3.90/1.82	YES
3.90/1.83	proof of /export/starexec/sandbox/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3.90/1.83	# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty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, 167 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	 > 	~ 	Gt: (Integer, Integer) -> Boolean
3.90/1.83	 + 	~ 	Add: (Integer, Integer) -> Integer
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	f(TRUE, x, y) -> f(x > y && y > 2, 1 + x, 2 * y)
3.90/1.83	The set Q consists of the following terms:
3.90/1.83	f(TRUE, x0, x1)
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
3.90/1.83	 ^ 	~ 	Bwxor: (Integer, Integer) -> Integer
3.90/1.83	 % 	~ 	Mod: (Integer, Integer) -> Integer
3.90/1.83	 > 	~ 	Gt: (Integer, Integer) -> Boolean
3.90/1.83	 + 	~ 	Add: (Integer, Integer) -> Integer
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	
3.90/1.83	The following domains are used:
3.90/1.83	   Boolean, Integer
3.90/1.83	
3.90/1.83	The ITRS R consists of the following rules:
3.90/1.83	f(TRUE, x, y) -> f(x > y && y > 2, 1 + x, 2 * y)
3.90/1.83	
3.90/1.83	The integer pair graph contains the following rules and edges:
3.90/1.83	(0): F(TRUE, x[0], y[0]) -> F(x[0] > y[0] && y[0] > 2, 1 + x[0], 2 * y[0])
3.90/1.83	
3.90/1.83	   (0) -> (0), if (x[0] > y[0] && y[0] > 2  & 1 + x[0] ->^* x[0]' & 2 * y[0] ->^* y[0]')
3.90/1.83	
3.90/1.83	The set Q consists of the following terms:
3.90/1.83	f(TRUE, x0, x1)
3.90/1.83	
3.90/1.83	----------------------------------------
3.90/1.83	
3.90/1.83	(3) UsableRulesProof (EQUIVALENT)
3.90/1.83	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.
3.90/1.83	----------------------------------------
3.90/1.83	
3.90/1.83	(4)
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
3.90/1.83	 ^ 	~ 	Bwxor: (Integer, Integer) -> Integer
3.90/1.83	 % 	~ 	Mod: (Integer, Integer) -> Integer
3.90/1.83	 > 	~ 	Gt: (Integer, Integer) -> Boolean
3.90/1.83	 + 	~ 	Add: (Integer, Integer) -> Integer
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	
3.90/1.83	The following domains are used:
3.90/1.83	   Boolean, Integer
3.90/1.83	
3.90/1.83	R is empty.
3.90/1.83	
3.90/1.83	The integer pair graph contains the following rules and edges:
3.90/1.83	(0): F(TRUE, x[0], y[0]) -> F(x[0] > y[0] && y[0] > 2, 1 + x[0], 2 * y[0])
3.90/1.83	
3.90/1.83	   (0) -> (0), if (x[0] > y[0] && y[0] > 2  & 1 + x[0] ->^* x[0]' & 2 * y[0] ->^* y[0]')
3.90/1.83	
3.90/1.83	The set Q consists of the following terms:
3.90/1.83	f(TRUE, x0, x1)
3.90/1.83	
3.90/1.83	----------------------------------------
3.90/1.83	
3.90/1.83	(5) IDPNonInfProof (SOUND)
3.90/1.83	Used the following options for this NonInfProof:
3.90/1.83	
3.90/1.83	IDPGPoloSolver:
3.90/1.83	Range: [(-1,2)]
3.90/1.83	IsNat: false
3.90/1.83	Interpretation Shape Heuristic: aprove.DPFramework.IDPProblem.Processors.nonInf.poly.IdpDefaultShapeHeuristic@2c4174b7
3.90/1.83	Constraint Generator: NonInfConstraintGenerator:
3.90/1.83	PathGenerator: MetricPathGenerator:
3.90/1.83	Max Left Steps: 1
3.90/1.83	Max Right Steps: 1
3.90/1.83	
3.90/1.83	
3.90/1.83	
3.90/1.83	The constraints were generated the following way:
3.90/1.83	
3.90/1.83	The DP Problem is simplified using the Induction Calculus [NONINF] with the following steps:
3.90/1.83	
3.90/1.83	Note that final constraints are written in bold face.
