3.78/1.79	YES
3.81/1.80	proof of /export/starexec/sandbox2/benchmark/theBenchmark.itrs
3.81/1.80	# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty
3.81/1.80	
3.81/1.80	
3.81/1.80	Termination of the given ITRS could be proven:
3.81/1.80	
3.81/1.80	(0) ITRS
3.81/1.80	(1) ITRStoIDPProof [EQUIVALENT, 0 ms]
3.81/1.80	(2) IDP
3.81/1.80	(3) UsableRulesProof [EQUIVALENT, 0 ms]
3.81/1.80	(4) IDP
3.81/1.80	(5) IDPNonInfProof [SOUND, 148 ms]
3.81/1.80	(6) IDP
3.81/1.80	(7) IDependencyGraphProof [EQUIVALENT, 0 ms]
3.81/1.80	(8) TRUE
3.81/1.80	
3.81/1.80	
3.81/1.80	----------------------------------------
3.81/1.80	
3.81/1.80	(0)
3.81/1.80	Obligation:
3.81/1.80	ITRS problem:
3.81/1.80	
3.81/1.80	The following function symbols are pre-defined:
3.81/1.80	<<<
3.81/1.80	 & 	~ 	Bwand: (Integer, Integer) -> Integer
3.81/1.80	 >= 	~ 	Ge: (Integer, Integer) -> Boolean
3.81/1.80	 | 	~ 	Bwor: (Integer, Integer) -> Integer
3.81/1.80	 / 	~ 	Div: (Integer, Integer) -> Integer
3.81/1.80	 != 	~ 	Neq: (Integer, Integer) -> Boolean
3.81/1.80	 && 	~ 	Land: (Boolean, Boolean) -> Boolean
3.81/1.80	 ! 	~ 	Lnot: (Boolean) -> Boolean
3.81/1.80	 = 	~ 	Eq: (Integer, Integer) -> Boolean
3.81/1.80	 <= 	~ 	Le: (Integer, Integer) -> Boolean
3.81/1.80	 ^ 	~ 	Bwxor: (Integer, Integer) -> Integer
3.81/1.80	 % 	~ 	Mod: (Integer, Integer) -> Integer
3.81/1.80	 > 	~ 	Gt: (Integer, Integer) -> Boolean
3.81/1.80	 + 	~ 	Add: (Integer, Integer) -> Integer
3.81/1.80	 -1 	~ 	UnaryMinus: (Integer) -> Integer
3.81/1.80	 < 	~ 	Lt: (Integer, Integer) -> Boolean
3.81/1.80	 || 	~ 	Lor: (Boolean, Boolean) -> Boolean
3.81/1.80	 - 	~ 	Sub: (Integer, Integer) -> Integer
3.81/1.80	 ~ 	~ 	Bwnot: (Integer) -> Integer
3.81/1.80	 * 	~ 	Mul: (Integer, Integer) -> Integer
3.81/1.80	>>>
3.81/1.80	
3.81/1.80	The TRS R consists of the following rules:
3.81/1.80	eval(x, y) -> Cond_eval(x > y, x, y)
3.81/1.80	Cond_eval(TRUE, x, y) -> eval(x - 1, y)
3.81/1.80	The set Q consists of the following terms:
3.81/1.80	eval(x0, x1)
3.81/1.80	Cond_eval(TRUE, x0, x1)
3.81/1.80	
3.81/1.80	----------------------------------------
3.81/1.80	
3.81/1.80	(1) ITRStoIDPProof (EQUIVALENT)
