4.06/1.95	 ~ 	~ 	Bwnot: (Integer) -> Integer
4.06/1.95	 * 	~ 	Mul: (Integer, Integer) -> Integer
4.06/1.95	>>>
4.06/1.95	
4.06/1.95	The TRS R consists of the following rules:
4.06/1.95	log(x, y) -> lif(x >= y && y > 1, x, y)
4.06/1.95	lif(TRUE, x, y) -> 1 + log(x / y, y)
4.06/1.95	lif(FALSE, x, y) -> 0
4.06/1.95	The set Q consists of the following terms:
4.06/1.95	log(x0, x1)
4.06/1.95	lif(TRUE, x0, x1)
4.06/1.95	lif(FALSE, x0, x1)
4.06/1.95	
4.06/1.95	----------------------------------------
4.06/1.95	
4.06/1.95	(1) ITRStoIDPProof (EQUIVALENT)
4.06/1.95	Added dependency pairs
4.06/1.95	----------------------------------------
4.06/1.95	
4.06/1.95	(2)
4.06/1.95	Obligation:
4.06/1.95	IDP problem:
4.06/1.95	The following function symbols are pre-defined:
4.06/1.95	<<<
4.06/1.95	 & 	~ 	Bwand: (Integer, Integer) -> Integer
4.06/1.95	 >= 	~ 	Ge: (Integer, Integer) -> Boolean
4.06/1.95	 / 	~ 	Div: (Integer, Integer) -> Integer
4.06/1.95	 | 	~ 	Bwor: (Integer, Integer) -> Integer
4.06/1.95	 != 	~ 	Neq: (Integer, Integer) -> Boolean
4.06/1.95	 && 	~ 	Land: (Boolean, Boolean) -> Boolean
4.06/1.95	 ! 	~ 	Lnot: (Boolean) -> Boolean
4.06/1.95	 = 	~ 	Eq: (Integer, Integer) -> Boolean
4.06/1.95	 <= 	~ 	Le: (Integer, Integer) -> Boolean
4.06/1.95	 ^ 	~ 	Bwxor: (Integer, Integer) -> Integer
4.06/1.95	 % 	~ 	Mod: (Integer, Integer) -> Integer
4.06/1.95	 > 	~ 	Gt: (Integer, Integer) -> Boolean
4.06/1.95	 + 	~ 	Add: (Integer, Integer) -> Integer
4.06/1.95	 -1 	~ 	UnaryMinus: (Integer) -> Integer
4.06/1.95	 < 	~ 	Lt: (Integer, Integer) -> Boolean
4.06/1.95	 || 	~ 	Lor: (Boolean, Boolean) -> Boolean
4.06/1.95	 - 	~ 	Sub: (Integer, Integer) -> Integer
4.06/1.95	 ~ 	~ 	Bwnot: (Integer) -> Integer
4.06/1.95	 * 	~ 	Mul: (Integer, Integer) -> Integer
4.06/1.95	>>>
4.06/1.95	
4.06/1.95	
4.06/1.95	The following domains are used:
4.06/1.95	   Boolean, Integer
4.06/1.95	
4.06/1.95	The ITRS R consists of the following rules:
4.06/1.95	log(x, y) -> lif(x >= y && y > 1, x, y)
4.06/1.95	lif(TRUE, x, y) -> 1 + log(x / y, y)
4.06/1.95	lif(FALSE, x, y) -> 0
4.06/1.95	
4.06/1.95	The integer pair graph contains the following rules and edges:
4.06/1.95	(0): LOG(x[0], y[0]) -> LIF(x[0] >= y[0] && y[0] > 1, x[0], y[0])
4.06/1.95	(1): LIF(TRUE, x[1], y[1]) -> LOG(x[1] / y[1], y[1])
4.06/1.95	
4.06/1.95	   (0) -> (1), if (x[0] >= y[0] && y[0] > 1  & x[0] ->^* x[1] & y[0] ->^* y[1])
4.06/1.95	   (1) -> (0), if (x[1] / y[1] ->^* x[0] & y[1] ->^* y[0])
4.06/1.95	
4.06/1.95	The set Q consists of the following terms:
4.06/1.95	log(x0, x1)
4.06/1.95	lif(TRUE, x0, x1)
4.06/1.95	lif(FALSE, x0, x1)
4.06/1.95	
4.06/1.95	----------------------------------------
4.06/1.95	
4.06/1.95	(3) UsableRulesProof (EQUIVALENT)
4.06/1.95	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.