3.90/1.83	
3.90/1.83	
3.90/1.83	
3.90/1.83	For Pair F(TRUE, x, y) -> F(&&(>(x, y), >(y, 2)), +(1, x), *(2, y)) the following chains were created:
3.90/1.83	*We consider the chain F(TRUE, x[0], y[0]) -> F(&&(>(x[0], y[0]), >(y[0], 2)), +(1, x[0]), *(2, y[0])), F(TRUE, x[0], y[0]) -> F(&&(>(x[0], y[0]), >(y[0], 2)), +(1, x[0]), *(2, y[0])), F(TRUE, x[0], y[0]) -> F(&&(>(x[0], y[0]), >(y[0], 2)), +(1, x[0]), *(2, y[0])) which results in the following constraint:
3.90/1.83	
3.90/1.83	(1)    (&&(>(x[0], y[0]), >(y[0], 2))=TRUE & +(1, x[0])=x[0]1 & *(2, y[0])=y[0]1 & &&(>(x[0]1, y[0]1), >(y[0]1, 2))=TRUE & +(1, x[0]1)=x[0]2 & *(2, y[0]1)=y[0]2  ==>  F(TRUE, x[0]1, y[0]1)_>=_NonInfC & F(TRUE, x[0]1, y[0]1)_>=_F(&&(>(x[0]1, y[0]1), >(y[0]1, 2)), +(1, x[0]1), *(2, y[0]1)) & (U^Increasing(F(&&(>(x[0]1, y[0]1), >(y[0]1, 2)), +(1, x[0]1), *(2, y[0]1))), >=))
3.90/1.83	
3.90/1.83	
3.90/1.83	
3.90/1.83	We simplified constraint (1) using rules (III), (IV), (IDP_BOOLEAN) which results in the following new constraint:
3.90/1.83	
3.90/1.83	(2)    (>(x[0], y[0])=TRUE & >(y[0], 2)=TRUE & >(+(1, x[0]), *(2, y[0]))=TRUE & >(*(2, y[0]), 2)=TRUE  ==>  F(TRUE, +(1, x[0]), *(2, y[0]))_>=_NonInfC & F(TRUE, +(1, x[0]), *(2, y[0]))_>=_F(&&(>(+(1, x[0]), *(2, y[0])), >(*(2, y[0]), 2)), +(1, +(1, x[0])), *(2, *(2, y[0]))) & (U^Increasing(F(&&(>(x[0]1, y[0]1), >(y[0]1, 2)), +(1, x[0]1), *(2, y[0]1))), >=))
3.90/1.83	
3.90/1.83	
3.90/1.83	
3.90/1.83	We simplified constraint (2) using rule (POLY_CONSTRAINTS) which results in the following new constraint:
3.90/1.83	
3.90/1.83	(3)    (x[0] + [-1] + [-1]y[0] >= 0 & y[0] + [-3] >= 0 & x[0] + [-2]y[0] >= 0 & [2]y[0] + [-3] >= 0  ==>  (U^Increasing(F(&&(>(x[0]1, y[0]1), >(y[0]1, 2)), +(1, x[0]1), *(2, y[0]1))), >=) & [(3)bni_9 + (-1)Bound*bni_9] + [(-2)bni_9]y[0] + [bni_9]x[0] >= 0 & [-2 + (-1)bso_10] + [2]y[0] >= 0)
3.90/1.83	
3.90/1.83	
3.90/1.83	
3.90/1.83	We simplified constraint (3) using rule (IDP_POLY_SIMPLIFY) which results in the following new constraint:
3.90/1.83	
3.90/1.83	(4)    (x[0] + [-1] + [-1]y[0] >= 0 & y[0] + [-3] >= 0 & x[0] + [-2]y[0] >= 0 & [2]y[0] + [-3] >= 0  ==>  (U^Increasing(F(&&(>(x[0]1, y[0]1), >(y[0]1, 2)), +(1, x[0]1), *(2, y[0]1))), >=) & [(3)bni_9 + (-1)Bound*bni_9] + [(-2)bni_9]y[0] + [bni_9]x[0] >= 0 & [-2 + (-1)bso_10] + [2]y[0] >= 0)
3.90/1.83	
3.90/1.83	
3.90/1.83	
3.90/1.83	We simplified constraint (4) using rule (POLY_REMOVE_MIN_MAX) which results in the following new constraint:
3.90/1.83	
3.90/1.83	(5)    (x[0] + [-1] + [-1]y[0] >= 0 & y[0] + [-3] >= 0 & x[0] + [-2]y[0] >= 0 & [2]y[0] + [-3] >= 0  ==>  (U^Increasing(F(&&(>(x[0]1, y[0]1), >(y[0]1, 2)), +(1, x[0]1), *(2, y[0]1))), >=) & [(3)bni_9 + (-1)Bound*bni_9] + [(-2)bni_9]y[0] + [bni_9]x[0] >= 0 & [-2 + (-1)bso_10] + [2]y[0] >= 0)
3.90/1.83	
3.90/1.83	
3.90/1.83	
3.90/1.83	
3.90/1.83	
3.90/1.83	
3.90/1.83	
3.90/1.83	
3.90/1.83	To summarize, we get the following constraints P__>=_ for the following pairs.
