3.73/1.89	YES
3.73/1.90	proof of /export/starexec/sandbox/benchmark/theBenchmark.itrs
3.73/1.90	# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty
3.73/1.90	
3.73/1.90	
3.73/1.90	Termination of the given ITRS could be proven:
3.73/1.90	
3.73/1.90	(0) ITRS
3.73/1.90	(1) ITRStoIDPProof [EQUIVALENT, 0 ms]
3.73/1.90	(2) IDP
3.73/1.90	(3) UsableRulesProof [EQUIVALENT, 0 ms]
3.73/1.90	(4) IDP
3.73/1.90	(5) IDPNonInfProof [SOUND, 170 ms]
3.73/1.90	(6) IDP
3.73/1.90	(7) PisEmptyProof [EQUIVALENT, 0 ms]
3.73/1.90	(8) YES
3.73/1.90	
3.73/1.90	
3.73/1.90	----------------------------------------
3.73/1.90	
3.73/1.90	(0)
3.73/1.90	Obligation:
3.73/1.90	ITRS problem:
3.73/1.90	
3.73/1.90	The following function symbols are pre-defined:
3.73/1.90	<<<
3.73/1.90	 & 	~ 	Bwand: (Integer, Integer) -> Integer
3.73/1.90	 >= 	~ 	Ge: (Integer, Integer) -> Boolean
3.73/1.90	 | 	~ 	Bwor: (Integer, Integer) -> Integer
3.73/1.90	 / 	~ 	Div: (Integer, Integer) -> Integer
3.73/1.90	 != 	~ 	Neq: (Integer, Integer) -> Boolean
3.73/1.90	 && 	~ 	Land: (Boolean, Boolean) -> Boolean
3.73/1.90	 ! 	~ 	Lnot: (Boolean) -> Boolean
3.73/1.90	 = 	~ 	Eq: (Integer, Integer) -> Boolean
3.73/1.90	 <= 	~ 	Le: (Integer, Integer) -> Boolean
3.73/1.90	 ^ 	~ 	Bwxor: (Integer, Integer) -> Integer
3.73/1.90	 % 	~ 	Mod: (Integer, Integer) -> Integer
3.73/1.90	 > 	~ 	Gt: (Integer, Integer) -> Boolean
3.73/1.90	 + 	~ 	Add: (Integer, Integer) -> Integer
3.73/1.90	 -1 	~ 	UnaryMinus: (Integer) -> Integer
3.73/1.90	 < 	~ 	Lt: (Integer, Integer) -> Boolean
3.73/1.90	 || 	~ 	Lor: (Boolean, Boolean) -> Boolean
3.73/1.90	 - 	~ 	Sub: (Integer, Integer) -> Integer
3.73/1.90	 ~ 	~ 	Bwnot: (Integer) -> Integer
3.73/1.90	 * 	~ 	Mul: (Integer, Integer) -> Integer
3.73/1.90	>>>
3.73/1.90	
3.73/1.90	The TRS R consists of the following rules:
3.73/1.90	f(TRUE, x) -> f(x > y && x >= 0, y)
3.73/1.90	The set Q consists of the following terms:
3.73/1.90	f(TRUE, x0)
3.73/1.90	
3.73/1.90	----------------------------------------
3.73/1.90	
3.73/1.90	(1) ITRStoIDPProof (EQUIVALENT)
3.73/1.90	Added dependency pairs
3.73/1.90	----------------------------------------
3.73/1.90	
3.73/1.90	(2)
3.73/1.90	Obligation:
3.73/1.90	IDP problem:
3.73/1.90	The following function symbols are pre-defined:
3.73/1.90	<<<
3.73/1.90	 & 	~ 	Bwand: (Integer, Integer) -> Integer
3.73/1.90	 >= 	~ 	Ge: (Integer, Integer) -> Boolean
3.73/1.90	 | 	~ 	Bwor: (Integer, Integer) -> Integer
3.73/1.90	 / 	~ 	Div: (Integer, Integer) -> Integer
3.73/1.90	 != 	~ 	Neq: (Integer, Integer) -> Boolean
3.73/1.90	 && 	~ 	Land: (Boolean, Boolean) -> Boolean
3.73/1.90	 ! 	~ 	Lnot: (Boolean) -> Boolean
