4.50/2.04	YES
4.50/2.06	proof of /export/starexec/sandbox2/benchmark/theBenchmark.itrs
4.50/2.06	# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty
4.50/2.06	
4.50/2.06	
4.50/2.06	Termination of the given ITRS could be proven:
4.50/2.06	
4.50/2.06	(0) ITRS
4.50/2.06	(1) ITRStoIDPProof [EQUIVALENT, 0 ms]
4.50/2.06	(2) IDP
4.50/2.06	(3) UsableRulesProof [EQUIVALENT, 0 ms]
4.50/2.06	(4) IDP
4.50/2.06	(5) IDependencyGraphProof [EQUIVALENT, 0 ms]
4.50/2.06	(6) IDP
4.50/2.06	(7) IDPNonInfProof [SOUND, 196 ms]
4.50/2.06	(8) AND
4.50/2.06	    (9) IDP
4.50/2.06	        (10) IDependencyGraphProof [EQUIVALENT, 0 ms]
4.50/2.06	        (11) TRUE
4.50/2.06	    (12) IDP
4.50/2.06	        (13) IDependencyGraphProof [EQUIVALENT, 0 ms]
4.50/2.06	        (14) TRUE
4.50/2.06	
4.50/2.06	
4.50/2.06	----------------------------------------
4.50/2.06	
4.50/2.06	(0)
4.50/2.06	Obligation:
4.50/2.06	ITRS problem:
4.50/2.06	
4.50/2.06	The following function symbols are pre-defined:
4.50/2.06	<<<
4.50/2.06	 & 	~ 	Bwand: (Integer, Integer) -> Integer
4.50/2.06	 >= 	~ 	Ge: (Integer, Integer) -> Boolean
4.50/2.06	 | 	~ 	Bwor: (Integer, Integer) -> Integer
4.50/2.06	 / 	~ 	Div: (Integer, Integer) -> Integer
4.50/2.06	 != 	~ 	Neq: (Integer, Integer) -> Boolean
4.50/2.06	 && 	~ 	Land: (Boolean, Boolean) -> Boolean
4.50/2.06	 ! 	~ 	Lnot: (Boolean) -> Boolean
4.50/2.06	 = 	~ 	Eq: (Integer, Integer) -> Boolean
4.50/2.06	 <= 	~ 	Le: (Integer, Integer) -> Boolean
4.50/2.06	 ^ 	~ 	Bwxor: (Integer, Integer) -> Integer
4.50/2.06	 % 	~ 	Mod: (Integer, Integer) -> Integer
4.50/2.06	 + 	~ 	Add: (Integer, Integer) -> Integer
4.50/2.06	 > 	~ 	Gt: (Integer, Integer) -> Boolean
4.50/2.06	 -1 	~ 	UnaryMinus: (Integer) -> Integer
4.50/2.06	 < 	~ 	Lt: (Integer, Integer) -> Boolean
4.50/2.06	 || 	~ 	Lor: (Boolean, Boolean) -> Boolean
4.50/2.06	 - 	~ 	Sub: (Integer, Integer) -> Integer
4.50/2.06	 ~ 	~ 	Bwnot: (Integer) -> Integer
4.50/2.06	 * 	~ 	Mul: (Integer, Integer) -> Integer
4.50/2.06	>>>
4.50/2.06	
4.50/2.06	The TRS R consists of the following rules:
4.50/2.06	sqrt(x) -> f(x, 0, 1, 1)
4.50/2.06	f(x, z, u, w) -> Cond_f(x >= w && u >= 0, x, z, u, w)
4.50/2.06	Cond_f(TRUE, x, z, u, w) -> f(x, z + 1, u + 2, w + u + 2)
4.50/2.06	f(x, z, u, w) -> Cond_f1(w > x, x, z, u, w)
4.50/2.06	Cond_f1(TRUE, x, z, u, w) -> z
4.50/2.06	The set Q consists of the following terms:
4.50/2.06	sqrt(x0)
4.50/2.06	f(x0, x1, x2, x3)
4.50/2.06	Cond_f(TRUE, x0, x1, x2, x3)
4.50/2.06	Cond_f1(TRUE, x0, x1, x2, x3)
4.50/2.06	
4.50/2.06	----------------------------------------
4.50/2.06	
4.50/2.06	(1) ITRStoIDPProof (EQUIVALENT)
4.50/2.06	Added dependency pairs
4.50/2.06	----------------------------------------
4.50/2.06	
4.50/2.06	(2)
4.50/2.06	Obligation:
4.50/2.06	IDP problem:
4.50/2.06	The following function symbols are pre-defined:
4.50/2.06	<<<
4.50/2.06	 & 	~ 	Bwand: (Integer, Integer) -> Integer
4.50/2.06	 >= 	~ 	Ge: (Integer, Integer) -> Boolean
4.50/2.06	 | 	~ 	Bwor: (Integer, Integer) -> Integer
4.50/2.06	 / 	~ 	Div: (Integer, Integer) -> Integer
4.50/2.06	 != 	~ 	Neq: (Integer, Integer) -> Boolean
4.50/2.06	 && 	~ 	Land: (Boolean, Boolean) -> Boolean
4.50/2.06	 ! 	~ 	Lnot: (Boolean) -> Boolean
4.50/2.06	 = 	~ 	Eq: (Integer, Integer) -> Boolean
4.50/2.06	 <= 	~ 	Le: (Integer, Integer) -> Boolean
4.50/2.06	 ^ 	~ 	Bwxor: (Integer, Integer) -> Integer
4.50/2.06	 % 	~ 	Mod: (Integer, Integer) -> Integer
4.50/2.06	 + 	~ 	Add: (Integer, Integer) -> Integer
