3.88/1.78	YES
3.88/1.79	proof of /export/starexec/sandbox2/benchmark/theBenchmark.itrs
3.88/1.79	# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty
3.88/1.79	
3.88/1.79	
3.88/1.79	Termination of the given ITRS could be proven:
3.88/1.79	
3.88/1.79	(0) ITRS
3.88/1.79	(1) ITRStoIDPProof [EQUIVALENT, 0 ms]
3.88/1.79	(2) IDP
3.88/1.79	(3) UsableRulesProof [EQUIVALENT, 0 ms]
3.88/1.79	(4) IDP
3.88/1.79	(5) IDPNonInfProof [SOUND, 169 ms]
3.88/1.79	(6) IDP
3.88/1.79	(7) PisEmptyProof [EQUIVALENT, 0 ms]
3.88/1.79	(8) YES
3.88/1.79	
3.88/1.79	
3.88/1.79	----------------------------------------
3.88/1.79	
3.88/1.79	(0)
3.88/1.79	Obligation:
3.88/1.79	ITRS problem:
3.88/1.79	
3.88/1.79	The following function symbols are pre-defined:
3.88/1.79	<<<
3.88/1.79	 & 	~ 	Bwand: (Integer, Integer) -> Integer
3.88/1.79	 >= 	~ 	Ge: (Integer, Integer) -> Boolean
3.88/1.79	 | 	~ 	Bwor: (Integer, Integer) -> Integer
3.88/1.79	 / 	~ 	Div: (Integer, Integer) -> Integer
3.88/1.79	 != 	~ 	Neq: (Integer, Integer) -> Boolean
3.88/1.79	 && 	~ 	Land: (Boolean, Boolean) -> Boolean
3.88/1.79	 ! 	~ 	Lnot: (Boolean) -> Boolean
3.88/1.79	 = 	~ 	Eq: (Integer, Integer) -> Boolean
3.88/1.79	 <= 	~ 	Le: (Integer, Integer) -> Boolean
3.88/1.79	 ^ 	~ 	Bwxor: (Integer, Integer) -> Integer
3.88/1.79	 % 	~ 	Mod: (Integer, Integer) -> Integer
3.88/1.79	 + 	~ 	Add: (Integer, Integer) -> Integer
3.88/1.79	 > 	~ 	Gt: (Integer, Integer) -> Boolean
3.88/1.79	 -1 	~ 	UnaryMinus: (Integer) -> Integer
3.88/1.79	 < 	~ 	Lt: (Integer, Integer) -> Boolean
3.88/1.79	 || 	~ 	Lor: (Boolean, Boolean) -> Boolean
3.88/1.79	 - 	~ 	Sub: (Integer, Integer) -> Integer
3.88/1.79	 ~ 	~ 	Bwnot: (Integer) -> Integer
3.88/1.79	 * 	~ 	Mul: (Integer, Integer) -> Integer
3.88/1.79	>>>
3.88/1.79	
3.88/1.79	The TRS R consists of the following rules:
3.88/1.79	f(x) -> Cond_f(x * x < 0, x)
3.88/1.79	Cond_f(TRUE, x) -> f(x)
3.88/1.79	The set Q consists of the following terms:
3.88/1.79	f(x0)
3.88/1.79	Cond_f(TRUE, x0)
3.88/1.79	
3.88/1.79	----------------------------------------
3.88/1.79	
3.88/1.79	(1) ITRStoIDPProof (EQUIVALENT)
3.88/1.79	Added dependency pairs
3.88/1.79	----------------------------------------
3.88/1.79	
3.88/1.79	(2)
3.88/1.79	Obligation:
3.88/1.79	IDP problem:
3.88/1.79	The following function symbols are pre-defined:
3.88/1.79	<<<
3.88/1.79	 & 	~ 	Bwand: (Integer, Integer) -> Integer
3.88/1.79	 >= 	~ 	Ge: (Integer, Integer) -> Boolean
3.88/1.79	 | 	~ 	Bwor: (Integer, Integer) -> Integer
3.88/1.79	 / 	~ 	Div: (Integer, Integer) -> Integer
3.88/1.79	 != 	~ 	Neq: (Integer, Integer) -> Boolean
3.88/1.79	 && 	~ 	Land: (Boolean, Boolean) -> Boolean
3.88/1.79	 ! 	~ 	Lnot: (Boolean) -> Boolean
3.88/1.79	 = 	~ 	Eq: (Integer, Integer) -> Boolean
3.88/1.79	 <= 	~ 	Le: (Integer, Integer) -> Boolean
3.88/1.79	 ^ 	~ 	Bwxor: (Integer, Integer) -> Integer
3.88/1.79	 % 	~ 	Mod: (Integer, Integer) -> Integer
3.88/1.79	 + 	~ 	Add: (Integer, Integer) -> Integer
3.88/1.79	 > 	~ 	Gt: (Integer, Integer) -> Boolean
3.88/1.79	 -1 	~ 	UnaryMinus: (Integer) -> Integer
3.88/1.79	 < 	~ 	Lt: (Integer, Integer) -> Boolean
3.88/1.79	 || 	~ 	Lor: (Boolean, Boolean) -> Boolean
3.88/1.79	 - 	~ 	Sub: (Integer, Integer) -> Integer
3.88/1.79	 ~ 	~ 	Bwnot: (Integer) -> Integer
3.88/1.79	 * 	~ 	Mul: (Integer, Integer) -> Integer
3.88/1.79	>>>
3.88/1.79	
3.88/1.79	
3.88/1.79	The following domains are used:
3.88/1.79	   Integer
3.88/1.79	
3.88/1.79	The ITRS R consists of the following rules:
3.88/1.79	f(x) -> Cond_f(x * x < 0, x)
3.88/1.79	Cond_f(TRUE, x) -> f(x)
3.88/1.79	
3.88/1.79	The integer pair graph contains the following rules and edges:
3.88/1.79	(0): F(x[0]) -> COND_F(x[0] * x[0] < 0, x[0])
3.88/1.79	(1): COND_F(TRUE, x[1]) -> F(x[1])
3.88/1.79	
3.88/1.79	   (0) -> (1), if (x[0] * x[0] < 0  & x[0] ->^* x[1])
3.88/1.79	   (1) -> (0), if (x[1] ->^* x[0])
3.88/1.79	
3.88/1.79	The set Q consists of the following terms:
3.88/1.79	f(x0)
3.88/1.79	Cond_f(TRUE, x0)
3.88/1.79	
3.88/1.79	----------------------------------------
3.88/1.79	
3.88/1.79	(3) UsableRulesProof (EQUIVALENT)
3.88/1.79	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.