3.73/1.90	 = 	~ 	Eq: (Integer, Integer) -> Boolean
3.73/1.90	 <= 	~ 	Le: (Integer, Integer) -> Boolean
3.73/1.90	 ^ 	~ 	Bwxor: (Integer, Integer) -> Integer
3.73/1.90	 % 	~ 	Mod: (Integer, Integer) -> Integer
3.73/1.90	 > 	~ 	Gt: (Integer, Integer) -> Boolean
3.73/1.90	 + 	~ 	Add: (Integer, Integer) -> Integer
3.73/1.90	 -1 	~ 	UnaryMinus: (Integer) -> Integer
3.73/1.90	 < 	~ 	Lt: (Integer, Integer) -> Boolean
3.73/1.90	 || 	~ 	Lor: (Boolean, Boolean) -> Boolean
3.73/1.90	 - 	~ 	Sub: (Integer, Integer) -> Integer
3.73/1.90	 ~ 	~ 	Bwnot: (Integer) -> Integer
3.73/1.90	 * 	~ 	Mul: (Integer, Integer) -> Integer
3.73/1.90	>>>
3.73/1.90	
3.73/1.90	
3.73/1.90	The following domains are used:
3.73/1.90	   Boolean, Integer
3.73/1.90	
3.73/1.90	The ITRS R consists of the following rules:
3.73/1.90	f(TRUE, x) -> f(x > y && x >= 0, y)
3.73/1.90	
3.73/1.90	The integer pair graph contains the following rules and edges:
3.73/1.90	(0): F(TRUE, x[0]) -> F(x[0] > y[0] && x[0] >= 0, y[0])
3.73/1.90	
3.73/1.90	   (0) -> (0), if (x[0] > y[0] && x[0] >= 0  & y[0] ->^* x[0]')
3.73/1.90	
3.73/1.90	The set Q consists of the following terms:
3.73/1.90	f(TRUE, x0)
3.73/1.90	
3.73/1.90	----------------------------------------
3.73/1.90	
3.73/1.90	(3) UsableRulesProof (EQUIVALENT)
3.73/1.90	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.
3.73/1.90	----------------------------------------
3.73/1.90	
3.73/1.90	(4)
3.73/1.90	Obligation:
3.73/1.90	IDP problem:
3.73/1.90	The following function symbols are pre-defined:
3.73/1.90	<<<
3.73/1.90	 & 	~ 	Bwand: (Integer, Integer) -> Integer
3.73/1.90	 >= 	~ 	Ge: (Integer, Integer) -> Boolean
3.73/1.90	 | 	~ 	Bwor: (Integer, Integer) -> Integer
3.73/1.90	 / 	~ 	Div: (Integer, Integer) -> Integer
3.73/1.90	 != 	~ 	Neq: (Integer, Integer) -> Boolean
3.73/1.90	 && 	~ 	Land: (Boolean, Boolean) -> Boolean
3.73/1.90	 ! 	~ 	Lnot: (Boolean) -> Boolean
3.73/1.90	 = 	~ 	Eq: (Integer, Integer) -> Boolean
3.73/1.90	 <= 	~ 	Le: (Integer, Integer) -> Boolean
3.73/1.90	 ^ 	~ 	Bwxor: (Integer, Integer) -> Integer
3.73/1.90	 % 	~ 	Mod: (Integer, Integer) -> Integer
3.73/1.90	 > 	~ 	Gt: (Integer, Integer) -> Boolean
3.73/1.90	 + 	~ 	Add: (Integer, Integer) -> Integer
3.73/1.90	 -1 	~ 	UnaryMinus: (Integer) -> Integer
3.73/1.90	 < 	~ 	Lt: (Integer, Integer) -> Boolean
3.73/1.90	 || 	~ 	Lor: (Boolean, Boolean) -> Boolean
3.73/1.90	 - 	~ 	Sub: (Integer, Integer) -> Integer
3.73/1.90	 ~ 	~ 	Bwnot: (Integer) -> Integer
3.73/1.90	 * 	~ 	Mul: (Integer, Integer) -> Integer
3.73/1.90	>>>
3.73/1.90	
3.73/1.90	
3.73/1.90	The following domains are used:
3.73/1.90	   Boolean, Integer
3.73/1.90	
3.73/1.90	R is empty.