3.81/1.80	Added dependency pairs
3.81/1.80	----------------------------------------
3.81/1.80	
3.81/1.80	(2)
3.81/1.80	Obligation:
3.81/1.80	IDP problem:
3.81/1.80	The following function symbols are pre-defined:
3.81/1.80	<<<
3.81/1.80	 & 	~ 	Bwand: (Integer, Integer) -> Integer
3.81/1.80	 >= 	~ 	Ge: (Integer, Integer) -> Boolean
3.81/1.80	 | 	~ 	Bwor: (Integer, Integer) -> Integer
3.81/1.80	 / 	~ 	Div: (Integer, Integer) -> Integer
3.81/1.80	 != 	~ 	Neq: (Integer, Integer) -> Boolean
3.81/1.80	 && 	~ 	Land: (Boolean, Boolean) -> Boolean
3.81/1.80	 ! 	~ 	Lnot: (Boolean) -> Boolean
3.81/1.80	 = 	~ 	Eq: (Integer, Integer) -> Boolean
3.81/1.80	 <= 	~ 	Le: (Integer, Integer) -> Boolean
3.81/1.80	 ^ 	~ 	Bwxor: (Integer, Integer) -> Integer
3.81/1.80	 % 	~ 	Mod: (Integer, Integer) -> Integer
3.81/1.80	 > 	~ 	Gt: (Integer, Integer) -> Boolean
3.81/1.80	 + 	~ 	Add: (Integer, Integer) -> Integer
3.81/1.80	 -1 	~ 	UnaryMinus: (Integer) -> Integer
3.81/1.80	 < 	~ 	Lt: (Integer, Integer) -> Boolean
3.81/1.80	 || 	~ 	Lor: (Boolean, Boolean) -> Boolean
3.81/1.80	 - 	~ 	Sub: (Integer, Integer) -> Integer
3.81/1.80	 ~ 	~ 	Bwnot: (Integer) -> Integer
3.81/1.80	 * 	~ 	Mul: (Integer, Integer) -> Integer
3.81/1.80	>>>
3.81/1.80	
3.81/1.80	
3.81/1.80	The following domains are used:
3.81/1.80	   Integer
3.81/1.80	
3.81/1.80	The ITRS R consists of the following rules:
3.81/1.80	eval(x, y) -> Cond_eval(x > y, x, y)
3.81/1.80	Cond_eval(TRUE, x, y) -> eval(x - 1, y)
3.81/1.80	
3.81/1.80	The integer pair graph contains the following rules and edges:
3.81/1.80	(0): EVAL(x[0], y[0]) -> COND_EVAL(x[0] > y[0], x[0], y[0])
3.81/1.80	(1): COND_EVAL(TRUE, x[1], y[1]) -> EVAL(x[1] - 1, y[1])
3.81/1.80	
3.81/1.80	   (0) -> (1), if (x[0] > y[0]  & x[0] ->^* x[1] & y[0] ->^* y[1])
3.81/1.80	   (1) -> (0), if (x[1] - 1 ->^* x[0] & y[1] ->^* y[0])
3.81/1.80	
3.81/1.80	The set Q consists of the following terms:
3.81/1.80	eval(x0, x1)
3.81/1.80	Cond_eval(TRUE, x0, x1)
3.81/1.80	
3.81/1.80	----------------------------------------
3.81/1.80	
3.81/1.80	(3) UsableRulesProof (EQUIVALENT)
3.81/1.80	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.