4.50/2.06	 > 	~ 	Gt: (Integer, Integer) -> Boolean
4.50/2.06	 -1 	~ 	UnaryMinus: (Integer) -> Integer
4.50/2.06	 < 	~ 	Lt: (Integer, Integer) -> Boolean
4.50/2.06	 || 	~ 	Lor: (Boolean, Boolean) -> Boolean
4.50/2.06	 - 	~ 	Sub: (Integer, Integer) -> Integer
4.50/2.06	 ~ 	~ 	Bwnot: (Integer) -> Integer
4.50/2.06	 * 	~ 	Mul: (Integer, Integer) -> Integer
4.50/2.06	>>>
4.50/2.06	
4.50/2.06	
4.50/2.06	The following domains are used:
4.50/2.06	   Boolean, Integer
4.50/2.06	
4.50/2.06	The ITRS R consists of the following rules:
4.50/2.06	sqrt(x) -> f(x, 0, 1, 1)
4.50/2.06	f(x, z, u, w) -> Cond_f(x >= w && u >= 0, x, z, u, w)
4.50/2.06	Cond_f(TRUE, x, z, u, w) -> f(x, z + 1, u + 2, w + u + 2)
4.50/2.06	f(x, z, u, w) -> Cond_f1(w > x, x, z, u, w)
4.50/2.06	Cond_f1(TRUE, x, z, u, w) -> z
4.50/2.06	
4.50/2.06	The integer pair graph contains the following rules and edges:
4.50/2.06	(0): SQRT(x[0]) -> F(x[0], 0, 1, 1)
4.50/2.06	(1): F(x[1], z[1], u[1], w[1]) -> COND_F(x[1] >= w[1] && u[1] >= 0, x[1], z[1], u[1], w[1])
4.50/2.06	(2): COND_F(TRUE, x[2], z[2], u[2], w[2]) -> F(x[2], z[2] + 1, u[2] + 2, w[2] + u[2] + 2)
4.50/2.06	(3): F(x[3], z[3], u[3], w[3]) -> COND_F1(w[3] > x[3], x[3], z[3], u[3], w[3])
4.50/2.06	
4.50/2.06	   (0) -> (1), if (x[0] ->^* x[1] & 0 ->^* z[1] & 1 ->^* u[1] & 1 ->^* w[1])
4.50/2.06	   (0) -> (3), if (x[0] ->^* x[3] & 0 ->^* z[3] & 1 ->^* u[3] & 1 ->^* w[3])
4.50/2.06	   (1) -> (2), if (x[1] >= w[1] && u[1] >= 0  & x[1] ->^* x[2] & z[1] ->^* z[2] & u[1] ->^* u[2] & w[1] ->^* w[2])
4.50/2.06	   (2) -> (1), if (x[2] ->^* x[1] & z[2] + 1 ->^* z[1] & u[2] + 2 ->^* u[1] & w[2] + u[2] + 2 ->^* w[1])
4.50/2.06	   (2) -> (3), if (x[2] ->^* x[3] & z[2] + 1 ->^* z[3] & u[2] + 2 ->^* u[3] & w[2] + u[2] + 2 ->^* w[3])
4.50/2.06	
4.50/2.06	The set Q consists of the following terms:
4.50/2.06	sqrt(x0)
4.50/2.06	f(x0, x1, x2, x3)
4.50/2.06	Cond_f(TRUE, x0, x1, x2, x3)
4.50/2.06	Cond_f1(TRUE, x0, x1, x2, x3)
4.50/2.06	
4.50/2.06	----------------------------------------
4.50/2.06	
4.50/2.06	(3) UsableRulesProof (EQUIVALENT)
4.50/2.06	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.
4.50/2.06	----------------------------------------
4.50/2.06	
4.50/2.06	(4)
4.50/2.06	Obligation:
4.50/2.06	IDP problem:
4.50/2.06	The following function symbols are pre-defined:
4.50/2.06	<<<
4.50/2.06	 & 	~ 	Bwand: (Integer, Integer) -> Integer
4.50/2.06	 >= 	~ 	Ge: (Integer, Integer) -> Boolean
4.50/2.06	 | 	~ 	Bwor: (Integer, Integer) -> Integer
4.50/2.06	 / 	~ 	Div: (Integer, Integer) -> Integer
4.50/2.06	 != 	~ 	Neq: (Integer, Integer) -> Boolean
4.50/2.06	 && 	~ 	Land: (Boolean, Boolean) -> Boolean
4.50/2.06	 ! 	~ 	Lnot: (Boolean) -> Boolean
4.50/2.06	 = 	~ 	Eq: (Integer, Integer) -> Boolean
4.50/2.06	 <= 	~ 	Le: (Integer, Integer) -> Boolean
4.50/2.06	 ^ 	~ 	Bwxor: (Integer, Integer) -> Integer
4.50/2.06	 % 	~ 	Mod: (Integer, Integer) -> Integer
4.50/2.06	 + 	~ 	Add: (Integer, Integer) -> Integer
4.50/2.06	 > 	~ 	Gt: (Integer, Integer) -> Boolean
4.50/2.06	 -1 	~ 	UnaryMinus: (Integer) -> Integer
4.50/2.06	 < 	~ 	Lt: (Integer, Integer) -> Boolean
4.50/2.06	 || 	~ 	Lor: (Boolean, Boolean) -> Boolean
4.50/2.06	 - 	~ 	Sub: (Integer, Integer) -> Integer
4.50/2.06	 ~ 	~ 	Bwnot: (Integer) -> Integer
4.50/2.06	 * 	~ 	Mul: (Integer, Integer) -> Integer
4.50/2.06	>>>
4.50/2.06	
4.50/2.06	
4.50/2.06	The following domains are used:
4.50/2.06	   Boolean, Integer
4.50/2.06	
4.50/2.06	R is empty.