4.06/1.95	----------------------------------------
4.06/1.95	
4.06/1.95	(4)
4.06/1.95	Obligation:
4.06/1.95	IDP problem:
4.06/1.95	The following function symbols are pre-defined:
4.06/1.95	<<<
4.06/1.95	 & 	~ 	Bwand: (Integer, Integer) -> Integer
4.06/1.95	 >= 	~ 	Ge: (Integer, Integer) -> Boolean
4.06/1.95	 / 	~ 	Div: (Integer, Integer) -> Integer
4.06/1.95	 | 	~ 	Bwor: (Integer, Integer) -> Integer
4.06/1.95	 != 	~ 	Neq: (Integer, Integer) -> Boolean
4.06/1.95	 && 	~ 	Land: (Boolean, Boolean) -> Boolean
4.06/1.95	 ! 	~ 	Lnot: (Boolean) -> Boolean
4.06/1.95	 = 	~ 	Eq: (Integer, Integer) -> Boolean
4.06/1.95	 <= 	~ 	Le: (Integer, Integer) -> Boolean
4.06/1.95	 ^ 	~ 	Bwxor: (Integer, Integer) -> Integer
4.06/1.95	 % 	~ 	Mod: (Integer, Integer) -> Integer
4.06/1.95	 > 	~ 	Gt: (Integer, Integer) -> Boolean
4.06/1.95	 + 	~ 	Add: (Integer, Integer) -> Integer
4.06/1.95	 -1 	~ 	UnaryMinus: (Integer) -> Integer
4.06/1.95	 < 	~ 	Lt: (Integer, Integer) -> Boolean
4.06/1.95	 || 	~ 	Lor: (Boolean, Boolean) -> Boolean
4.06/1.95	 - 	~ 	Sub: (Integer, Integer) -> Integer
4.06/1.95	 ~ 	~ 	Bwnot: (Integer) -> Integer
4.06/1.95	 * 	~ 	Mul: (Integer, Integer) -> Integer
4.06/1.95	>>>
4.06/1.95	
4.06/1.95	
4.06/1.95	The following domains are used:
4.06/1.95	   Boolean, Integer
4.06/1.95	
4.06/1.95	R is empty.
4.06/1.95	
4.06/1.95	The integer pair graph contains the following rules and edges:
4.06/1.95	(0): LOG(x[0], y[0]) -> LIF(x[0] >= y[0] && y[0] > 1, x[0], y[0])
4.06/1.95	(1): LIF(TRUE, x[1], y[1]) -> LOG(x[1] / y[1], y[1])
4.06/1.95	
4.06/1.95	   (0) -> (1), if (x[0] >= y[0] && y[0] > 1  & x[0] ->^* x[1] & y[0] ->^* y[1])
4.06/1.95	   (1) -> (0), if (x[1] / y[1] ->^* x[0] & y[1] ->^* y[0])
4.06/1.95	
4.06/1.95	The set Q consists of the following terms:
4.06/1.95	log(x0, x1)
4.06/1.95	lif(TRUE, x0, x1)
4.06/1.95	lif(FALSE, x0, x1)
4.06/1.95	
4.06/1.95	----------------------------------------
4.06/1.95	
4.06/1.95	(5) IDPNonInfProof (SOUND)
4.06/1.95	Used the following options for this NonInfProof:
4.06/1.95	
4.06/1.95	IDPGPoloSolver:
4.06/1.95	Range: [(-1,2)]
4.06/1.95	IsNat: false
4.06/1.95	Interpretation Shape Heuristic: aprove.DPFramework.IDPProblem.Processors.nonInf.poly.IdpDefaultShapeHeuristic@36ed0024
4.06/1.95	Constraint Generator: NonInfConstraintGenerator:
4.06/1.95	PathGenerator: MetricPathGenerator:
4.06/1.95	Max Left Steps: 1
4.06/1.95	Max Right Steps: 1
4.06/1.95	
4.06/1.95	
4.06/1.95	
4.06/1.95	The constraints were generated the following way:
4.06/1.95	
4.06/1.95	The DP Problem is simplified using the Induction Calculus [NONINF] with the following steps:
4.06/1.95	
4.06/1.95	Note that final constraints are written in bold face.
