3.70/1.87	YES
3.88/1.89	proof of /export/starexec/sandbox/benchmark/theBenchmark.itrs
3.88/1.89	# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty
3.88/1.89	
3.88/1.89	
3.88/1.89	Termination of the given ITRS could be proven:
3.88/1.89	
3.88/1.89	(0) ITRS
3.88/1.89	(1) ITRStoIDPProof [EQUIVALENT, 0 ms]
3.88/1.89	(2) IDP
3.88/1.89	(3) UsableRulesProof [EQUIVALENT, 0 ms]
3.88/1.89	(4) IDP
3.88/1.89	(5) IDPNonInfProof [SOUND, 143 ms]
3.88/1.89	(6) IDP
3.88/1.89	(7) IDependencyGraphProof [EQUIVALENT, 0 ms]
3.88/1.89	(8) TRUE
3.88/1.89	
3.88/1.89	
3.88/1.89	----------------------------------------
3.88/1.89	
3.88/1.89	(0)
3.88/1.89	Obligation:
3.88/1.89	ITRS problem:
3.88/1.89	
3.88/1.89	The following function symbols are pre-defined:
3.88/1.89	<<<
3.88/1.89	 & 	~ 	Bwand: (Integer, Integer) -> Integer
3.88/1.89	 >= 	~ 	Ge: (Integer, Integer) -> Boolean
3.88/1.89	 | 	~ 	Bwor: (Integer, Integer) -> Integer
3.88/1.89	 / 	~ 	Div: (Integer, Integer) -> Integer
3.88/1.89	 != 	~ 	Neq: (Integer, Integer) -> Boolean
3.88/1.89	 && 	~ 	Land: (Boolean, Boolean) -> Boolean
3.88/1.89	 ! 	~ 	Lnot: (Boolean) -> Boolean
3.88/1.89	 = 	~ 	Eq: (Integer, Integer) -> Boolean
3.88/1.89	 <= 	~ 	Le: (Integer, Integer) -> Boolean
3.88/1.89	 ^ 	~ 	Bwxor: (Integer, Integer) -> Integer
3.88/1.89	 % 	~ 	Mod: (Integer, Integer) -> Integer
3.88/1.89	 > 	~ 	Gt: (Integer, Integer) -> Boolean
3.88/1.89	 + 	~ 	Add: (Integer, Integer) -> Integer
3.88/1.89	 -1 	~ 	UnaryMinus: (Integer) -> Integer
3.88/1.89	 < 	~ 	Lt: (Integer, Integer) -> Boolean
3.88/1.89	 || 	~ 	Lor: (Boolean, Boolean) -> Boolean
3.88/1.89	 - 	~ 	Sub: (Integer, Integer) -> Integer
3.88/1.89	 ~ 	~ 	Bwnot: (Integer) -> Integer
3.88/1.89	 * 	~ 	Mul: (Integer, Integer) -> Integer
3.88/1.89	>>>
3.88/1.89	
3.88/1.89	The TRS R consists of the following rules:
3.88/1.89	eval(x) -> Cond_eval(x % 2 = 0 && x > 0, x)
3.88/1.89	Cond_eval(TRUE, x) -> eval(x - 1)
3.88/1.89	The set Q consists of the following terms:
3.88/1.89	eval(x0)
3.88/1.89	Cond_eval(TRUE, x0)
3.88/1.89	
3.88/1.89	----------------------------------------
3.88/1.89	
3.88/1.89	(1) ITRStoIDPProof (EQUIVALENT)
3.88/1.89	Added dependency pairs
3.88/1.89	----------------------------------------
3.88/1.89	
3.88/1.89	(2)
3.88/1.89	Obligation:
3.88/1.89	IDP problem:
3.88/1.89	The following function symbols are pre-defined:
3.88/1.89	<<<
3.88/1.89	 & 	~ 	Bwand: (Integer, Integer) -> Integer
3.88/1.89	 >= 	~ 	Ge: (Integer, Integer) -> Boolean
3.88/1.89	 | 	~ 	Bwor: (Integer, Integer) -> Integer
3.88/1.89	 / 	~ 	Div: (Integer, Integer) -> Integer
3.88/1.89	 != 	~ 	Neq: (Integer, Integer) -> Boolean
3.88/1.89	 && 	~ 	Land: (Boolean, Boolean) -> Boolean
3.88/1.89	 ! 	~ 	Lnot: (Boolean) -> Boolean
3.88/1.89	 = 	~ 	Eq: (Integer, Integer) -> Boolean