3.73/1.90	
3.73/1.90	The integer pair graph contains the following rules and edges:
3.73/1.90	(0): F(TRUE, x[0]) -> F(x[0] > y[0] && x[0] >= 0, y[0])
3.73/1.90	
3.73/1.90	   (0) -> (0), if (x[0] > y[0] && x[0] >= 0  & y[0] ->^* x[0]')
3.73/1.90	
3.73/1.90	The set Q consists of the following terms:
3.73/1.90	f(TRUE, x0)
3.73/1.90	
3.73/1.90	----------------------------------------
3.73/1.90	
3.73/1.90	(5) IDPNonInfProof (SOUND)
3.73/1.90	Used the following options for this NonInfProof:
3.73/1.90	
3.73/1.90	IDPGPoloSolver:
3.73/1.90	Range: [(-1,2)]
3.73/1.90	IsNat: false
3.73/1.90	Interpretation Shape Heuristic: aprove.DPFramework.IDPProblem.Processors.nonInf.poly.IdpDefaultShapeHeuristic@590630b9
3.73/1.90	Constraint Generator: NonInfConstraintGenerator:
3.73/1.90	PathGenerator: MetricPathGenerator:
3.73/1.90	Max Left Steps: 1
3.73/1.90	Max Right Steps: 1
3.73/1.90	
3.73/1.90	
3.73/1.90	
3.73/1.90	The constraints were generated the following way:
3.73/1.90	
3.73/1.90	The DP Problem is simplified using the Induction Calculus [NONINF] with the following steps:
3.73/1.90	
3.73/1.90	Note that final constraints are written in bold face.
3.73/1.90	
3.73/1.90	
3.73/1.90	
3.73/1.90	For Pair F(TRUE, x) -> F(&&(>(x, y), >=(x, 0)), y) the following chains were created:
3.73/1.90	*We consider the chain F(TRUE, x[0]) -> F(&&(>(x[0], y[0]), >=(x[0], 0)), y[0]), F(TRUE, x[0]) -> F(&&(>(x[0], y[0]), >=(x[0], 0)), y[0]), F(TRUE, x[0]) -> F(&&(>(x[0], y[0]), >=(x[0], 0)), y[0]) which results in the following constraint:
3.73/1.90	
3.73/1.90	(1)    (&&(>(x[0], y[0]), >=(x[0], 0))=TRUE & y[0]=x[0]1 & &&(>(x[0]1, y[0]1), >=(x[0]1, 0))=TRUE & y[0]1=x[0]2  ==>  F(TRUE, x[0]1)_>=_NonInfC & F(TRUE, x[0]1)_>=_F(&&(>(x[0]1, y[0]1), >=(x[0]1, 0)), y[0]1) & (U^Increasing(F(&&(>(x[0]1, y[0]1), >=(x[0]1, 0)), y[0]1)), >=))
3.73/1.90	
3.73/1.90	
3.73/1.90	
3.73/1.90	We simplified constraint (1) using rules (III), (IV), (IDP_BOOLEAN) which results in the following new constraint:
3.73/1.90	
3.73/1.90	(2)    (>(x[0], y[0])=TRUE & >=(x[0], 0)=TRUE & >(y[0], y[0]1)=TRUE & >=(y[0], 0)=TRUE  ==>  F(TRUE, y[0])_>=_NonInfC & F(TRUE, y[0])_>=_F(&&(>(y[0], y[0]1), >=(y[0], 0)), y[0]1) & (U^Increasing(F(&&(>(x[0]1, y[0]1), >=(x[0]1, 0)), y[0]1)), >=))
3.73/1.90	
3.73/1.90	
3.73/1.90	
3.73/1.90	We simplified constraint (2) using rule (POLY_CONSTRAINTS) which results in the following new constraint:
3.73/1.90	
3.73/1.90	(3)    (x[0] + [-1] + [-1]y[0] >= 0 & x[0] >= 0 & y[0] + [-1] + [-1]y[0]1 >= 0 & y[0] >= 0  ==>  (U^Increasing(F(&&(>(x[0]1, y[0]1), >=(x[0]1, 0)), y[0]1)), >=) & [(-1)Bound*bni_9] + [bni_9]y[0] >= 0 & [(-1)bso_10] + y[0] + [-1]y[0]1 >= 0)