3.81/1.80	----------------------------------------
3.81/1.80	
3.81/1.80	(4)
3.81/1.80	Obligation:
3.81/1.80	IDP problem:
3.81/1.80	The following function symbols are pre-defined:
3.81/1.80	<<<
3.81/1.80	 & 	~ 	Bwand: (Integer, Integer) -> Integer
3.81/1.80	 >= 	~ 	Ge: (Integer, Integer) -> Boolean
3.81/1.80	 | 	~ 	Bwor: (Integer, Integer) -> Integer
3.81/1.80	 / 	~ 	Div: (Integer, Integer) -> Integer
3.81/1.80	 != 	~ 	Neq: (Integer, Integer) -> Boolean
3.81/1.80	 && 	~ 	Land: (Boolean, Boolean) -> Boolean
3.81/1.80	 ! 	~ 	Lnot: (Boolean) -> Boolean
3.81/1.80	 = 	~ 	Eq: (Integer, Integer) -> Boolean
3.81/1.80	 <= 	~ 	Le: (Integer, Integer) -> Boolean
3.81/1.80	 ^ 	~ 	Bwxor: (Integer, Integer) -> Integer
3.81/1.80	 % 	~ 	Mod: (Integer, Integer) -> Integer
3.81/1.80	 > 	~ 	Gt: (Integer, Integer) -> Boolean
3.81/1.80	 + 	~ 	Add: (Integer, Integer) -> Integer
3.81/1.80	 -1 	~ 	UnaryMinus: (Integer) -> Integer
3.81/1.80	 < 	~ 	Lt: (Integer, Integer) -> Boolean
3.81/1.80	 || 	~ 	Lor: (Boolean, Boolean) -> Boolean
3.81/1.80	 - 	~ 	Sub: (Integer, Integer) -> Integer
3.81/1.80	 ~ 	~ 	Bwnot: (Integer) -> Integer
3.81/1.80	 * 	~ 	Mul: (Integer, Integer) -> Integer
3.81/1.80	>>>
3.81/1.80	
3.81/1.80	
3.81/1.80	The following domains are used:
3.81/1.80	   Integer
3.81/1.80	
3.81/1.80	R is empty.
3.81/1.80	
3.81/1.80	The integer pair graph contains the following rules and edges:
3.81/1.80	(0): EVAL(x[0], y[0]) -> COND_EVAL(x[0] > y[0], x[0], y[0])
3.81/1.80	(1): COND_EVAL(TRUE, x[1], y[1]) -> EVAL(x[1] - 1, y[1])
3.81/1.80	
3.81/1.80	   (0) -> (1), if (x[0] > y[0]  & x[0] ->^* x[1] & y[0] ->^* y[1])
3.81/1.80	   (1) -> (0), if (x[1] - 1 ->^* x[0] & y[1] ->^* y[0])
3.81/1.80	
3.81/1.80	The set Q consists of the following terms:
3.81/1.80	eval(x0, x1)
3.81/1.80	Cond_eval(TRUE, x0, x1)
3.81/1.80	
3.81/1.80	----------------------------------------
3.81/1.80	
3.81/1.80	(5) IDPNonInfProof (SOUND)
3.81/1.80	Used the following options for this NonInfProof:
3.81/1.80	
3.81/1.80	IDPGPoloSolver:
3.81/1.80	Range: [(-1,2)]
3.81/1.80	IsNat: false
3.81/1.80	Interpretation Shape Heuristic: aprove.DPFramework.IDPProblem.Processors.nonInf.poly.IdpDefaultShapeHeuristic@65d89150
3.81/1.80	Constraint Generator: NonInfConstraintGenerator:
3.81/1.80	PathGenerator: MetricPathGenerator:
3.81/1.80	Max Left Steps: 1
3.81/1.80	Max Right Steps: 1
3.81/1.80	
3.81/1.80	
3.81/1.80	
3.81/1.80	The constraints were generated the following way:
3.81/1.80	
3.81/1.80	The DP Problem is simplified using the Induction Calculus [NONINF] with the following steps:
3.81/1.80	
3.81/1.80	Note that final constraints are written in bold face.