3.88/1.89	 <= 	~ 	Le: (Integer, Integer) -> Boolean
3.88/1.89	 ^ 	~ 	Bwxor: (Integer, Integer) -> Integer
3.88/1.89	 % 	~ 	Mod: (Integer, Integer) -> Integer
3.88/1.89	 > 	~ 	Gt: (Integer, Integer) -> Boolean
3.88/1.89	 + 	~ 	Add: (Integer, Integer) -> Integer
3.88/1.89	 -1 	~ 	UnaryMinus: (Integer) -> Integer
3.88/1.89	 < 	~ 	Lt: (Integer, Integer) -> Boolean
3.88/1.89	 || 	~ 	Lor: (Boolean, Boolean) -> Boolean
3.88/1.89	 - 	~ 	Sub: (Integer, Integer) -> Integer
3.88/1.89	 ~ 	~ 	Bwnot: (Integer) -> Integer
3.88/1.89	 * 	~ 	Mul: (Integer, Integer) -> Integer
3.88/1.89	>>>
3.88/1.89	
3.88/1.89	
3.88/1.89	The following domains are used:
3.88/1.89	   Boolean, Integer
3.88/1.89	
3.88/1.89	The ITRS R consists of the following rules:
3.88/1.89	eval(x) -> Cond_eval(x % 2 = 0 && x > 0, x)
3.88/1.89	Cond_eval(TRUE, x) -> eval(x - 1)
3.88/1.89	
3.88/1.89	The integer pair graph contains the following rules and edges:
3.88/1.89	(0): EVAL(x[0]) -> COND_EVAL(x[0] % 2 = 0 && x[0] > 0, x[0])
3.88/1.89	(1): COND_EVAL(TRUE, x[1]) -> EVAL(x[1] - 1)
3.88/1.89	
3.88/1.89	   (0) -> (1), if (x[0] % 2 = 0 && x[0] > 0  & x[0] ->^* x[1])
3.88/1.89	   (1) -> (0), if (x[1] - 1 ->^* x[0])
3.88/1.89	
3.88/1.89	The set Q consists of the following terms:
3.88/1.89	eval(x0)
3.88/1.89	Cond_eval(TRUE, x0)
3.88/1.89	
3.88/1.89	----------------------------------------
3.88/1.89	
3.88/1.89	(3) UsableRulesProof (EQUIVALENT)
3.88/1.89	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.
3.88/1.89	----------------------------------------
3.88/1.89	
3.88/1.89	(4)
3.88/1.89	Obligation:
3.88/1.89	IDP problem:
3.88/1.89	The following function symbols are pre-defined:
3.88/1.89	<<<
3.88/1.89	 & 	~ 	Bwand: (Integer, Integer) -> Integer
3.88/1.89	 >= 	~ 	Ge: (Integer, Integer) -> Boolean
3.88/1.89	 | 	~ 	Bwor: (Integer, Integer) -> Integer
3.88/1.89	 / 	~ 	Div: (Integer, Integer) -> Integer
3.88/1.89	 != 	~ 	Neq: (Integer, Integer) -> Boolean
3.88/1.89	 && 	~ 	Land: (Boolean, Boolean) -> Boolean
3.88/1.89	 ! 	~ 	Lnot: (Boolean) -> Boolean
3.88/1.89	 = 	~ 	Eq: (Integer, Integer) -> Boolean
3.88/1.89	 <= 	~ 	Le: (Integer, Integer) -> Boolean
3.88/1.89	 ^ 	~ 	Bwxor: (Integer, Integer) -> Integer
3.88/1.89	 % 	~ 	Mod: (Integer, Integer) -> Integer
3.88/1.89	 > 	~ 	Gt: (Integer, Integer) -> Boolean
3.88/1.89	 + 	~ 	Add: (Integer, Integer) -> Integer
3.88/1.89	 -1 	~ 	UnaryMinus: (Integer) -> Integer
3.88/1.89	 < 	~ 	Lt: (Integer, Integer) -> Boolean
3.88/1.89	 || 	~ 	Lor: (Boolean, Boolean) -> Boolean
3.88/1.89	 - 	~ 	Sub: (Integer, Integer) -> Integer
3.88/1.89	 ~ 	~ 	Bwnot: (Integer) -> Integer
3.88/1.89	 * 	~ 	Mul: (Integer, Integer) -> Integer
3.88/1.89	>>>
3.88/1.89	
3.88/1.89	
3.88/1.89	The following domains are used:
3.88/1.89	   Boolean, Integer
3.88/1.89	
3.88/1.89	R is empty.