4.06/1.95	
4.06/1.95	
4.06/1.95	
4.06/1.95	For Pair LOG(x, y) -> LIF(&&(>=(x, y), >(y, 1)), x, y) the following chains were created:
4.06/1.95	*We consider the chain LOG(x[0], y[0]) -> LIF(&&(>=(x[0], y[0]), >(y[0], 1)), x[0], y[0]), LIF(TRUE, x[1], y[1]) -> LOG(/(x[1], y[1]), y[1]) which results in the following constraint:
4.06/1.95	
4.06/1.95	(1)    (&&(>=(x[0], y[0]), >(y[0], 1))=TRUE & x[0]=x[1] & y[0]=y[1]  ==>  LOG(x[0], y[0])_>=_NonInfC & LOG(x[0], y[0])_>=_LIF(&&(>=(x[0], y[0]), >(y[0], 1)), x[0], y[0]) & (U^Increasing(LIF(&&(>=(x[0], y[0]), >(y[0], 1)), x[0], y[0])), >=))
4.06/1.95	
4.06/1.95	
4.06/1.95	
4.06/1.95	We simplified constraint (1) using rules (IV), (IDP_BOOLEAN) which results in the following new constraint:
4.06/1.95	
4.06/1.95	(2)    (>=(x[0], y[0])=TRUE & >(y[0], 1)=TRUE  ==>  LOG(x[0], y[0])_>=_NonInfC & LOG(x[0], y[0])_>=_LIF(&&(>=(x[0], y[0]), >(y[0], 1)), x[0], y[0]) & (U^Increasing(LIF(&&(>=(x[0], y[0]), >(y[0], 1)), x[0], y[0])), >=))
4.06/1.95	
4.06/1.95	
4.06/1.95	
4.06/1.95	We simplified constraint (2) using rule (POLY_CONSTRAINTS) which results in the following new constraint:
4.06/1.95	
4.06/1.95	(3)    (x[0] + [-1]y[0] >= 0 & y[0] + [-2] >= 0  ==>  (U^Increasing(LIF(&&(>=(x[0], y[0]), >(y[0], 1)), x[0], y[0])), >=) & [(-1)bni_13 + (-1)Bound*bni_13] + [(-1)bni_13]y[0] + [bni_13]x[0] >= 0 & [(-1)bso_14] >= 0)
4.06/1.95	
4.06/1.95	
4.06/1.95	
4.06/1.95	We simplified constraint (3) using rule (IDP_POLY_SIMPLIFY) which results in the following new constraint:
4.06/1.95	
4.06/1.95	(4)    (x[0] + [-1]y[0] >= 0 & y[0] + [-2] >= 0  ==>  (U^Increasing(LIF(&&(>=(x[0], y[0]), >(y[0], 1)), x[0], y[0])), >=) & [(-1)bni_13 + (-1)Bound*bni_13] + [(-1)bni_13]y[0] + [bni_13]x[0] >= 0 & [(-1)bso_14] >= 0)
4.06/1.95	
4.06/1.95	
4.06/1.95	
4.06/1.95	We simplified constraint (4) using rule (POLY_REMOVE_MIN_MAX) which results in the following new constraint:
4.06/1.95	
4.06/1.95	(5)    (x[0] + [-1]y[0] >= 0 & y[0] + [-2] >= 0  ==>  (U^Increasing(LIF(&&(>=(x[0], y[0]), >(y[0], 1)), x[0], y[0])), >=) & [(-1)bni_13 + (-1)Bound*bni_13] + [(-1)bni_13]y[0] + [bni_13]x[0] >= 0 & [(-1)bso_14] >= 0)
4.06/1.95	
4.06/1.95	
4.06/1.95	
4.06/1.95	
4.06/1.95	
4.06/1.95	
4.06/1.95	
4.06/1.95	
4.06/1.95	For Pair LIF(TRUE, x, y) -> LOG(/(x, y), y) the following chains were created:
4.06/1.95	*We consider the chain LOG(x[0], y[0]) -> LIF(&&(>=(x[0], y[0]), >(y[0], 1)), x[0], y[0]), LIF(TRUE, x[1], y[1]) -> LOG(/(x[1], y[1]), y[1]), LOG(x[0], y[0]) -> LIF(&&(>=(x[0], y[0]), >(y[0], 1)), x[0], y[0]) which results in the following constraint:
4.06/1.95	
4.06/1.95	(1)    (&&(>=(x[0], y[0]), >(y[0], 1))=TRUE & x[0]=x[1] & y[0]=y[1] & /(x[1], y[1])=x[0]1 & y[1]=y[0]1  ==>  LIF(TRUE, x[1], y[1])_>=_NonInfC & LIF(TRUE, x[1], y[1])_>=_LOG(/(x[1], y[1]), y[1]) & (U^Increasing(LOG(/(x[1], y[1]), y[1])), >=))
4.06/1.95	
4.06/1.95	
4.06/1.95	
4.06/1.95	We simplified constraint (1) using rules (III), (IV), (IDP_BOOLEAN) which results in the following new constraint:
4.06/1.95	
4.06/1.95	(2)    (>=(x[0], y[0])=TRUE & >(y[0], 1)=TRUE  ==>  LIF(TRUE, x[0], y[0])_>=_NonInfC & LIF(TRUE, x[0], y[0])_>=_LOG(/(x[0], y[0]), y[0]) & (U^Increasing(LOG(/(x[1], y[1]), y[1])), >=))
4.06/1.95	
4.06/1.95	
4.06/1.95	
4.06/1.95	We simplified constraint (2) using rule (POLY_CONSTRAINTS) which results in the following new constraint:
4.06/1.95	
4.06/1.95	(3)    (x[0] + [-1]y[0] >= 0 & y[0] + [-2] >= 0  ==>  (U^Increasing(LOG(/(x[1], y[1]), y[1])), >=) & [(-1)bni_15 + (-1)Bound*bni_15] + [(-1)bni_15]y[0] + [bni_15]x[0] >= 0 & [(-1)bso_19] + x[0] + [-1]max{x[0], [-1]x[0]} + min{max{y[0], [-1]y[0]} + [-1], max{x[0], [-1]x[0]}} >= 0)
4.06/1.95	
4.06/1.95	
4.06/1.95	
4.06/1.95	We simplified constraint (3) using rule (IDP_POLY_SIMPLIFY) which results in the following new constraint:
4.06/1.95	
4.06/1.95	(4)    (x[0] + [-1]y[0] >= 0 & y[0] + [-2] >= 0  ==>  (U^Increasing(LOG(/(x[1], y[1]), y[1])), >=) & [(-1)bni_15 + (-1)Bound*bni_15] + [(-1)bni_15]y[0] + [bni_15]x[0] >= 0 & [(-1)bso_19] + x[0] + [-1]max{x[0], [-1]x[0]} + min{max{y[0], [-1]y[0]} + [-1], max{x[0], [-1]x[0]}} >= 0)
4.06/1.95	
4.06/1.95	
4.06/1.95	
4.06/1.95	We simplified constraint (4) using rule (POLY_REMOVE_MIN_MAX) which results in the following new constraint:
4.06/1.95	
4.06/1.95	(5)    (x[0] + [-1]y[0] >= 0 & y[0] + [-2] >= 0 & [2]x[0] >= 0 & [2]y[0] >= 0  ==>  (U^Increasing(LOG(/(x[1], y[1]), y[1])), >=) & [(-1)bni_15 + (-1)Bound*bni_15] + [(-1)bni_15]y[0] + [bni_15]x[0] >= 0 & [-1 + (-1)bso_19] + y[0] >= 0)
4.06/1.95	
4.06/1.95	
4.06/1.95	
4.06/1.95	We simplified constraint (5) using rule (IDP_POLY_GCD) which results in the following new constraint:
4.06/1.95	
4.06/1.95	(6)    (x[0] + [-1]y[0] >= 0 & y[0] + [-2] >= 0 & x[0] >= 0 & y[0] >= 0  ==>  (U^Increasing(LOG(/(x[1], y[1]), y[1])), >=) & [(-1)bni_15 + (-1)Bound*bni_15] + [(-1)bni_15]y[0] + [bni_15]x[0] >= 0 & [-1 + (-1)bso_19] + y[0] >= 0)
4.06/1.95	
4.06/1.95	
4.06/1.95	
4.06/1.95	
4.06/1.95	
4.06/1.95	
4.06/1.95	
4.06/1.95	
4.06/1.95	To summarize, we get the following constraints P__>=_ for the following pairs.