4.50/2.06	
4.50/2.06	The integer pair graph contains the following rules and edges:
4.50/2.06	(0): SQRT(x[0]) -> F(x[0], 0, 1, 1)
4.50/2.06	(1): F(x[1], z[1], u[1], w[1]) -> COND_F(x[1] >= w[1] && u[1] >= 0, x[1], z[1], u[1], w[1])
4.50/2.06	(2): COND_F(TRUE, x[2], z[2], u[2], w[2]) -> F(x[2], z[2] + 1, u[2] + 2, w[2] + u[2] + 2)
4.50/2.06	(3): F(x[3], z[3], u[3], w[3]) -> COND_F1(w[3] > x[3], x[3], z[3], u[3], w[3])
4.50/2.06	
4.50/2.06	   (0) -> (1), if (x[0] ->^* x[1] & 0 ->^* z[1] & 1 ->^* u[1] & 1 ->^* w[1])
4.50/2.06	   (0) -> (3), if (x[0] ->^* x[3] & 0 ->^* z[3] & 1 ->^* u[3] & 1 ->^* w[3])
4.50/2.06	   (1) -> (2), if (x[1] >= w[1] && u[1] >= 0  & x[1] ->^* x[2] & z[1] ->^* z[2] & u[1] ->^* u[2] & w[1] ->^* w[2])
4.50/2.06	   (2) -> (1), if (x[2] ->^* x[1] & z[2] + 1 ->^* z[1] & u[2] + 2 ->^* u[1] & w[2] + u[2] + 2 ->^* w[1])
4.50/2.06	   (2) -> (3), if (x[2] ->^* x[3] & z[2] + 1 ->^* z[3] & u[2] + 2 ->^* u[3] & w[2] + u[2] + 2 ->^* w[3])
4.50/2.06	
4.50/2.06	The set Q consists of the following terms:
4.50/2.06	sqrt(x0)
4.50/2.06	f(x0, x1, x2, x3)
4.50/2.06	Cond_f(TRUE, x0, x1, x2, x3)
4.50/2.06	Cond_f1(TRUE, x0, x1, x2, x3)
4.50/2.06	
4.50/2.06	----------------------------------------
4.50/2.06	
4.50/2.06	(5) IDependencyGraphProof (EQUIVALENT)
4.50/2.06	The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 2 less nodes.
4.50/2.06	----------------------------------------
4.50/2.06	
4.50/2.06	(6)
4.50/2.06	Obligation:
4.50/2.06	IDP problem:
4.50/2.06	The following function symbols are pre-defined:
4.50/2.06	<<<
4.50/2.06	 & 	~ 	Bwand: (Integer, Integer) -> Integer
4.50/2.06	 >= 	~ 	Ge: (Integer, Integer) -> Boolean
4.50/2.06	 | 	~ 	Bwor: (Integer, Integer) -> Integer
4.50/2.06	 / 	~ 	Div: (Integer, Integer) -> Integer
4.50/2.06	 != 	~ 	Neq: (Integer, Integer) -> Boolean
4.50/2.06	 && 	~ 	Land: (Boolean, Boolean) -> Boolean
4.50/2.06	 ! 	~ 	Lnot: (Boolean) -> Boolean
4.50/2.06	 = 	~ 	Eq: (Integer, Integer) -> Boolean
4.50/2.06	 <= 	~ 	Le: (Integer, Integer) -> Boolean
4.50/2.06	 ^ 	~ 	Bwxor: (Integer, Integer) -> Integer
4.50/2.06	 % 	~ 	Mod: (Integer, Integer) -> Integer
4.50/2.06	 + 	~ 	Add: (Integer, Integer) -> Integer
4.50/2.06	 > 	~ 	Gt: (Integer, Integer) -> Boolean
4.50/2.06	 -1 	~ 	UnaryMinus: (Integer) -> Integer
4.50/2.06	 < 	~ 	Lt: (Integer, Integer) -> Boolean
4.50/2.06	 || 	~ 	Lor: (Boolean, Boolean) -> Boolean
4.50/2.06	 - 	~ 	Sub: (Integer, Integer) -> Integer
4.50/2.06	 ~ 	~ 	Bwnot: (Integer) -> Integer
4.50/2.06	 * 	~ 	Mul: (Integer, Integer) -> Integer
4.50/2.06	>>>
4.50/2.06	
4.50/2.06	
4.50/2.06	The following domains are used:
4.50/2.06	   Integer, Boolean
4.50/2.06	
4.50/2.06	R is empty.