3.73/1.90	
3.73/1.90	
3.73/1.90	
3.73/1.90	We simplified constraint (3) using rule (IDP_POLY_SIMPLIFY) which results in the following new constraint:
3.73/1.90	
3.73/1.90	(4)    (x[0] + [-1] + [-1]y[0] >= 0 & x[0] >= 0 & y[0] + [-1] + [-1]y[0]1 >= 0 & y[0] >= 0  ==>  (U^Increasing(F(&&(>(x[0]1, y[0]1), >=(x[0]1, 0)), y[0]1)), >=) & [(-1)Bound*bni_9] + [bni_9]y[0] >= 0 & [(-1)bso_10] + y[0] + [-1]y[0]1 >= 0)
3.73/1.90	
3.73/1.90	
3.73/1.90	
3.73/1.90	We simplified constraint (4) using rule (POLY_REMOVE_MIN_MAX) which results in the following new constraint:
3.73/1.90	
3.73/1.90	(5)    (x[0] + [-1] + [-1]y[0] >= 0 & x[0] >= 0 & y[0] + [-1] + [-1]y[0]1 >= 0 & y[0] >= 0  ==>  (U^Increasing(F(&&(>(x[0]1, y[0]1), >=(x[0]1, 0)), y[0]1)), >=) & [(-1)Bound*bni_9] + [bni_9]y[0] >= 0 & [(-1)bso_10] + y[0] + [-1]y[0]1 >= 0)
3.73/1.90	
3.73/1.90	
3.73/1.90	
3.73/1.90	We simplified constraint (5) using rule (IDP_SMT_SPLIT) which results in the following new constraints:
3.73/1.90	
3.73/1.90	(6)    (x[0] + [-1] + [-1]y[0] >= 0 & x[0] >= 0 & y[0] + [-1] + y[0]1 >= 0 & y[0] >= 0 & y[0]1 >= 0  ==>  (U^Increasing(F(&&(>(x[0]1, y[0]1), >=(x[0]1, 0)), y[0]1)), >=) & [(-1)Bound*bni_9] + [bni_9]y[0] >= 0 & [(-1)bso_10] + y[0] + y[0]1 >= 0)
3.73/1.90	
3.73/1.90	(7)    (x[0] + [-1] + [-1]y[0] >= 0 & x[0] >= 0 & y[0] + [-1] + [-1]y[0]1 >= 0 & y[0] >= 0 & y[0]1 >= 0  ==>  (U^Increasing(F(&&(>(x[0]1, y[0]1), >=(x[0]1, 0)), y[0]1)), >=) & [(-1)Bound*bni_9] + [bni_9]y[0] >= 0 & [(-1)bso_10] + y[0] + [-1]y[0]1 >= 0)
3.73/1.90	
3.73/1.90	
3.73/1.90	
3.73/1.90	
3.73/1.90	
3.73/1.90	
3.73/1.90	
3.73/1.90	
3.73/1.90	To summarize, we get the following constraints P__>=_ for the following pairs.
3.73/1.90	
3.73/1.90	*F(TRUE, x) -> F(&&(>(x, y), >=(x, 0)), y)
3.73/1.90	
3.73/1.90	*(x[0] + [-1] + [-1]y[0] >= 0 & x[0] >= 0 & y[0] + [-1] + y[0]1 >= 0 & y[0] >= 0 & y[0]1 >= 0  ==>  (U^Increasing(F(&&(>(x[0]1, y[0]1), >=(x[0]1, 0)), y[0]1)), >=) & [(-1)Bound*bni_9] + [bni_9]y[0] >= 0 & [(-1)bso_10] + y[0] + y[0]1 >= 0)
3.73/1.90	
3.73/1.90	
3.73/1.90	*(x[0] + [-1] + [-1]y[0] >= 0 & x[0] >= 0 & y[0] + [-1] + [-1]y[0]1 >= 0 & y[0] >= 0 & y[0]1 >= 0  ==>  (U^Increasing(F(&&(>(x[0]1, y[0]1), >=(x[0]1, 0)), y[0]1)), >=) & [(-1)Bound*bni_9] + [bni_9]y[0] >= 0 & [(-1)bso_10] + y[0] + [-1]y[0]1 >= 0)
3.73/1.90	
3.73/1.90	
3.73/1.90	
3.73/1.90	
3.73/1.90	
3.73/1.90	
3.73/1.90	
3.73/1.90	
3.73/1.90	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. 