4.06/1.95	
4.06/1.95	*LOG(x, y) -> LIF(&&(>=(x, y), >(y, 1)), x, y)
4.06/1.95	
4.06/1.95	*(x[0] + [-1]y[0] >= 0 & y[0] + [-2] >= 0  ==>  (U^Increasing(LIF(&&(>=(x[0], y[0]), >(y[0], 1)), x[0], y[0])), >=) & [(-1)bni_13 + (-1)Bound*bni_13] + [(-1)bni_13]y[0] + [bni_13]x[0] >= 0 & [(-1)bso_14] >= 0)
4.06/1.95	
4.06/1.95	
4.06/1.95	
4.06/1.95	
4.06/1.95	*LIF(TRUE, x, y) -> LOG(/(x, y), y)
4.06/1.95	
4.06/1.95	*(x[0] + [-1]y[0] >= 0 & y[0] + [-2] >= 0 & x[0] >= 0 & y[0] >= 0  ==>  (U^Increasing(LOG(/(x[1], y[1]), y[1])), >=) & [(-1)bni_15 + (-1)Bound*bni_15] + [(-1)bni_15]y[0] + [bni_15]x[0] >= 0 & [-1 + (-1)bso_19] + y[0] >= 0)
4.06/1.95	
4.06/1.95	
4.06/1.95	
4.06/1.95	
4.06/1.95	
4.06/1.95	
4.06/1.95	
4.06/1.95	
4.06/1.95	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. 
4.06/1.95	
4.06/1.95	Using the following integer polynomial ordering the  resulting constraints can be solved 
4.06/1.95	
4.06/1.95	Polynomial interpretation over integers[POLO]:
4.06/1.95	
4.06/1.95	   POL(TRUE) = 0
4.06/1.95	   POL(FALSE) = [1]
4.06/1.95	   POL(LOG(x_1, x_2)) = [-1] + [-1]x_2 + x_1
4.06/1.95	   POL(LIF(x_1, x_2, x_3)) = [-1] + [-1]x_3 + x_2 + [-1]x_1
4.06/1.95	   POL(&&(x_1, x_2)) = 0
4.06/1.95	   POL(>=(x_1, x_2)) = [-1]
4.06/1.95	   POL(>(x_1, x_2)) = [-1]
4.06/1.95	   POL(1) = [1]
4.06/1.95	   
4.06/1.95	   Polynomial Interpretations with Context Sensitive Arithemetic Replacement
4.06/1.95	   POL(Term^CSAR-Mode @ Context)
4.06/1.95	   
4.06/1.95	   POL(/(x_1, y[0])^1 @ {LOG_2/0}) = max{x_1, [-1]x_1} + [-1]min{max{x_2, [-1]x_2} + [-1], max{x_1, [-1]x_1}}
4.06/1.95	
4.06/1.95	
4.06/1.95	The following pairs are in P_>:
4.06/1.95	
4.06/1.95	
4.06/1.95	   LIF(TRUE, x[1], y[1]) -> LOG(/(x[1], y[1]), y[1])
4.06/1.95	
4.06/1.95	
4.06/1.95	The following pairs are in P_bound:
4.06/1.95	
4.06/1.95	
4.06/1.95	   LOG(x[0], y[0]) -> LIF(&&(>=(x[0], y[0]), >(y[0], 1)), x[0], y[0])
4.06/1.95	   LIF(TRUE, x[1], y[1]) -> LOG(/(x[1], y[1]), y[1])
4.06/1.95	
4.06/1.95	
4.06/1.95	The following pairs are in P_>=:
4.06/1.95	
4.06/1.95	
4.06/1.95	   LOG(x[0], y[0]) -> LIF(&&(>=(x[0], y[0]), >(y[0], 1)), x[0], y[0])
4.06/1.95	
4.06/1.95	