4.50/2.06	
4.50/2.06	The integer pair graph contains the following rules and edges:
4.50/2.06	(2): COND_F(TRUE, x[2], z[2], u[2], w[2]) -> F(x[2], z[2] + 1, u[2] + 2, w[2] + u[2] + 2)
4.50/2.06	(1): F(x[1], z[1], u[1], w[1]) -> COND_F(x[1] >= w[1] && u[1] >= 0, x[1], z[1], u[1], w[1])
4.50/2.06	
4.50/2.06	   (2) -> (1), if (x[2] ->^* x[1] & z[2] + 1 ->^* z[1] & u[2] + 2 ->^* u[1] & w[2] + u[2] + 2 ->^* w[1])
4.50/2.06	   (1) -> (2), if (x[1] >= w[1] && u[1] >= 0  & x[1] ->^* x[2] & z[1] ->^* z[2] & u[1] ->^* u[2] & w[1] ->^* w[2])
4.50/2.06	
4.50/2.06	The set Q consists of the following terms:
4.50/2.06	sqrt(x0)
4.50/2.06	f(x0, x1, x2, x3)
4.50/2.06	Cond_f(TRUE, x0, x1, x2, x3)
4.50/2.06	Cond_f1(TRUE, x0, x1, x2, x3)
4.50/2.06	
4.50/2.06	----------------------------------------
4.50/2.06	
4.50/2.06	(7) IDPNonInfProof (SOUND)
4.50/2.06	Used the following options for this NonInfProof:
4.50/2.06	
4.50/2.06	IDPGPoloSolver:
4.50/2.06	Range: [(-1,2)]
4.50/2.06	IsNat: false
4.50/2.06	Interpretation Shape Heuristic: aprove.DPFramework.IDPProblem.Processors.nonInf.poly.IdpDefaultShapeHeuristic@5c9256b6
4.50/2.06	Constraint Generator: NonInfConstraintGenerator:
4.50/2.06	PathGenerator: MetricPathGenerator:
4.50/2.06	Max Left Steps: 1
4.50/2.06	Max Right Steps: 1
4.50/2.06	
4.50/2.06	
4.50/2.06	
4.50/2.06	The constraints were generated the following way:
4.50/2.06	
4.50/2.06	The DP Problem is simplified using the Induction Calculus [NONINF] with the following steps:
4.50/2.06	
4.50/2.06	Note that final constraints are written in bold face.
4.50/2.06	
4.50/2.06	
4.50/2.06	
4.50/2.06	For Pair COND_F(TRUE, x[2], z[2], u[2], w[2]) -> F(x[2], +(z[2], 1), +(u[2], 2), +(+(w[2], u[2]), 2)) the following chains were created:
4.50/2.06	*We consider the chain F(x[1], z[1], u[1], w[1]) -> COND_F(&&(>=(x[1], w[1]), >=(u[1], 0)), x[1], z[1], u[1], w[1]), COND_F(TRUE, x[2], z[2], u[2], w[2]) -> F(x[2], +(z[2], 1), +(u[2], 2), +(+(w[2], u[2]), 2)), F(x[1], z[1], u[1], w[1]) -> COND_F(&&(>=(x[1], w[1]), >=(u[1], 0)), x[1], z[1], u[1], w[1]) which results in the following constraint:
4.50/2.06	
4.50/2.06	(1)    (&&(>=(x[1], w[1]), >=(u[1], 0))=TRUE & x[1]=x[2] & z[1]=z[2] & u[1]=u[2] & w[1]=w[2] & x[2]=x[1]1 & +(z[2], 1)=z[1]1 & +(u[2], 2)=u[1]1 & +(+(w[2], u[2]), 2)=w[1]1  ==>  COND_F(TRUE, x[2], z[2], u[2], w[2])_>=_NonInfC & COND_F(TRUE, x[2], z[2], u[2], w[2])_>=_F(x[2], +(z[2], 1), +(u[2], 2), +(+(w[2], u[2]), 2)) & (U^Increasing(F(x[2], +(z[2], 1), +(u[2], 2), +(+(w[2], u[2]), 2))), >=))
4.50/2.06	
4.50/2.06	
4.50/2.06	
4.50/2.06	We simplified constraint (1) using rules (III), (IV), (IDP_BOOLEAN) which results in the following new constraint:
4.50/2.06	
4.50/2.06	(2)    (>=(x[1], w[1])=TRUE & >=(u[1], 0)=TRUE  ==>  COND_F(TRUE, x[1], z[1], u[1], w[1])_>=_NonInfC & COND_F(TRUE, x[1], z[1], u[1], w[1])_>=_F(x[1], +(z[1], 1), +(u[1], 2), +(+(w[1], u[1]), 2)) & (U^Increasing(F(x[2], +(z[2], 1), +(u[2], 2), +(+(w[2], u[2]), 2))), >=))
4.50/2.06	
4.50/2.06	
4.50/2.06	
4.50/2.06	We simplified constraint (2) using rule (POLY_CONSTRAINTS) which results in the following new constraint:
4.50/2.06	