3.73/1.90	
3.73/1.90	Using the following integer polynomial ordering the  resulting constraints can be solved 
3.73/1.90	
3.73/1.90	Polynomial interpretation over integers[POLO]:
3.73/1.90	
3.73/1.90	   POL(TRUE) = [2]
3.73/1.90	   POL(FALSE) = [3]
3.73/1.90	   POL(F(x_1, x_2)) = [2] + x_2 + [-1]x_1
3.73/1.90	   POL(&&(x_1, x_2)) = [2]
3.73/1.90	   POL(>(x_1, x_2)) = [-1]
3.73/1.90	   POL(>=(x_1, x_2)) = [-1]
3.73/1.90	   POL(0) = 0
3.73/1.90	
3.73/1.90	
3.73/1.90	The following pairs are in P_>:
3.73/1.90	
3.73/1.90	
3.73/1.90	   F(TRUE, x[0]) -> F(&&(>(x[0], y[0]), >=(x[0], 0)), y[0])
3.73/1.90	
3.73/1.90	
3.73/1.90	The following pairs are in P_bound:
3.73/1.90	
3.73/1.90	
3.73/1.90	   F(TRUE, x[0]) -> F(&&(>(x[0], y[0]), >=(x[0], 0)), y[0])
3.73/1.90	
3.73/1.90	
3.73/1.90	The following pairs are in P_>=:
3.73/1.90	
3.73/1.90	none
3.73/1.90	
3.73/1.90	
3.73/1.90	At least the following rules have been oriented under context sensitive arithmetic replacement:
3.73/1.90	
3.73/1.90	   TRUE^1 -> &&(TRUE, TRUE)^1
3.73/1.90	   FALSE^1 -> &&(TRUE, FALSE)^1
3.73/1.90	   FALSE^1 -> &&(FALSE, TRUE)^1
3.73/1.90	   FALSE^1 -> &&(FALSE, FALSE)^1
3.73/1.90	
3.73/1.90	----------------------------------------
3.73/1.90	
3.73/1.90	(6)
3.73/1.90	Obligation:
3.73/1.90	IDP problem:
3.73/1.90	The following function symbols are pre-defined:
3.73/1.90	<<<
3.73/1.90	 & 	~ 	Bwand: (Integer, Integer) -> Integer
3.73/1.90	 >= 	~ 	Ge: (Integer, Integer) -> Boolean
3.73/1.90	 | 	~ 	Bwor: (Integer, Integer) -> Integer
3.73/1.90	 / 	~ 	Div: (Integer, Integer) -> Integer
3.73/1.90	 != 	~ 	Neq: (Integer, Integer) -> Boolean
3.73/1.90	 && 	~ 	Land: (Boolean, Boolean) -> Boolean
3.73/1.90	 ! 	~ 	Lnot: (Boolean) -> Boolean
3.73/1.90	 = 	~ 	Eq: (Integer, Integer) -> Boolean
3.73/1.90	 <= 	~ 	Le: (Integer, Integer) -> Boolean
3.73/1.90	 ^ 	~ 	Bwxor: (Integer, Integer) -> Integer
3.73/1.90	 % 	~ 	Mod: (Integer, Integer) -> Integer
3.73/1.90	 > 	~ 	Gt: (Integer, Integer) -> Boolean
3.73/1.90	 + 	~ 	Add: (Integer, Integer) -> Integer
3.73/1.90	 -1 	~ 	UnaryMinus: (Integer) -> Integer
3.73/1.90	 < 	~ 	Lt: (Integer, Integer) -> Boolean
3.73/1.90	 || 	~ 	Lor: (Boolean, Boolean) -> Boolean
3.73/1.90	 - 	~ 	Sub: (Integer, Integer) -> Integer
3.73/1.90	 ~ 	~ 	Bwnot: (Integer) -> Integer
3.73/1.90	 * 	~ 	Mul: (Integer, Integer) -> Integer
3.73/1.90	>>>
3.73/1.90	
3.73/1.90	
3.73/1.90	The following domains are used:
3.73/1.90	none
3.73/1.90	
3.73/1.90	R is empty.
3.73/1.90	
3.73/1.90	The integer pair graph is empty.
3.73/1.90	
3.73/1.90	The set Q consists of the following terms:
3.73/1.90	f(TRUE, x0)
3.73/1.90	
3.73/1.90	----------------------------------------
3.73/1.90	
3.73/1.90	(7) PisEmptyProof (EQUIVALENT)
3.73/1.90	The TRS P is empty. Hence, there is no (P,Q,R) chain.
3.73/1.90	----------------------------------------
3.73/1.90	
3.73/1.90	(8)
3.73/1.90	YES
3.87/1.91	EOF