3.90/1.82	YES
3.90/1.83	proof of /export/starexec/sandbox2/benchmark/theBenchmark.itrs
3.90/1.83	# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty
3.90/1.83	
3.90/1.83	
3.90/1.83	Termination of the given ITRS could be proven:
3.90/1.83	
3.90/1.83	(0) ITRS
3.90/1.83	(1) ITRStoIDPProof [EQUIVALENT, 0 ms]
3.90/1.83	(2) IDP
3.90/1.83	(3) UsableRulesProof [EQUIVALENT, 0 ms]
3.90/1.83	(4) IDP
3.90/1.83	(5) IDPNonInfProof [SOUND, 117 ms]
3.90/1.83	(6) IDP
3.90/1.83	(7) PisEmptyProof [EQUIVALENT, 0 ms]
3.90/1.83	(8) YES
3.90/1.83	
3.90/1.83	
3.90/1.83	----------------------------------------
3.90/1.83	
3.90/1.83	(0)
3.90/1.83	Obligation:
3.90/1.83	ITRS problem:
3.90/1.83	
3.90/1.83	The following function symbols are pre-defined:
3.90/1.83	<<<
3.90/1.83	 & 	~ 	Bwand: (Integer, Integer) -> Integer
3.90/1.83	 >= 	~ 	Ge: (Integer, Integer) -> Boolean
3.90/1.83	 | 	~ 	Bwor: (Integer, Integer) -> Integer
3.90/1.83	 / 	~ 	Div: (Integer, Integer) -> Integer
3.90/1.83	 != 	~ 	Neq: (Integer, Integer) -> Boolean
3.90/1.83	 && 	~ 	Land: (Boolean, Boolean) -> Boolean
3.90/1.83	 ! 	~ 	Lnot: (Boolean) -> Boolean
3.90/1.83	 = 	~ 	Eq: (Integer, Integer) -> Boolean
3.90/1.83	 <= 	~ 	Le: (Integer, Integer) -> Boolean
3.90/1.83	 ^ 	~ 	Bwxor: (Integer, Integer) -> Integer
3.90/1.83	 % 	~ 	Mod: (Integer, Integer) -> Integer
3.90/1.83	 + 	~ 	Add: (Integer, Integer) -> Integer
3.90/1.83	 > 	~ 	Gt: (Integer, Integer) -> Boolean
3.90/1.83	 -1 	~ 	UnaryMinus: (Integer) -> Integer
3.90/1.83	 < 	~ 	Lt: (Integer, Integer) -> Boolean
3.90/1.83	 || 	~ 	Lor: (Boolean, Boolean) -> Boolean
3.90/1.83	 - 	~ 	Sub: (Integer, Integer) -> Integer
3.90/1.83	 ~ 	~ 	Bwnot: (Integer) -> Integer
3.90/1.83	 * 	~ 	Mul: (Integer, Integer) -> Integer
3.90/1.83	>>>
3.90/1.83	
3.90/1.83	The TRS R consists of the following rules:
3.90/1.83	cu(TRUE, x) -> cu(x < 100000, x + 1)
3.90/1.83	The set Q consists of the following terms:
3.90/1.83	cu(TRUE, x0)
3.90/1.83	
3.90/1.83	----------------------------------------
3.90/1.83	
3.90/1.83	(1) ITRStoIDPProof (EQUIVALENT)
3.90/1.83	Added dependency pairs
3.90/1.83	----------------------------------------
3.90/1.83	
3.90/1.83	(2)
3.90/1.83	Obligation:
3.90/1.83	IDP problem:
3.90/1.83	The following function symbols are pre-defined:
3.90/1.83	<<<
3.90/1.83	 & 	~ 	Bwand: (Integer, Integer) -> Integer
3.90/1.83	 >= 	~ 	Ge: (Integer, Integer) -> Boolean
3.90/1.83	 | 	~ 	Bwor: (Integer, Integer) -> Integer
3.90/1.83	 / 	~ 	Div: (Integer, Integer) -> Integer
3.90/1.83	 != 	~ 	Neq: (Integer, Integer) -> Boolean
3.90/1.83	 && 	~ 	Land: (Boolean, Boolean) -> Boolean
3.90/1.83	 ! 	~ 	Lnot: (Boolean) -> Boolean