3.81/1.80	
3.81/1.80	
3.81/1.80	
3.81/1.80	For Pair EVAL(x, y) -> COND_EVAL(>(x, y), x, y) the following chains were created:
3.81/1.80	*We consider the chain EVAL(x[0], y[0]) -> COND_EVAL(>(x[0], y[0]), x[0], y[0]), COND_EVAL(TRUE, x[1], y[1]) -> EVAL(-(x[1], 1), y[1]) which results in the following constraint:
3.81/1.80	
3.81/1.80	(1)    (>(x[0], y[0])=TRUE & x[0]=x[1] & y[0]=y[1]  ==>  EVAL(x[0], y[0])_>=_NonInfC & EVAL(x[0], y[0])_>=_COND_EVAL(>(x[0], y[0]), x[0], y[0]) & (U^Increasing(COND_EVAL(>(x[0], y[0]), x[0], y[0])), >=))
3.81/1.80	
3.81/1.80	
3.81/1.80	
3.81/1.80	We simplified constraint (1) using rule (IV) which results in the following new constraint:
3.81/1.80	
3.81/1.80	(2)    (>(x[0], y[0])=TRUE  ==>  EVAL(x[0], y[0])_>=_NonInfC & EVAL(x[0], y[0])_>=_COND_EVAL(>(x[0], y[0]), x[0], y[0]) & (U^Increasing(COND_EVAL(>(x[0], y[0]), x[0], y[0])), >=))
3.81/1.80	
3.81/1.80	
3.81/1.80	
3.81/1.80	We simplified constraint (2) using rule (POLY_CONSTRAINTS) which results in the following new constraint:
3.81/1.80	
3.81/1.80	(3)    (x[0] + [-1] + [-1]y[0] >= 0  ==>  (U^Increasing(COND_EVAL(>(x[0], y[0]), x[0], y[0])), >=) & [(-1)bni_11 + (-1)Bound*bni_11] + [(-1)bni_11]y[0] + [bni_11]x[0] >= 0 & [(-1)bso_12] >= 0)
3.81/1.80	
3.81/1.80	
3.81/1.80	
3.81/1.80	We simplified constraint (3) using rule (IDP_POLY_SIMPLIFY) which results in the following new constraint:
3.81/1.80	
3.81/1.80	(4)    (x[0] + [-1] + [-1]y[0] >= 0  ==>  (U^Increasing(COND_EVAL(>(x[0], y[0]), x[0], y[0])), >=) & [(-1)bni_11 + (-1)Bound*bni_11] + [(-1)bni_11]y[0] + [bni_11]x[0] >= 0 & [(-1)bso_12] >= 0)
3.81/1.80	
3.81/1.80	
3.81/1.80	
3.81/1.80	We simplified constraint (4) using rule (POLY_REMOVE_MIN_MAX) which results in the following new constraint:
3.81/1.80	
3.81/1.80	(5)    (x[0] + [-1] + [-1]y[0] >= 0  ==>  (U^Increasing(COND_EVAL(>(x[0], y[0]), x[0], y[0])), >=) & [(-1)bni_11 + (-1)Bound*bni_11] + [(-1)bni_11]y[0] + [bni_11]x[0] >= 0 & [(-1)bso_12] >= 0)
3.81/1.80	
3.81/1.80	
3.81/1.80	
3.81/1.80	We simplified constraint (5) using rule (IDP_SMT_SPLIT) which results in the following new constraint:
3.81/1.80	
3.81/1.80	(6)    (x[0] >= 0  ==>  (U^Increasing(COND_EVAL(>(x[0], y[0]), x[0], y[0])), >=) & [(-1)Bound*bni_11] + [bni_11]x[0] >= 0 & [(-1)bso_12] >= 0)
3.81/1.80	
3.81/1.80	
3.81/1.80	
3.81/1.80	We simplified constraint (6) using rule (IDP_SMT_SPLIT) which results in the following new constraints:
3.81/1.80	
3.81/1.80	(7)    (x[0] >= 0 & y[0] >= 0  ==>  (U^Increasing(COND_EVAL(>(x[0], y[0]), x[0], y[0])), >=) & [(-1)Bound*bni_11] + [bni_11]x[0] >= 0 & [(-1)bso_12] >= 0)
3.81/1.80	
3.81/1.80	(8)    (x[0] >= 0 & y[0] >= 0  ==>  (U^Increasing(COND_EVAL(>(x[0], y[0]), x[0], y[0])), >=) & [(-1)Bound*bni_11] + [bni_11]x[0] >= 0 & [(-1)bso_12] >= 0)
3.81/1.80	
3.81/1.80	
3.81/1.80	
3.81/1.80	
3.81/1.80	
3.81/1.80	
3.81/1.80	
3.81/1.80	