3.88/1.89	
3.88/1.89	The integer pair graph contains the following rules and edges:
3.88/1.89	(0): EVAL(x[0]) -> COND_EVAL(x[0] % 2 = 0 && x[0] > 0, x[0])
3.88/1.89	(1): COND_EVAL(TRUE, x[1]) -> EVAL(x[1] - 1)
3.88/1.89	
3.88/1.89	   (0) -> (1), if (x[0] % 2 = 0 && x[0] > 0  & x[0] ->^* x[1])
3.88/1.89	   (1) -> (0), if (x[1] - 1 ->^* x[0])
3.88/1.89	
3.88/1.89	The set Q consists of the following terms:
3.88/1.89	eval(x0)
3.88/1.89	Cond_eval(TRUE, x0)
3.88/1.89	
3.88/1.89	----------------------------------------
3.88/1.89	
3.88/1.89	(5) IDPNonInfProof (SOUND)
3.88/1.89	Used the following options for this NonInfProof:
3.88/1.89	
3.88/1.89	IDPGPoloSolver:
3.88/1.89	Range: [(-1,2)]
3.88/1.89	IsNat: false
3.88/1.89	Interpretation Shape Heuristic: aprove.DPFramework.IDPProblem.Processors.nonInf.poly.IdpDefaultShapeHeuristic@1ccfbc04
3.88/1.89	Constraint Generator: NonInfConstraintGenerator:
3.88/1.89	PathGenerator: MetricPathGenerator:
3.88/1.89	Max Left Steps: 1
3.88/1.89	Max Right Steps: 1
3.88/1.89	
3.88/1.89	
3.88/1.89	
3.88/1.89	The constraints were generated the following way:
3.88/1.89	
3.88/1.89	The DP Problem is simplified using the Induction Calculus [NONINF] with the following steps:
3.88/1.89	
3.88/1.89	Note that final constraints are written in bold face.
3.88/1.89	
3.88/1.89	
3.88/1.89	
3.88/1.89	For Pair EVAL(x) -> COND_EVAL(&&(=(%(x, 2), 0), >(x, 0)), x) the following chains were created:
3.88/1.89	*We consider the chain EVAL(x[0]) -> COND_EVAL(&&(=(%(x[0], 2), 0), >(x[0], 0)), x[0]), COND_EVAL(TRUE, x[1]) -> EVAL(-(x[1], 1)) which results in the following constraint:
3.88/1.89	
3.88/1.89	(1)    (&&(=(%(x[0], 2), 0), >(x[0], 0))=TRUE & x[0]=x[1]  ==>  EVAL(x[0])_>=_NonInfC & EVAL(x[0])_>=_COND_EVAL(&&(=(%(x[0], 2), 0), >(x[0], 0)), x[0]) & (U^Increasing(COND_EVAL(&&(=(%(x[0], 2), 0), >(x[0], 0)), x[0])), >=))
3.88/1.89	
3.88/1.89	
3.88/1.89	
3.88/1.89	We simplified constraint (1) using rules (IV), (IDP_BOOLEAN) which results in the following new constraint:
3.88/1.89	
3.88/1.89	(2)    (>(x[0], 0)=TRUE & >=(%(x[0], 2), 0)=TRUE & <=(%(x[0], 2), 0)=TRUE  ==>  EVAL(x[0])_>=_NonInfC & EVAL(x[0])_>=_COND_EVAL(&&(=(%(x[0], 2), 0), >(x[0], 0)), x[0]) & (U^Increasing(COND_EVAL(&&(=(%(x[0], 2), 0), >(x[0], 0)), x[0])), >=))
3.88/1.89	
3.88/1.89	
3.88/1.89	
3.88/1.89	We simplified constraint (2) using rule (POLY_CONSTRAINTS) which results in the following new constraint:
3.88/1.89	
3.88/1.89	(3)    (x[0] + [-1] >= 0 & max{[2], [-2]} >= 0 & [-1]min{[2], [-2]} >= 0  ==>  (U^Increasing(COND_EVAL(&&(=(%(x[0], 2), 0), >(x[0], 0)), x[0])), >=) & [(-1)bni_11 + (-1)Bound*bni_11] + [bni_11]x[0] >= 0 & [(-1)bso_12] >= 0)
3.88/1.89	
3.88/1.89	
3.88/1.89	