4.50/2.06	(3)    (x[1] + [-1]w[1] >= 0 & u[1] >= 0  ==>  (U^Increasing(F(x[2], +(z[2], 1), +(u[2], 2), +(+(w[2], u[2]), 2))), >=) & [(-1)Bound*bni_16] + [(-1)bni_16]w[1] + [(-1)bni_16]u[1] + [bni_16]x[1] >= 0 & [1 + (-1)bso_17] >= 0)
4.50/2.06	
4.50/2.06	
4.50/2.06	
4.50/2.06	We simplified constraint (3) using rule (IDP_POLY_SIMPLIFY) which results in the following new constraint:
4.50/2.06	
4.50/2.06	(4)    (x[1] + [-1]w[1] >= 0 & u[1] >= 0  ==>  (U^Increasing(F(x[2], +(z[2], 1), +(u[2], 2), +(+(w[2], u[2]), 2))), >=) & [(-1)Bound*bni_16] + [(-1)bni_16]w[1] + [(-1)bni_16]u[1] + [bni_16]x[1] >= 0 & [1 + (-1)bso_17] >= 0)
4.50/2.06	
4.50/2.06	
4.50/2.06	
4.50/2.06	We simplified constraint (4) using rule (POLY_REMOVE_MIN_MAX) which results in the following new constraint:
4.50/2.06	
4.50/2.06	(5)    (x[1] + [-1]w[1] >= 0 & u[1] >= 0  ==>  (U^Increasing(F(x[2], +(z[2], 1), +(u[2], 2), +(+(w[2], u[2]), 2))), >=) & [(-1)Bound*bni_16] + [(-1)bni_16]w[1] + [(-1)bni_16]u[1] + [bni_16]x[1] >= 0 & [1 + (-1)bso_17] >= 0)
4.50/2.06	
4.50/2.06	
4.50/2.06	
4.50/2.06	We simplified constraint (5) using rule (IDP_UNRESTRICTED_VARS) which results in the following new constraint:
4.50/2.06	
4.50/2.06	(6)    (x[1] + [-1]w[1] >= 0 & u[1] >= 0  ==>  (U^Increasing(F(x[2], +(z[2], 1), +(u[2], 2), +(+(w[2], u[2]), 2))), >=) & 0 = 0 & [(-1)Bound*bni_16] + [(-1)bni_16]w[1] + [(-1)bni_16]u[1] + [bni_16]x[1] >= 0 & [1 + (-1)bso_17] >= 0)
4.50/2.06	
4.50/2.06	
4.50/2.06	
4.50/2.06	We simplified constraint (6) using rule (IDP_SMT_SPLIT) which results in the following new constraint:
4.50/2.06	
4.50/2.06	(7)    (x[1] >= 0 & u[1] >= 0  ==>  (U^Increasing(F(x[2], +(z[2], 1), +(u[2], 2), +(+(w[2], u[2]), 2))), >=) & 0 = 0 & [(-1)Bound*bni_16] + [(-1)bni_16]u[1] + [bni_16]x[1] >= 0 & [1 + (-1)bso_17] >= 0)
4.50/2.06	
4.50/2.06	
4.50/2.06	
4.50/2.06	We simplified constraint (7) using rule (IDP_SMT_SPLIT) which results in the following new constraints:
4.50/2.06	
4.50/2.06	(8)    (x[1] >= 0 & u[1] >= 0 & w[1] >= 0  ==>  (U^Increasing(F(x[2], +(z[2], 1), +(u[2], 2), +(+(w[2], u[2]), 2))), >=) & 0 = 0 & [(-1)Bound*bni_16] + [(-1)bni_16]u[1] + [bni_16]x[1] >= 0 & [1 + (-1)bso_17] >= 0)
4.50/2.06	
4.50/2.06	(9)    (x[1] >= 0 & u[1] >= 0 & w[1] >= 0  ==>  (U^Increasing(F(x[2], +(z[2], 1), +(u[2], 2), +(+(w[2], u[2]), 2))), >=) & 0 = 0 & [(-1)Bound*bni_16] + [(-1)bni_16]u[1] + [bni_16]x[1] >= 0 & [1 + (-1)bso_17] >= 0)
4.50/2.06	
4.50/2.06	
4.50/2.06	
4.50/2.06	
4.50/2.06	
4.50/2.06	
4.50/2.06	
4.50/2.06	
4.50/2.06	For Pair F(x[1], z[1], u[1], w[1]) -> COND_F(&&(>=(x[1], w[1]), >=(u[1], 0)), x[1], z[1], u[1], w[1]) the following chains were created:
4.50/2.06	*We consider the chain F(x[1], z[1], u[1], w[1]) -> COND_F(&&(>=(x[1], w[1]), >=(u[1], 0)), x[1], z[1], u[1], w[1]), COND_F(TRUE, x[2], z[2], u[2], w[2]) -> F(x[2], +(z[2], 1), +(u[2], 2), +(+(w[2], u[2]), 2)) which results in the following constraint:
4.50/2.06	
4.50/2.06	(1)    (&&(>=(x[1], w[1]), >=(u[1], 0))=TRUE & x[1]=x[2] & z[1]=z[2] & u[1]=u[2] & w[1]=w[2]  ==>  F(x[1], z[1], u[1], w[1])_>=_NonInfC & F(x[1], z[1], u[1], w[1])_>=_COND_F(&&(>=(x[1], w[1]), >=(u[1], 0)), x[1], z[1], u[1], w[1]) & (U^Increasing(COND_F(&&(>=(x[1], w[1]), >=(u[1], 0)), x[1], z[1], u[1], w[1])), >=))
4.50/2.06	
4.50/2.06	