3.90/1.83	 = 	~ 	Eq: (Integer, Integer) -> Boolean
3.90/1.83	 <= 	~ 	Le: (Integer, Integer) -> Boolean
3.90/1.84	 ^ 	~ 	Bwxor: (Integer, Integer) -> Integer
3.90/1.84	 % 	~ 	Mod: (Integer, Integer) -> Integer
3.90/1.84	 + 	~ 	Add: (Integer, Integer) -> Integer
3.90/1.84	 > 	~ 	Gt: (Integer, Integer) -> Boolean
3.90/1.84	 -1 	~ 	UnaryMinus: (Integer) -> Integer
3.90/1.84	 < 	~ 	Lt: (Integer, Integer) -> Boolean
3.90/1.84	 || 	~ 	Lor: (Boolean, Boolean) -> Boolean
3.90/1.84	 - 	~ 	Sub: (Integer, Integer) -> Integer
3.90/1.84	 ~ 	~ 	Bwnot: (Integer) -> Integer
3.90/1.84	 * 	~ 	Mul: (Integer, Integer) -> Integer
3.90/1.84	>>>
3.90/1.84	
3.90/1.84	
3.90/1.84	The following domains are used:
3.90/1.84	   Integer
3.90/1.84	
3.90/1.84	The ITRS R consists of the following rules:
3.90/1.84	cu(TRUE, x) -> cu(x < 100000, x + 1)
3.90/1.84	
3.90/1.84	The integer pair graph contains the following rules and edges:
3.90/1.84	(0): CU(TRUE, x[0]) -> CU(x[0] < 100000, x[0] + 1)
3.90/1.84	
3.90/1.84	   (0) -> (0), if (x[0] < 100000  & x[0] + 1 ->^* x[0]')
3.90/1.84	
3.90/1.84	The set Q consists of the following terms:
3.90/1.84	cu(TRUE, x0)
3.90/1.84	
3.90/1.84	----------------------------------------
3.90/1.84	
3.90/1.84	(3) UsableRulesProof (EQUIVALENT)
3.90/1.84	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.
3.90/1.84	----------------------------------------
3.90/1.84	
3.90/1.84	(4)
3.90/1.84	Obligation:
3.90/1.84	IDP problem:
3.90/1.84	The following function symbols are pre-defined:
3.90/1.84	<<<
3.90/1.84	 & 	~ 	Bwand: (Integer, Integer) -> Integer
3.90/1.84	 >= 	~ 	Ge: (Integer, Integer) -> Boolean
3.90/1.84	 | 	~ 	Bwor: (Integer, Integer) -> Integer
3.90/1.84	 / 	~ 	Div: (Integer, Integer) -> Integer
3.90/1.84	 != 	~ 	Neq: (Integer, Integer) -> Boolean
3.90/1.84	 && 	~ 	Land: (Boolean, Boolean) -> Boolean
3.90/1.84	 ! 	~ 	Lnot: (Boolean) -> Boolean
3.90/1.84	 = 	~ 	Eq: (Integer, Integer) -> Boolean
3.90/1.84	 <= 	~ 	Le: (Integer, Integer) -> Boolean
3.90/1.84	 ^ 	~ 	Bwxor: (Integer, Integer) -> Integer
3.90/1.84	 % 	~ 	Mod: (Integer, Integer) -> Integer
3.90/1.84	 + 	~ 	Add: (Integer, Integer) -> Integer
3.90/1.84	 > 	~ 	Gt: (Integer, Integer) -> Boolean
3.90/1.84	 -1 	~ 	UnaryMinus: (Integer) -> Integer
3.90/1.84	 < 	~ 	Lt: (Integer, Integer) -> Boolean
3.90/1.84	 || 	~ 	Lor: (Boolean, Boolean) -> Boolean
3.90/1.84	 - 	~ 	Sub: (Integer, Integer) -> Integer
3.90/1.84	 ~ 	~ 	Bwnot: (Integer) -> Integer
3.90/1.84	 * 	~ 	Mul: (Integer, Integer) -> Integer
3.90/1.84	>>>
3.90/1.84	
3.90/1.84	
3.90/1.84	The following domains are used:
3.90/1.84	   Integer
3.90/1.84	
3.90/1.84	R is empty.