3.81/1.80	For Pair COND_EVAL(TRUE, x, y) -> EVAL(-(x, 1), y) the following chains were created:
3.81/1.80	*We consider the chain EVAL(x[0], y[0]) -> COND_EVAL(>(x[0], y[0]), x[0], y[0]), COND_EVAL(TRUE, x[1], y[1]) -> EVAL(-(x[1], 1), y[1]), EVAL(x[0], y[0]) -> COND_EVAL(>(x[0], y[0]), x[0], y[0]) which results in the following constraint:
3.81/1.80	
3.81/1.80	(1)    (>(x[0], y[0])=TRUE & x[0]=x[1] & y[0]=y[1] & -(x[1], 1)=x[0]1 & y[1]=y[0]1  ==>  COND_EVAL(TRUE, x[1], y[1])_>=_NonInfC & COND_EVAL(TRUE, x[1], y[1])_>=_EVAL(-(x[1], 1), y[1]) & (U^Increasing(EVAL(-(x[1], 1), y[1])), >=))
3.81/1.80	
3.81/1.80	
3.81/1.80	
3.81/1.80	We simplified constraint (1) using rules (III), (IV) which results in the following new constraint:
3.81/1.80	
3.81/1.80	(2)    (>(x[0], y[0])=TRUE  ==>  COND_EVAL(TRUE, x[0], y[0])_>=_NonInfC & COND_EVAL(TRUE, x[0], y[0])_>=_EVAL(-(x[0], 1), y[0]) & (U^Increasing(EVAL(-(x[1], 1), y[1])), >=))
3.81/1.80	
3.81/1.80	
3.81/1.80	
3.81/1.80	We simplified constraint (2) using rule (POLY_CONSTRAINTS) which results in the following new constraint:
3.81/1.80	
3.81/1.80	(3)    (x[0] + [-1] + [-1]y[0] >= 0  ==>  (U^Increasing(EVAL(-(x[1], 1), y[1])), >=) & [(-1)bni_13 + (-1)Bound*bni_13] + [(-1)bni_13]y[0] + [bni_13]x[0] >= 0 & [1 + (-1)bso_14] >= 0)
3.81/1.80	
3.81/1.80	
3.81/1.80	
3.81/1.80	We simplified constraint (3) using rule (IDP_POLY_SIMPLIFY) which results in the following new constraint:
3.81/1.80	
3.81/1.80	(4)    (x[0] + [-1] + [-1]y[0] >= 0  ==>  (U^Increasing(EVAL(-(x[1], 1), y[1])), >=) & [(-1)bni_13 + (-1)Bound*bni_13] + [(-1)bni_13]y[0] + [bni_13]x[0] >= 0 & [1 + (-1)bso_14] >= 0)
3.81/1.80	
3.81/1.80	
3.81/1.80	
3.81/1.80	We simplified constraint (4) using rule (POLY_REMOVE_MIN_MAX) which results in the following new constraint:
3.81/1.80	
3.81/1.80	(5)    (x[0] + [-1] + [-1]y[0] >= 0  ==>  (U^Increasing(EVAL(-(x[1], 1), y[1])), >=) & [(-1)bni_13 + (-1)Bound*bni_13] + [(-1)bni_13]y[0] + [bni_13]x[0] >= 0 & [1 + (-1)bso_14] >= 0)
3.81/1.80	
3.81/1.80	
3.81/1.80	
3.81/1.80	We simplified constraint (5) using rule (IDP_SMT_SPLIT) which results in the following new constraint:
3.81/1.80	
3.81/1.80	(6)    (x[0] >= 0  ==>  (U^Increasing(EVAL(-(x[1], 1), y[1])), >=) & [(-1)Bound*bni_13] + [bni_13]x[0] >= 0 & [1 + (-1)bso_14] >= 0)
3.81/1.80	
3.81/1.80	
3.81/1.80	
3.81/1.80	We simplified constraint (6) using rule (IDP_SMT_SPLIT) which results in the following new constraints:
3.81/1.80	
3.81/1.80	(7)    (x[0] >= 0 & y[0] >= 0  ==>  (U^Increasing(EVAL(-(x[1], 1), y[1])), >=) & [(-1)Bound*bni_13] + [bni_13]x[0] >= 0 & [1 + (-1)bso_14] >= 0)
3.81/1.80	
3.81/1.80	(8)    (x[0] >= 0 & y[0] >= 0  ==>  (U^Increasing(EVAL(-(x[1], 1), y[1])), >=) & [(-1)Bound*bni_13] + [bni_13]x[0] >= 0 & [1 + (-1)bso_14] >= 0)
3.81/1.80	
3.81/1.80	
3.81/1.80	
3.81/1.80	
3.81/1.80	
3.81/1.80	
3.81/1.80	
3.81/1.80	
3.81/1.80	To summarize, we get the following constraints P__>=_ for the following pairs.