4.50/2.06	
4.50/2.06	We simplified constraint (1) using rules (IV), (IDP_BOOLEAN) which results in the following new constraint:
4.50/2.06	
4.50/2.06	(2)    (>=(x[1], w[1])=TRUE & >=(u[1], 0)=TRUE  ==>  F(x[1], z[1], u[1], w[1])_>=_NonInfC & F(x[1], z[1], u[1], w[1])_>=_COND_F(&&(>=(x[1], w[1]), >=(u[1], 0)), x[1], z[1], u[1], w[1]) & (U^Increasing(COND_F(&&(>=(x[1], w[1]), >=(u[1], 0)), x[1], z[1], u[1], w[1])), >=))
4.50/2.06	
4.50/2.06	
4.50/2.06	
4.50/2.06	We simplified constraint (2) using rule (POLY_CONSTRAINTS) which results in the following new constraint:
4.50/2.06	
4.50/2.06	(3)    (x[1] + [-1]w[1] >= 0 & u[1] >= 0  ==>  (U^Increasing(COND_F(&&(>=(x[1], w[1]), >=(u[1], 0)), x[1], z[1], u[1], w[1])), >=) & [bni_18 + (-1)Bound*bni_18] + [(-1)bni_18]w[1] + [bni_18]x[1] >= 0 & [(-1)bso_19] + u[1] >= 0)
4.50/2.06	
4.50/2.06	
4.50/2.06	
4.50/2.06	We simplified constraint (3) using rule (IDP_POLY_SIMPLIFY) which results in the following new constraint:
4.50/2.06	
4.50/2.06	(4)    (x[1] + [-1]w[1] >= 0 & u[1] >= 0  ==>  (U^Increasing(COND_F(&&(>=(x[1], w[1]), >=(u[1], 0)), x[1], z[1], u[1], w[1])), >=) & [bni_18 + (-1)Bound*bni_18] + [(-1)bni_18]w[1] + [bni_18]x[1] >= 0 & [(-1)bso_19] + u[1] >= 0)
4.50/2.06	
4.50/2.06	
4.50/2.06	
4.50/2.06	We simplified constraint (4) using rule (POLY_REMOVE_MIN_MAX) which results in the following new constraint:
4.50/2.06	
4.50/2.06	(5)    (x[1] + [-1]w[1] >= 0 & u[1] >= 0  ==>  (U^Increasing(COND_F(&&(>=(x[1], w[1]), >=(u[1], 0)), x[1], z[1], u[1], w[1])), >=) & [bni_18 + (-1)Bound*bni_18] + [(-1)bni_18]w[1] + [bni_18]x[1] >= 0 & [(-1)bso_19] + u[1] >= 0)
4.50/2.06	
4.50/2.06	
4.50/2.06	
4.50/2.06	We simplified constraint (5) using rule (IDP_UNRESTRICTED_VARS) which results in the following new constraint:
4.50/2.06	
4.50/2.06	(6)    (x[1] + [-1]w[1] >= 0 & u[1] >= 0  ==>  (U^Increasing(COND_F(&&(>=(x[1], w[1]), >=(u[1], 0)), x[1], z[1], u[1], w[1])), >=) & 0 = 0 & [bni_18 + (-1)Bound*bni_18] + [(-1)bni_18]w[1] + [bni_18]x[1] >= 0 & [(-1)bso_19] + u[1] >= 0)
4.50/2.06	
4.50/2.06	
4.50/2.06	
4.50/2.06	We simplified constraint (6) using rule (IDP_SMT_SPLIT) which results in the following new constraint:
4.50/2.06	
4.50/2.06	(7)    (x[1] >= 0 & u[1] >= 0  ==>  (U^Increasing(COND_F(&&(>=(x[1], w[1]), >=(u[1], 0)), x[1], z[1], u[1], w[1])), >=) & 0 = 0 & [bni_18 + (-1)Bound*bni_18] + [bni_18]x[1] >= 0 & [(-1)bso_19] + u[1] >= 0)
4.50/2.06	
4.50/2.06	
4.50/2.06	
4.50/2.06	We simplified constraint (7) using rule (IDP_SMT_SPLIT) which results in the following new constraints:
4.50/2.06	
4.50/2.06	(8)    (x[1] >= 0 & u[1] >= 0 & w[1] >= 0  ==>  (U^Increasing(COND_F(&&(>=(x[1], w[1]), >=(u[1], 0)), x[1], z[1], u[1], w[1])), >=) & 0 = 0 & [bni_18 + (-1)Bound*bni_18] + [bni_18]x[1] >= 0 & [(-1)bso_19] + u[1] >= 0)
4.50/2.06	
4.50/2.06	(9)    (x[1] >= 0 & u[1] >= 0 & w[1] >= 0  ==>  (U^Increasing(COND_F(&&(>=(x[1], w[1]), >=(u[1], 0)), x[1], z[1], u[1], w[1])), >=) & 0 = 0 & [bni_18 + (-1)Bound*bni_18] + [bni_18]x[1] >= 0 & [(-1)bso_19] + u[1] >= 0)
4.50/2.06	
4.50/2.06	
4.50/2.06	
4.50/2.06	
4.50/2.06	
4.50/2.06	
4.50/2.06	
4.50/2.06	
4.50/2.06	To summarize, we get the following constraints P__>=_ for the following pairs.