3.90/1.84	
3.90/1.84	The integer pair graph contains the following rules and edges:
3.90/1.84	(0): CU(TRUE, x[0]) -> CU(x[0] < 100000, x[0] + 1)
3.90/1.84	
3.90/1.84	   (0) -> (0), if (x[0] < 100000  & x[0] + 1 ->^* x[0]')
3.90/1.84	
3.90/1.84	The set Q consists of the following terms:
3.90/1.84	cu(TRUE, x0)
3.90/1.84	
3.90/1.84	----------------------------------------
3.90/1.84	
3.90/1.84	(5) IDPNonInfProof (SOUND)
3.90/1.84	Used the following options for this NonInfProof:
3.90/1.84	
3.90/1.84	IDPGPoloSolver:
3.90/1.84	Range: [(-1,2)]
3.90/1.84	IsNat: false
3.90/1.84	Interpretation Shape Heuristic: aprove.DPFramework.IDPProblem.Processors.nonInf.poly.IdpDefaultShapeHeuristic@6012b267
3.90/1.84	Constraint Generator: NonInfConstraintGenerator:
3.90/1.84	PathGenerator: MetricPathGenerator:
3.90/1.84	Max Left Steps: 1
3.90/1.84	Max Right Steps: 1
3.90/1.84	
3.90/1.84	
3.90/1.84	
3.90/1.84	The constraints were generated the following way:
3.90/1.84	
3.90/1.84	The DP Problem is simplified using the Induction Calculus [NONINF] with the following steps:
3.90/1.84	
3.90/1.84	Note that final constraints are written in bold face.
3.90/1.84	
3.90/1.84	
3.90/1.84	
3.90/1.84	For Pair CU(TRUE, x) -> CU(<(x, 100000), +(x, 1)) the following chains were created:
3.90/1.84	*We consider the chain CU(TRUE, x[0]) -> CU(<(x[0], 100000), +(x[0], 1)), CU(TRUE, x[0]) -> CU(<(x[0], 100000), +(x[0], 1)), CU(TRUE, x[0]) -> CU(<(x[0], 100000), +(x[0], 1)) which results in the following constraint:
3.90/1.84	
3.90/1.84	(1)    (<(x[0], 100000)=TRUE & +(x[0], 1)=x[0]1 & <(x[0]1, 100000)=TRUE & +(x[0]1, 1)=x[0]2  ==>  CU(TRUE, x[0]1)_>=_NonInfC & CU(TRUE, x[0]1)_>=_CU(<(x[0]1, 100000), +(x[0]1, 1)) & (U^Increasing(CU(<(x[0]1, 100000), +(x[0]1, 1))), >=))
3.90/1.84	
3.90/1.84	
3.90/1.84	
3.90/1.84	We simplified constraint (1) using rules (III), (IV) which results in the following new constraint:
3.90/1.84	
3.90/1.84	(2)    (<(x[0], 100000)=TRUE & <(+(x[0], 1), 100000)=TRUE  ==>  CU(TRUE, +(x[0], 1))_>=_NonInfC & CU(TRUE, +(x[0], 1))_>=_CU(<(+(x[0], 1), 100000), +(+(x[0], 1), 1)) & (U^Increasing(CU(<(x[0]1, 100000), +(x[0]1, 1))), >=))
3.90/1.84	
3.90/1.84	
3.90/1.84	
3.90/1.84	We simplified constraint (2) using rule (POLY_CONSTRAINTS) which results in the following new constraint:
3.90/1.84	