3.81/1.80	
3.81/1.80	*EVAL(x, y) -> COND_EVAL(>(x, y), x, y)
3.81/1.80	
3.81/1.80	*(x[0] >= 0 & y[0] >= 0  ==>  (U^Increasing(COND_EVAL(>(x[0], y[0]), x[0], y[0])), >=) & [(-1)Bound*bni_11] + [bni_11]x[0] >= 0 & [(-1)bso_12] >= 0)
3.81/1.80	
3.81/1.80	
3.81/1.80	*(x[0] >= 0 & y[0] >= 0  ==>  (U^Increasing(COND_EVAL(>(x[0], y[0]), x[0], y[0])), >=) & [(-1)Bound*bni_11] + [bni_11]x[0] >= 0 & [(-1)bso_12] >= 0)
3.81/1.80	
3.81/1.80	
3.81/1.80	
3.81/1.80	
3.81/1.80	*COND_EVAL(TRUE, x, y) -> EVAL(-(x, 1), y)
3.81/1.80	
3.81/1.80	*(x[0] >= 0 & y[0] >= 0  ==>  (U^Increasing(EVAL(-(x[1], 1), y[1])), >=) & [(-1)Bound*bni_13] + [bni_13]x[0] >= 0 & [1 + (-1)bso_14] >= 0)
3.81/1.80	
3.81/1.80	
3.81/1.80	*(x[0] >= 0 & y[0] >= 0  ==>  (U^Increasing(EVAL(-(x[1], 1), y[1])), >=) & [(-1)Bound*bni_13] + [bni_13]x[0] >= 0 & [1 + (-1)bso_14] >= 0)
3.81/1.80	
3.81/1.80	
3.81/1.80	
3.81/1.80	
3.81/1.80	
3.81/1.80	
3.81/1.80	
3.81/1.80	
3.81/1.80	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. 