4.50/2.06	
4.50/2.06	*COND_F(TRUE, x[2], z[2], u[2], w[2]) -> F(x[2], +(z[2], 1), +(u[2], 2), +(+(w[2], u[2]), 2))
4.50/2.06	
4.50/2.06	*(x[1] >= 0 & u[1] >= 0 & w[1] >= 0  ==>  (U^Increasing(F(x[2], +(z[2], 1), +(u[2], 2), +(+(w[2], u[2]), 2))), >=) & 0 = 0 & [(-1)Bound*bni_16] + [(-1)bni_16]u[1] + [bni_16]x[1] >= 0 & [1 + (-1)bso_17] >= 0)
4.50/2.06	
4.50/2.06	
4.50/2.06	*(x[1] >= 0 & u[1] >= 0 & w[1] >= 0  ==>  (U^Increasing(F(x[2], +(z[2], 1), +(u[2], 2), +(+(w[2], u[2]), 2))), >=) & 0 = 0 & [(-1)Bound*bni_16] + [(-1)bni_16]u[1] + [bni_16]x[1] >= 0 & [1 + (-1)bso_17] >= 0)
4.50/2.06	
4.50/2.06	
4.50/2.06	
4.50/2.06	
4.50/2.06	*F(x[1], z[1], u[1], w[1]) -> COND_F(&&(>=(x[1], w[1]), >=(u[1], 0)), x[1], z[1], u[1], w[1])
4.50/2.06	
4.50/2.06	*(x[1] >= 0 & u[1] >= 0 & w[1] >= 0  ==>  (U^Increasing(COND_F(&&(>=(x[1], w[1]), >=(u[1], 0)), x[1], z[1], u[1], w[1])), >=) & 0 = 0 & [bni_18 + (-1)Bound*bni_18] + [bni_18]x[1] >= 0 & [(-1)bso_19] + u[1] >= 0)
4.50/2.06	
4.50/2.06	
4.50/2.06	*(x[1] >= 0 & u[1] >= 0 & w[1] >= 0  ==>  (U^Increasing(COND_F(&&(>=(x[1], w[1]), >=(u[1], 0)), x[1], z[1], u[1], w[1])), >=) & 0 = 0 & [bni_18 + (-1)Bound*bni_18] + [bni_18]x[1] >= 0 & [(-1)bso_19] + u[1] >= 0)
4.50/2.06	
4.50/2.06	
4.50/2.06	
4.50/2.06	
4.50/2.06	
4.50/2.06	
4.50/2.06	
4.50/2.06	
4.50/2.06	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. 
4.50/2.06	
4.50/2.06	Using the following integer polynomial ordering the  resulting constraints can be solved 
4.50/2.06	
4.50/2.06	Polynomial interpretation over integers[POLO]:
4.50/2.06	
4.50/2.06	   POL(TRUE) = 0
4.50/2.06	   POL(FALSE) = [1]
4.50/2.06	   POL(COND_F(x_1, x_2, x_3, x_4, x_5)) = [-1]x_5 + [-1]x_4 + x_2 + [-1]x_1
4.50/2.06	   POL(F(x_1, x_2, x_3, x_4)) = [1] + [-1]x_4 + x_1
4.50/2.06	   POL(+(x_1, x_2)) = x_1 + x_2
4.50/2.06	   POL(1) = [1]
4.50/2.06	   POL(2) = [2]
4.50/2.06	   POL(&&(x_1, x_2)) = [-1]
4.50/2.06	   POL(>=(x_1, x_2)) = [-1]
4.50/2.06	   POL(0) = 0
4.50/2.06	
4.50/2.06	
4.50/2.06	The following pairs are in P_>:
4.50/2.06	
4.50/2.06	
4.50/2.06	   COND_F(TRUE, x[2], z[2], u[2], w[2]) -> F(x[2], +(z[2], 1), +(u[2], 2), +(+(w[2], u[2]), 2))
4.50/2.06	
4.50/2.06	
4.50/2.06	The following pairs are in P_bound:
4.50/2.06	
4.50/2.06	
4.50/2.06	   F(x[1], z[1], u[1], w[1]) -> COND_F(&&(>=(x[1], w[1]), >=(u[1], 0)), x[1], z[1], u[1], w[1])
4.50/2.06	
4.50/2.06	
4.50/2.06	The following pairs are in P_>=:
4.50/2.06	
4.50/2.06	
4.50/2.06	   F(x[1], z[1], u[1], w[1]) -> COND_F(&&(>=(x[1], w[1]), >=(u[1], 0)), x[1], z[1], u[1], w[1])
4.50/2.06	
4.50/2.06	
4.50/2.06	At least the following rules have been oriented under context sensitive arithmetic replacement:
4.50/2.06	
4.50/2.06	   TRUE^1 -> &&(TRUE, TRUE)^1
4.50/2.06	   FALSE^1 -> &&(TRUE, FALSE)^1
4.50/2.06	   FALSE^1 -> &&(FALSE, TRUE)^1
4.50/2.06	   FALSE^1 -> &&(FALSE, FALSE)^1
4.50/2.06	
4.50/2.06	----------------------------------------
4.50/2.06	
4.50/2.06	(8)
4.50/2.06	Complex Obligation (AND)
4.50/2.06	
4.50/2.06	----------------------------------------
4.50/2.06	
4.50/2.06	(9)
4.50/2.06	Obligation:
4.50/2.06	IDP problem:
4.50/2.06	The following function symbols are pre-defined:
4.50/2.06	<<<
4.50/2.06	 & 	~ 	Bwand: (Integer, Integer) -> Integer
4.50/2.06	 >= 	~ 	Ge: (Integer, Integer) -> Boolean
4.50/2.06	 | 	~ 	Bwor: (Integer, Integer) -> Integer
4.50/2.06	 / 	~ 	Div: (Integer, Integer) -> Integer
4.50/2.06	 != 	~ 	Neq: (Integer, Integer) -> Boolean
4.50/2.06	 && 	~ 	Land: (Boolean, Boolean) -> Boolean
4.50/2.06	 ! 	~ 	Lnot: (Boolean) -> Boolean
4.50/2.06	 = 	~ 	Eq: (Integer, Integer) -> Boolean
4.50/2.06	 <= 	~ 	Le: (Integer, Integer) -> Boolean
4.50/2.06	 ^ 	~ 	Bwxor: (Integer, Integer) -> Integer
4.50/2.06	 % 	~ 	Mod: (Integer, Integer) -> Integer
4.50/2.06	 + 	~ 	Add: (Integer, Integer) -> Integer
4.50/2.06	 > 	~ 	Gt: (Integer, Integer) -> Boolean
4.50/2.06	 -1 	~ 	UnaryMinus: (Integer) -> Integer
4.50/2.06	 < 	~ 	Lt: (Integer, Integer) -> Boolean
4.50/2.06	 || 	~ 	Lor: (Boolean, Boolean) -> Boolean
4.50/2.06	 - 	~ 	Sub: (Integer, Integer) -> Integer
4.50/2.06	 ~ 	~ 	Bwnot: (Integer) -> Integer
4.50/2.06	 * 	~ 	Mul: (Integer, Integer) -> Integer
4.50/2.06	>>>
4.50/2.06	
4.50/2.06	
4.50/2.06	The following domains are used:
4.50/2.06	   Boolean, Integer
4.50/2.06	
4.50/2.06	R is empty.