3.90/1.84	(3)    ([99999] + [-1]x[0] >= 0 & [99998] + [-1]x[0] >= 0  ==>  (U^Increasing(CU(<(x[0]1, 100000), +(x[0]1, 1))), >=) & [bni_7 + (-1)Bound*bni_7] + [(-1)bni_7]x[0] >= 0 & [1 + (-1)bso_8] >= 0)
3.90/1.84	
3.90/1.84	
3.90/1.84	
3.90/1.84	We simplified constraint (3) using rule (IDP_POLY_SIMPLIFY) which results in the following new constraint:
3.90/1.84	
3.90/1.84	(4)    ([99999] + [-1]x[0] >= 0 & [99998] + [-1]x[0] >= 0  ==>  (U^Increasing(CU(<(x[0]1, 100000), +(x[0]1, 1))), >=) & [bni_7 + (-1)Bound*bni_7] + [(-1)bni_7]x[0] >= 0 & [1 + (-1)bso_8] >= 0)
3.90/1.84	
3.90/1.84	
3.90/1.84	
3.90/1.84	We simplified constraint (4) using rule (POLY_REMOVE_MIN_MAX) which results in the following new constraint:
3.90/1.84	
3.90/1.84	(5)    ([99999] + [-1]x[0] >= 0 & [99998] + [-1]x[0] >= 0  ==>  (U^Increasing(CU(<(x[0]1, 100000), +(x[0]1, 1))), >=) & [bni_7 + (-1)Bound*bni_7] + [(-1)bni_7]x[0] >= 0 & [1 + (-1)bso_8] >= 0)
3.90/1.84	
3.90/1.84	
3.90/1.84	
3.90/1.84	We simplified constraint (5) using rule (IDP_SMT_SPLIT) which results in the following new constraints:
3.90/1.84	
3.90/1.84	(6)    ([99999] + x[0] >= 0 & [99998] + x[0] >= 0 & x[0] >= 0  ==>  (U^Increasing(CU(<(x[0]1, 100000), +(x[0]1, 1))), >=) & [bni_7 + (-1)Bound*bni_7] + [bni_7]x[0] >= 0 & [1 + (-1)bso_8] >= 0)
3.90/1.84	
3.90/1.84	(7)    ([99999] + [-1]x[0] >= 0 & [99998] + [-1]x[0] >= 0 & x[0] >= 0  ==>  (U^Increasing(CU(<(x[0]1, 100000), +(x[0]1, 1))), >=) & [bni_7 + (-1)Bound*bni_7] + [(-1)bni_7]x[0] >= 0 & [1 + (-1)bso_8] >= 0)
3.90/1.84	
3.90/1.84	
3.90/1.84	
3.90/1.84	
3.90/1.84	
3.90/1.84	
3.90/1.84	
3.90/1.84	
3.90/1.84	To summarize, we get the following constraints P__>=_ for the following pairs.
3.90/1.84	
3.90/1.84	*CU(TRUE, x) -> CU(<(x, 100000), +(x, 1))
3.90/1.84	
3.90/1.84	*([99999] + x[0] >= 0 & [99998] + x[0] >= 0 & x[0] >= 0  ==>  (U^Increasing(CU(<(x[0]1, 100000), +(x[0]1, 1))), >=) & [bni_7 + (-1)Bound*bni_7] + [bni_7]x[0] >= 0 & [1 + (-1)bso_8] >= 0)
3.90/1.84	
3.90/1.84	
3.90/1.84	*([99999] + [-1]x[0] >= 0 & [99998] + [-1]x[0] >= 0 & x[0] >= 0  ==>  (U^Increasing(CU(<(x[0]1, 100000), +(x[0]1, 1))), >=) & [bni_7 + (-1)Bound*bni_7] + [(-1)bni_7]x[0] >= 0 & [1 + (-1)bso_8] >= 0)
3.90/1.84	
3.90/1.84	
3.90/1.84	
3.90/1.84	
3.90/1.84	
3.90/1.84	
3.90/1.84	
3.90/1.84	
3.90/1.84	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. 