4.06/1.95	At least the following rules have been oriented under context sensitive arithmetic replacement:
4.06/1.95	
4.06/1.95	   TRUE^1 -> &&(TRUE, TRUE)^1
4.06/1.95	   FALSE^1 -> &&(TRUE, FALSE)^1
4.06/1.95	   FALSE^1 -> &&(FALSE, TRUE)^1
4.06/1.95	   FALSE^1 -> &&(FALSE, FALSE)^1
4.06/1.95	   /^1 ->
4.06/1.95	
4.06/1.95	----------------------------------------
4.06/1.95	
4.06/1.95	(6)
4.06/1.95	Obligation:
4.06/1.95	IDP problem:
4.06/1.95	The following function symbols are pre-defined:
4.06/1.95	<<<
4.06/1.95	 & 	~ 	Bwand: (Integer, Integer) -> Integer
4.06/1.95	 >= 	~ 	Ge: (Integer, Integer) -> Boolean
4.06/1.95	 / 	~ 	Div: (Integer, Integer) -> Integer
4.06/1.95	 | 	~ 	Bwor: (Integer, Integer) -> Integer
4.06/1.95	 != 	~ 	Neq: (Integer, Integer) -> Boolean
4.06/1.95	 && 	~ 	Land: (Boolean, Boolean) -> Boolean
4.06/1.95	 ! 	~ 	Lnot: (Boolean) -> Boolean
4.06/1.95	 = 	~ 	Eq: (Integer, Integer) -> Boolean
4.06/1.95	 <= 	~ 	Le: (Integer, Integer) -> Boolean
4.06/1.95	 ^ 	~ 	Bwxor: (Integer, Integer) -> Integer
4.06/1.95	 % 	~ 	Mod: (Integer, Integer) -> Integer
4.06/1.95	 > 	~ 	Gt: (Integer, Integer) -> Boolean
4.06/1.95	 + 	~ 	Add: (Integer, Integer) -> Integer
4.06/1.95	 -1 	~ 	UnaryMinus: (Integer) -> Integer
4.06/1.95	 < 	~ 	Lt: (Integer, Integer) -> Boolean
4.06/1.95	 || 	~ 	Lor: (Boolean, Boolean) -> Boolean
4.06/1.95	 - 	~ 	Sub: (Integer, Integer) -> Integer
4.06/1.95	 ~ 	~ 	Bwnot: (Integer) -> Integer
4.06/1.95	 * 	~ 	Mul: (Integer, Integer) -> Integer
4.06/1.95	>>>
4.06/1.95	
4.06/1.95	
4.06/1.95	The following domains are used:
4.06/1.95	   Boolean, Integer
4.06/1.95	
4.06/1.95	R is empty.
4.06/1.95	
4.06/1.95	The integer pair graph contains the following rules and edges:
4.06/1.95	(0): LOG(x[0], y[0]) -> LIF(x[0] >= y[0] && y[0] > 1, x[0], y[0])
4.06/1.95	
4.06/1.95	
4.06/1.95	The set Q consists of the following terms:
4.06/1.95	log(x0, x1)
4.06/1.95	lif(TRUE, x0, x1)
4.06/1.95	lif(FALSE, x0, x1)
4.06/1.95	
4.06/1.95	----------------------------------------
4.06/1.95	
4.06/1.95	(7) IDependencyGraphProof (EQUIVALENT)
4.06/1.95	The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 1 less node.
4.06/1.95	----------------------------------------
4.06/1.95	
4.06/1.95	(8)
4.06/1.95	TRUE
4.14/1.98	EOF