4.50/2.06	
4.50/2.06	The integer pair graph contains the following rules and edges:
4.50/2.06	(1): F(x[1], z[1], u[1], w[1]) -> COND_F(x[1] >= w[1] && u[1] >= 0, x[1], z[1], u[1], w[1])
4.50/2.06	
4.50/2.06	
4.50/2.06	The set Q consists of the following terms:
4.50/2.06	sqrt(x0)
4.50/2.06	f(x0, x1, x2, x3)
4.50/2.06	Cond_f(TRUE, x0, x1, x2, x3)
4.50/2.06	Cond_f1(TRUE, x0, x1, x2, x3)
4.50/2.06	
4.50/2.06	----------------------------------------
4.50/2.06	
4.50/2.06	(10) IDependencyGraphProof (EQUIVALENT)
4.50/2.06	The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 1 less node.
4.50/2.06	----------------------------------------
4.50/2.06	
4.50/2.06	(11)
4.50/2.06	TRUE
4.50/2.06	
4.50/2.06	----------------------------------------
4.50/2.06	
4.50/2.06	(12)
4.50/2.06	Obligation:
4.50/2.06	IDP problem:
4.50/2.06	The following function symbols are pre-defined:
4.50/2.06	<<<
4.50/2.06	 & 	~ 	Bwand: (Integer, Integer) -> Integer
4.50/2.06	 >= 	~ 	Ge: (Integer, Integer) -> Boolean
4.50/2.06	 | 	~ 	Bwor: (Integer, Integer) -> Integer
4.50/2.06	 / 	~ 	Div: (Integer, Integer) -> Integer
4.50/2.06	 != 	~ 	Neq: (Integer, Integer) -> Boolean
4.50/2.06	 && 	~ 	Land: (Boolean, Boolean) -> Boolean
4.50/2.06	 ! 	~ 	Lnot: (Boolean) -> Boolean
4.50/2.06	 = 	~ 	Eq: (Integer, Integer) -> Boolean
4.50/2.06	 <= 	~ 	Le: (Integer, Integer) -> Boolean
4.50/2.06	 ^ 	~ 	Bwxor: (Integer, Integer) -> Integer
4.50/2.06	 % 	~ 	Mod: (Integer, Integer) -> Integer
4.50/2.06	 + 	~ 	Add: (Integer, Integer) -> Integer
4.50/2.06	 > 	~ 	Gt: (Integer, Integer) -> Boolean
4.50/2.06	 -1 	~ 	UnaryMinus: (Integer) -> Integer
4.50/2.06	 < 	~ 	Lt: (Integer, Integer) -> Boolean
4.50/2.06	 || 	~ 	Lor: (Boolean, Boolean) -> Boolean
4.50/2.06	 - 	~ 	Sub: (Integer, Integer) -> Integer
4.50/2.06	 ~ 	~ 	Bwnot: (Integer) -> Integer
4.50/2.06	 * 	~ 	Mul: (Integer, Integer) -> Integer
4.50/2.06	>>>
4.50/2.06	
4.50/2.06	
4.50/2.06	The following domains are used:
4.50/2.06	   Integer
4.50/2.06	
4.50/2.06	R is empty.
4.50/2.06	
4.50/2.06	The integer pair graph contains the following rules and edges:
4.50/2.06	(2): COND_F(TRUE, x[2], z[2], u[2], w[2]) -> F(x[2], z[2] + 1, u[2] + 2, w[2] + u[2] + 2)
4.50/2.06	
4.50/2.06	
4.50/2.06	The set Q consists of the following terms:
4.50/2.06	sqrt(x0)
4.50/2.06	f(x0, x1, x2, x3)
4.50/2.06	Cond_f(TRUE, x0, x1, x2, x3)
4.50/2.06	Cond_f1(TRUE, x0, x1, x2, x3)
4.50/2.06	
4.50/2.06	----------------------------------------
4.50/2.06	
4.50/2.06	(13) IDependencyGraphProof (EQUIVALENT)
4.50/2.06	The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 1 less node.
4.50/2.06	----------------------------------------
4.50/2.06	
4.50/2.06	(14)
4.50/2.06	TRUE
4.80/2.14	EOF