3.90/1.84	
3.90/1.84	Using the following integer polynomial ordering the  resulting constraints can be solved 
3.90/1.84	
3.90/1.84	Polynomial interpretation over integers[POLO]:
3.90/1.84	
3.90/1.84	   POL(TRUE) = 0
3.90/1.84	   POL(FALSE) = 0
3.90/1.84	   POL(CU(x_1, x_2)) = [2] + [-1]x_2
3.90/1.84	   POL(<(x_1, x_2)) = [-1]
3.90/1.84	   POL(100000) = [100000]
3.90/1.84	   POL(+(x_1, x_2)) = x_1 + x_2
3.90/1.84	   POL(1) = [1]
3.90/1.84	
3.90/1.84	
3.90/1.84	The following pairs are in P_>:
3.90/1.84	
3.90/1.84	
3.90/1.84	   CU(TRUE, x[0]) -> CU(<(x[0], 100000), +(x[0], 1))
3.90/1.84	
3.90/1.84	
3.90/1.84	The following pairs are in P_bound:
3.90/1.84	
3.90/1.84	
3.90/1.84	   CU(TRUE, x[0]) -> CU(<(x[0], 100000), +(x[0], 1))
3.90/1.84	
3.90/1.84	
3.90/1.84	The following pairs are in P_>=:
3.90/1.84	
3.90/1.84	none
3.90/1.84	
3.90/1.84	
3.90/1.84	There are no usable rules.
3.90/1.84	----------------------------------------
3.90/1.84	
3.90/1.84	(6)
3.90/1.84	Obligation:
3.90/1.84	IDP problem:
3.90/1.84	The following function symbols are pre-defined:
3.90/1.84	<<<
3.90/1.84	 & 	~ 	Bwand: (Integer, Integer) -> Integer
3.90/1.84	 >= 	~ 	Ge: (Integer, Integer) -> Boolean
3.90/1.84	 | 	~ 	Bwor: (Integer, Integer) -> Integer
3.90/1.84	 / 	~ 	Div: (Integer, Integer) -> Integer
3.90/1.84	 != 	~ 	Neq: (Integer, Integer) -> Boolean
3.90/1.84	 && 	~ 	Land: (Boolean, Boolean) -> Boolean
3.90/1.84	 ! 	~ 	Lnot: (Boolean) -> Boolean
3.90/1.84	 = 	~ 	Eq: (Integer, Integer) -> Boolean
3.90/1.84	 <= 	~ 	Le: (Integer, Integer) -> Boolean
3.90/1.84	 ^ 	~ 	Bwxor: (Integer, Integer) -> Integer
3.90/1.84	 % 	~ 	Mod: (Integer, Integer) -> Integer
3.90/1.84	 + 	~ 	Add: (Integer, Integer) -> Integer
3.90/1.84	 > 	~ 	Gt: (Integer, Integer) -> Boolean
3.90/1.84	 -1 	~ 	UnaryMinus: (Integer) -> Integer
3.90/1.84	 < 	~ 	Lt: (Integer, Integer) -> Boolean
3.90/1.84	 || 	~ 	Lor: (Boolean, Boolean) -> Boolean
3.90/1.84	 - 	~ 	Sub: (Integer, Integer) -> Integer
3.90/1.84	 ~ 	~ 	Bwnot: (Integer) -> Integer
3.90/1.84	 * 	~ 	Mul: (Integer, Integer) -> Integer
3.90/1.84	>>>
3.90/1.84	
3.90/1.84	
3.90/1.84	The following domains are used:
3.90/1.84	none
3.90/1.84	
3.90/1.84	R is empty.
3.90/1.84	
3.90/1.84	The integer pair graph is empty.
3.90/1.84	
3.90/1.84	The set Q consists of the following terms:
3.90/1.84	cu(TRUE, x0)
3.90/1.84	
3.90/1.84	----------------------------------------
3.90/1.84	
3.90/1.84	(7) PisEmptyProof (EQUIVALENT)
3.90/1.84	The TRS P is empty. Hence, there is no (P,Q,R) chain.
3.90/1.84	----------------------------------------
3.90/1.84	
3.90/1.84	(8)
3.90/1.84	YES
3.98/1.86	EOF