3.88/1.89	We simplified constraint (3) using rule (IDP_POLY_SIMPLIFY) which results in the following new constraint:
3.88/1.89	
3.88/1.89	(4)    (x[0] + [-1] >= 0 & max{[2], [-2]} >= 0 & [-1]min{[2], [-2]} >= 0  ==>  (U^Increasing(COND_EVAL(&&(=(%(x[0], 2), 0), >(x[0], 0)), x[0])), >=) & [(-1)bni_11 + (-1)Bound*bni_11] + [bni_11]x[0] >= 0 & [(-1)bso_12] >= 0)
3.88/1.89	
3.88/1.89	
3.88/1.89	
3.88/1.89	We simplified constraint (4) using rule (POLY_REMOVE_MIN_MAX) which results in the following new constraint:
3.88/1.89	
3.88/1.89	(5)    (x[0] + [-1] >= 0 & [4] >= 0 & [2] >= 0 & [2] >= 0  ==>  (U^Increasing(COND_EVAL(&&(=(%(x[0], 2), 0), >(x[0], 0)), x[0])), >=) & [(-1)bni_11 + (-1)Bound*bni_11] + [bni_11]x[0] >= 0 & [(-1)bso_12] >= 0)
3.88/1.89	
3.88/1.89	
3.88/1.89	
3.88/1.89	We simplified constraint (5) using rule (IDP_POLY_GCD) which results in the following new constraint:
3.88/1.89	
3.88/1.89	(6)    (x[0] + [-1] >= 0 & [1] >= 0 & [1] >= 0 & [1] >= 0  ==>  (U^Increasing(COND_EVAL(&&(=(%(x[0], 2), 0), >(x[0], 0)), x[0])), >=) & [(-1)bni_11 + (-1)Bound*bni_11] + [bni_11]x[0] >= 0 & [(-1)bso_12] >= 0)
3.88/1.89	
3.88/1.89	
3.88/1.89	
3.88/1.89	
3.88/1.89	
3.88/1.89	
3.88/1.89	
3.88/1.89	
3.88/1.89	For Pair COND_EVAL(TRUE, x) -> EVAL(-(x, 1)) the following chains were created:
3.88/1.89	*We consider the chain EVAL(x[0]) -> COND_EVAL(&&(=(%(x[0], 2), 0), >(x[0], 0)), x[0]), COND_EVAL(TRUE, x[1]) -> EVAL(-(x[1], 1)), EVAL(x[0]) -> COND_EVAL(&&(=(%(x[0], 2), 0), >(x[0], 0)), x[0]) which results in the following constraint:
3.88/1.89	
3.88/1.89	(1)    (&&(=(%(x[0], 2), 0), >(x[0], 0))=TRUE & x[0]=x[1] & -(x[1], 1)=x[0]1  ==>  COND_EVAL(TRUE, x[1])_>=_NonInfC & COND_EVAL(TRUE, x[1])_>=_EVAL(-(x[1], 1)) & (U^Increasing(EVAL(-(x[1], 1))), >=))
3.88/1.89	
3.88/1.89	
3.88/1.89	
3.88/1.89	We simplified constraint (1) using rules (III), (IV), (IDP_BOOLEAN) which results in the following new constraint:
3.88/1.89	
3.88/1.89	(2)    (>(x[0], 0)=TRUE & >=(%(x[0], 2), 0)=TRUE & <=(%(x[0], 2), 0)=TRUE  ==>  COND_EVAL(TRUE, x[0])_>=_NonInfC & COND_EVAL(TRUE, x[0])_>=_EVAL(-(x[0], 1)) & (U^Increasing(EVAL(-(x[1], 1))), >=))
3.88/1.89	
3.88/1.89	
3.88/1.89	
3.88/1.89	We simplified constraint (2) using rule (POLY_CONSTRAINTS) which results in the following new constraint:
3.88/1.89	
3.88/1.89	(3)    (x[0] + [-1] >= 0 & max{[2], [-2]} >= 0 & [-1]min{[2], [-2]} >= 0  ==>  (U^Increasing(EVAL(-(x[1], 1))), >=) & [(-1)bni_13 + (-1)Bound*bni_13] + [bni_13]x[0] >= 0 & [1 + (-1)bso_14] >= 0)
3.88/1.89	
3.88/1.89	
3.88/1.89	
3.88/1.89	We simplified constraint (3) using rule (IDP_POLY_SIMPLIFY) which results in the following new constraint:
3.88/1.89	