3.81/1.80	
3.81/1.80	Using the following integer polynomial ordering the  resulting constraints can be solved 
3.81/1.80	
3.81/1.80	Polynomial interpretation over integers[POLO]:
3.81/1.80	
3.81/1.80	   POL(TRUE) = 0
3.81/1.80	   POL(FALSE) = 0
3.81/1.80	   POL(EVAL(x_1, x_2)) = [-1] + [-1]x_2 + x_1
3.81/1.80	   POL(COND_EVAL(x_1, x_2, x_3)) = [-1] + [-1]x_3 + x_2
3.81/1.80	   POL(>(x_1, x_2)) = [-1]
3.81/1.80	   POL(-(x_1, x_2)) = x_1 + [-1]x_2
3.81/1.80	   POL(1) = [1]
3.81/1.80	
3.81/1.80	
3.81/1.80	The following pairs are in P_>:
3.81/1.80	
3.81/1.80	
3.81/1.80	   COND_EVAL(TRUE, x[1], y[1]) -> EVAL(-(x[1], 1), y[1])
3.81/1.80	
3.81/1.80	
3.81/1.80	The following pairs are in P_bound:
3.81/1.80	
3.81/1.80	
3.81/1.80	   EVAL(x[0], y[0]) -> COND_EVAL(>(x[0], y[0]), x[0], y[0])
3.81/1.80	   COND_EVAL(TRUE, x[1], y[1]) -> EVAL(-(x[1], 1), y[1])
3.81/1.80	
3.81/1.80	
3.81/1.80	The following pairs are in P_>=:
3.81/1.80	
3.81/1.80	
3.81/1.80	   EVAL(x[0], y[0]) -> COND_EVAL(>(x[0], y[0]), x[0], y[0])
3.81/1.80	
3.81/1.80	
3.81/1.80	There are no usable rules.
3.81/1.80	----------------------------------------
3.81/1.80	
3.81/1.80	(6)
3.81/1.80	Obligation:
3.81/1.80	IDP problem:
3.81/1.80	The following function symbols are pre-defined:
3.81/1.80	<<<
3.81/1.80	 & 	~ 	Bwand: (Integer, Integer) -> Integer
3.81/1.80	 >= 	~ 	Ge: (Integer, Integer) -> Boolean
3.81/1.80	 | 	~ 	Bwor: (Integer, Integer) -> Integer
3.81/1.80	 / 	~ 	Div: (Integer, Integer) -> Integer
3.81/1.80	 != 	~ 	Neq: (Integer, Integer) -> Boolean
3.81/1.80	 && 	~ 	Land: (Boolean, Boolean) -> Boolean
3.81/1.80	 ! 	~ 	Lnot: (Boolean) -> Boolean
3.81/1.80	 = 	~ 	Eq: (Integer, Integer) -> Boolean
3.81/1.80	 <= 	~ 	Le: (Integer, Integer) -> Boolean
3.81/1.80	 ^ 	~ 	Bwxor: (Integer, Integer) -> Integer
3.81/1.80	 % 	~ 	Mod: (Integer, Integer) -> Integer
3.81/1.80	 > 	~ 	Gt: (Integer, Integer) -> Boolean
3.81/1.80	 + 	~ 	Add: (Integer, Integer) -> Integer
3.81/1.80	 -1 	~ 	UnaryMinus: (Integer) -> Integer
3.81/1.80	 < 	~ 	Lt: (Integer, Integer) -> Boolean
3.81/1.80	 || 	~ 	Lor: (Boolean, Boolean) -> Boolean
3.81/1.80	 - 	~ 	Sub: (Integer, Integer) -> Integer
3.81/1.80	 ~ 	~ 	Bwnot: (Integer) -> Integer
3.81/1.80	 * 	~ 	Mul: (Integer, Integer) -> Integer
3.81/1.80	>>>
3.81/1.80	
3.81/1.80	
3.81/1.80	The following domains are used:
3.81/1.80	   Integer
3.81/1.80	
3.81/1.80	R is empty.
3.81/1.80	
3.81/1.80	The integer pair graph contains the following rules and edges:
3.81/1.80	(0): EVAL(x[0], y[0]) -> COND_EVAL(x[0] > y[0], x[0], y[0])
3.81/1.80	
3.81/1.80	
3.81/1.80	The set Q consists of the following terms:
3.81/1.80	eval(x0, x1)
3.81/1.80	Cond_eval(TRUE, x0, x1)
3.81/1.80	
3.81/1.80	----------------------------------------
3.81/1.80	
3.81/1.80	(7) IDependencyGraphProof (EQUIVALENT)
3.81/1.80	The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 1 less node.
3.81/1.80	----------------------------------------
3.81/1.80	
3.81/1.80	(8)
3.81/1.80	TRUE
3.84/1.82	EOF