3.90/1.83	
3.90/1.83	*F(TRUE, x, y) -> F(&&(>(x, y), >(y, 2)), +(1, x), *(2, y))
3.90/1.83	
3.90/1.83	*(x[0] + [-1] + [-1]y[0] >= 0 & y[0] + [-3] >= 0 & x[0] + [-2]y[0] >= 0 & [2]y[0] + [-3] >= 0  ==>  (U^Increasing(F(&&(>(x[0]1, y[0]1), >(y[0]1, 2)), +(1, x[0]1), *(2, y[0]1))), >=) & [(3)bni_9 + (-1)Bound*bni_9] + [(-2)bni_9]y[0] + [bni_9]x[0] >= 0 & [-2 + (-1)bso_10] + [2]y[0] >= 0)
3.90/1.83	
3.90/1.83	
3.90/1.83	
3.90/1.83	
3.90/1.83	
3.90/1.83	
3.90/1.83	
3.90/1.83	
3.90/1.83	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. 
3.90/1.83	
3.90/1.83	Using the following integer polynomial ordering the  resulting constraints can be solved 
3.90/1.83	
3.90/1.83	Polynomial interpretation over integers[POLO]:
3.90/1.83	
3.90/1.83	   POL(TRUE) = 0
3.90/1.83	   POL(FALSE) = [3]
3.90/1.83	   POL(F(x_1, x_2, x_3)) = [2] + [-1]x_3 + x_2 + [-1]x_1
3.90/1.83	   POL(&&(x_1, x_2)) = [-1]
3.90/1.83	   POL(>(x_1, x_2)) = [-1]
3.90/1.83	   POL(2) = [2]
3.90/1.83	   POL(+(x_1, x_2)) = x_1 + x_2
3.90/1.83	   POL(1) = [1]
3.90/1.83	   POL(*(x_1, x_2)) = x_1*x_2
3.90/1.83	
3.90/1.83	
3.90/1.83	The following pairs are in P_>:
3.90/1.83	
3.90/1.83	
3.90/1.83	   F(TRUE, x[0], y[0]) -> F(&&(>(x[0], y[0]), >(y[0], 2)), +(1, x[0]), *(2, y[0]))
3.90/1.83	
3.90/1.83	
3.90/1.83	The following pairs are in P_bound:
3.90/1.83	
3.90/1.83	
3.90/1.83	   F(TRUE, x[0], y[0]) -> F(&&(>(x[0], y[0]), >(y[0], 2)), +(1, x[0]), *(2, y[0]))
3.90/1.83	
3.90/1.83	
3.90/1.83	The following pairs are in P_>=:
3.90/1.83	
3.90/1.83	none
3.90/1.83	
3.90/1.83	
3.90/1.83	At least the following rules have been oriented under context sensitive arithmetic replacement:
3.90/1.83	
3.90/1.83	   TRUE^1 -> &&(TRUE, TRUE)^1
3.90/1.83	   FALSE^1 -> &&(TRUE, FALSE)^1
3.90/1.83	   FALSE^1 -> &&(FALSE, TRUE)^1
3.90/1.83	   FALSE^1 -> &&(FALSE, FALSE)^1
3.90/1.83	
3.90/1.83	----------------------------------------
3.90/1.83	
3.90/1.83	(6)
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
3.90/1.83	 ^ 	~ 	Bwxor: (Integer, Integer) -> Integer
3.90/1.83	 % 	~ 	Mod: (Integer, Integer) -> Integer
3.90/1.83	 > 	~ 	Gt: (Integer, Integer) -> Boolean
3.90/1.83	 + 	~ 	Add: (Integer, Integer) -> Integer
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	
3.90/1.83	The following domains are used:
3.90/1.83	none
3.90/1.83	
3.90/1.83	R is empty.
3.90/1.83	
3.90/1.83	The integer pair graph is empty.
3.90/1.83	
3.90/1.83	The set Q consists of the following terms:
3.90/1.83	f(TRUE, x0, x1)
3.90/1.83	
3.90/1.83	----------------------------------------
3.90/1.83	
3.90/1.83	(7) PisEmptyProof (EQUIVALENT)
3.90/1.83	The TRS P is empty. Hence, there is no (P,Q,R) chain.
3.90/1.83	----------------------------------------
3.90/1.83	
3.90/1.83	(8)
3.90/1.83	YES
3.90/1.84	EOF