3.88/1.89	(4)    (x[0] + [-1] >= 0 & max{[2], [-2]} >= 0 & [-1]min{[2], [-2]} >= 0  ==>  (U^Increasing(EVAL(-(x[1], 1))), >=) & [(-1)bni_13 + (-1)Bound*bni_13] + [bni_13]x[0] >= 0 & [1 + (-1)bso_14] >= 0)
3.88/1.89	
3.88/1.89	
3.88/1.89	
3.88/1.89	We simplified constraint (4) using rule (POLY_REMOVE_MIN_MAX) which results in the following new constraint:
3.88/1.89	
3.88/1.89	(5)    (x[0] + [-1] >= 0 & [4] >= 0 & [2] >= 0 & [2] >= 0  ==>  (U^Increasing(EVAL(-(x[1], 1))), >=) & [(-1)bni_13 + (-1)Bound*bni_13] + [bni_13]x[0] >= 0 & [1 + (-1)bso_14] >= 0)
3.88/1.89	
3.88/1.89	
3.88/1.89	
3.88/1.89	We simplified constraint (5) using rule (IDP_POLY_GCD) which results in the following new constraint:
3.88/1.89	
3.88/1.89	(6)    (x[0] + [-1] >= 0 & [1] >= 0 & [1] >= 0 & [1] >= 0  ==>  (U^Increasing(EVAL(-(x[1], 1))), >=) & [(-1)bni_13 + (-1)Bound*bni_13] + [bni_13]x[0] >= 0 & [1 + (-1)bso_14] >= 0)
3.88/1.89	
3.88/1.89	
3.88/1.89	
3.88/1.89	
3.88/1.89	
3.88/1.89	
3.88/1.89	
3.88/1.89	
3.88/1.89	To summarize, we get the following constraints P__>=_ for the following pairs.
3.88/1.89	
3.88/1.89	*EVAL(x) -> COND_EVAL(&&(=(%(x, 2), 0), >(x, 0)), x)
3.88/1.89	
3.88/1.89	*(x[0] + [-1] >= 0 & [1] >= 0 & [1] >= 0 & [1] >= 0  ==>  (U^Increasing(COND_EVAL(&&(=(%(x[0], 2), 0), >(x[0], 0)), x[0])), >=) & [(-1)bni_11 + (-1)Bound*bni_11] + [bni_11]x[0] >= 0 & [(-1)bso_12] >= 0)
3.88/1.89	
3.88/1.89	
3.88/1.89	
3.88/1.89	
3.88/1.89	*COND_EVAL(TRUE, x) -> EVAL(-(x, 1))
3.88/1.89	
3.88/1.89	*(x[0] + [-1] >= 0 & [1] >= 0 & [1] >= 0 & [1] >= 0  ==>  (U^Increasing(EVAL(-(x[1], 1))), >=) & [(-1)bni_13 + (-1)Bound*bni_13] + [bni_13]x[0] >= 0 & [1 + (-1)bso_14] >= 0)
3.88/1.89	
3.88/1.89	
3.88/1.89	
3.88/1.89	
3.88/1.89	
3.88/1.89	
3.88/1.89	
3.88/1.89	
3.88/1.89	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. 
3.88/1.89	
3.88/1.89	Using the following integer polynomial ordering the  resulting constraints can be solved 
3.88/1.89	
3.88/1.89	Polynomial interpretation over integers[POLO]:
3.88/1.89	
3.88/1.89	   POL(TRUE) = 0
3.88/1.89	   POL(FALSE) = [2]
3.88/1.89	   POL(EVAL(x_1)) = [-1] + x_1
3.88/1.89	   POL(COND_EVAL(x_1, x_2)) = [-1] + x_2 + [-1]x_1
3.88/1.89	   POL(&&(x_1, x_2)) = 0
3.88/1.89	   POL(=(x_1, x_2)) = [-1]
3.88/1.89	   POL(2) = [2]
3.88/1.89	   POL(0) = 0
3.88/1.89	   POL(>(x_1, x_2)) = [-1]
3.88/1.89	   POL(-(x_1, x_2)) = x_1 + [-1]x_2
3.88/1.89	   POL(1) = [1]
3.88/1.89	   
3.88/1.89	   Polynomial Interpretations with Context Sensitive Arithemetic Replacement
3.88/1.89	   POL(Term^CSAR-Mode @ Context)
3.88/1.89	   
3.88/1.89	   POL(%(x_1, 2)^1 @ {}) = max{x_2, [-1]x_2}
3.88/1.89	   POL(%(x_1, 2)^-1 @ {}) = min{x_2, [-1]x_2}
3.88/1.89	
3.88/1.89	
3.88/1.89	The following pairs are in P_>:
3.88/1.89	
3.88/1.89	
3.88/1.89	   COND_EVAL(TRUE, x[1]) -> EVAL(-(x[1], 1))
3.88/1.89	
3.88/1.89	
3.88/1.89	The following pairs are in P_bound:
3.88/1.89	
3.88/1.89	
3.88/1.89	   EVAL(x[0]) -> COND_EVAL(&&(=(%(x[0], 2), 0), >(x[0], 0)), x[0])
3.88/1.89	   COND_EVAL(TRUE, x[1]) -> EVAL(-(x[1], 1))
3.88/1.89	
3.88/1.89	
3.88/1.89	The following pairs are in P_>=:
3.88/1.89	
3.88/1.89	
3.88/1.89	   EVAL(x[0]) -> COND_EVAL(&&(=(%(x[0], 2), 0), >(x[0], 0)), x[0])
3.88/1.89	
3.88/1.89	
3.88/1.89	At least the following rules have been oriented under context sensitive arithmetic replacement:
3.88/1.89	
3.88/1.89	   TRUE^1 -> &&(TRUE, TRUE)^1
3.88/1.89	   FALSE^1 -> &&(TRUE, FALSE)^1
3.88/1.89	   FALSE^1 -> &&(FALSE, TRUE)^1
3.88/1.89	   FALSE^1 -> &&(FALSE, FALSE)^1
3.88/1.89	
3.88/1.89	----------------------------------------
3.88/1.89	
3.88/1.89	(6)
3.88/1.89	Obligation:
3.88/1.89	IDP problem:
3.88/1.89	The following function symbols are pre-defined:
3.88/1.89	<<<
3.88/1.89	 & 	~ 	Bwand: (Integer, Integer) -> Integer
3.88/1.89	 >= 	~ 	Ge: (Integer, Integer) -> Boolean
3.88/1.89	 | 	~ 	Bwor: (Integer, Integer) -> Integer
3.88/1.89	 / 	~ 	Div: (Integer, Integer) -> Integer
3.88/1.89	 != 	~ 	Neq: (Integer, Integer) -> Boolean
3.88/1.89	 && 	~ 	Land: (Boolean, Boolean) -> Boolean
3.88/1.89	 ! 	~ 	Lnot: (Boolean) -> Boolean
3.88/1.89	 = 	~ 	Eq: (Integer, Integer) -> Boolean
3.88/1.89	 <= 	~ 	Le: (Integer, Integer) -> Boolean
3.88/1.89	 ^ 	~ 	Bwxor: (Integer, Integer) -> Integer
3.88/1.89	 % 	~ 	Mod: (Integer, Integer) -> Integer
3.88/1.89	 > 	~ 	Gt: (Integer, Integer) -> Boolean
3.88/1.89	 + 	~ 	Add: (Integer, Integer) -> Integer
3.88/1.89	 -1 	~ 	UnaryMinus: (Integer) -> Integer
3.88/1.89	 < 	~ 	Lt: (Integer, Integer) -> Boolean
3.88/1.89	 || 	~ 	Lor: (Boolean, Boolean) -> Boolean
3.88/1.89	 - 	~ 	Sub: (Integer, Integer) -> Integer
3.88/1.89	 ~ 	~ 	Bwnot: (Integer) -> Integer
3.88/1.89	 * 	~ 	Mul: (Integer, Integer) -> Integer
3.88/1.89	>>>
3.88/1.89	
3.88/1.89	
3.88/1.89	The following domains are used:
3.88/1.89	   Boolean, Integer
3.88/1.89	
3.88/1.89	R is empty.
3.88/1.89	
3.88/1.89	The integer pair graph contains the following rules and edges:
3.88/1.89	(0): EVAL(x[0]) -> COND_EVAL(x[0] % 2 = 0 && x[0] > 0, x[0])
3.88/1.89	
3.88/1.89	
3.88/1.89	The set Q consists of the following terms:
3.88/1.89	eval(x0)
3.88/1.89	Cond_eval(TRUE, x0)
3.88/1.89	
3.88/1.89	----------------------------------------
3.88/1.89	
3.88/1.89	(7) IDependencyGraphProof (EQUIVALENT)
3.88/1.89	The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 1 less node.
3.88/1.89	----------------------------------------
3.88/1.89	
3.88/1.89	(8)
3.88/1.89	TRUE
3.88/1.91	EOF