3.88/1.79	----------------------------------------
3.88/1.79	
3.88/1.79	(4)
3.88/1.79	Obligation:
3.88/1.79	IDP problem:
3.88/1.79	The following function symbols are pre-defined:
3.88/1.79	<<<
3.88/1.79	 & 	~ 	Bwand: (Integer, Integer) -> Integer
3.88/1.79	 >= 	~ 	Ge: (Integer, Integer) -> Boolean
3.88/1.79	 | 	~ 	Bwor: (Integer, Integer) -> Integer
3.88/1.79	 / 	~ 	Div: (Integer, Integer) -> Integer
3.88/1.79	 != 	~ 	Neq: (Integer, Integer) -> Boolean
3.88/1.79	 && 	~ 	Land: (Boolean, Boolean) -> Boolean
3.88/1.79	 ! 	~ 	Lnot: (Boolean) -> Boolean
3.88/1.79	 = 	~ 	Eq: (Integer, Integer) -> Boolean
3.88/1.79	 <= 	~ 	Le: (Integer, Integer) -> Boolean
3.88/1.79	 ^ 	~ 	Bwxor: (Integer, Integer) -> Integer
3.88/1.79	 % 	~ 	Mod: (Integer, Integer) -> Integer
3.88/1.79	 + 	~ 	Add: (Integer, Integer) -> Integer
3.88/1.79	 > 	~ 	Gt: (Integer, Integer) -> Boolean
3.88/1.79	 -1 	~ 	UnaryMinus: (Integer) -> Integer
3.88/1.79	 < 	~ 	Lt: (Integer, Integer) -> Boolean
3.88/1.79	 || 	~ 	Lor: (Boolean, Boolean) -> Boolean
3.88/1.79	 - 	~ 	Sub: (Integer, Integer) -> Integer
3.88/1.79	 ~ 	~ 	Bwnot: (Integer) -> Integer
3.88/1.79	 * 	~ 	Mul: (Integer, Integer) -> Integer
3.88/1.79	>>>
3.88/1.79	
3.88/1.79	
3.88/1.79	The following domains are used:
3.88/1.79	   Integer
3.88/1.79	
3.88/1.79	R is empty.
3.88/1.79	
3.88/1.79	The integer pair graph contains the following rules and edges:
3.88/1.79	(0): F(x[0]) -> COND_F(x[0] * x[0] < 0, x[0])
3.88/1.79	(1): COND_F(TRUE, x[1]) -> F(x[1])
3.88/1.79	
3.88/1.79	   (0) -> (1), if (x[0] * x[0] < 0  & x[0] ->^* x[1])
3.88/1.79	   (1) -> (0), if (x[1] ->^* x[0])
3.88/1.79	
3.88/1.79	The set Q consists of the following terms:
3.88/1.79	f(x0)
3.88/1.79	Cond_f(TRUE, x0)
3.88/1.79	
3.88/1.79	----------------------------------------
3.88/1.79	
3.88/1.79	(5) IDPNonInfProof (SOUND)
3.88/1.79	Used the following options for this NonInfProof:
3.88/1.79	
3.88/1.79	IDPGPoloSolver:
3.88/1.79	Range: [(-1,2)]
3.88/1.79	IsNat: false
3.88/1.79	Interpretation Shape Heuristic: aprove.DPFramework.IDPProblem.Processors.nonInf.poly.IdpDefaultShapeHeuristic@5032ac69
3.88/1.79	Constraint Generator: NonInfConstraintGenerator:
3.88/1.79	PathGenerator: MetricPathGenerator:
3.88/1.79	Max Left Steps: 1
3.88/1.79	Max Right Steps: 1
3.88/1.79	
3.88/1.79	
3.88/1.79	
3.88/1.79	The constraints were generated the following way:
3.88/1.79	
3.88/1.79	The DP Problem is simplified using the Induction Calculus [NONINF] with the following steps:
3.88/1.79	
3.88/1.79	Note that final constraints are written in bold face.
3.88/1.79	
3.88/1.79	
3.88/1.79	
3.88/1.79	For Pair F(x) -> COND_F(<(*(x, x), 0), x) the following chains were created:
3.88/1.79	*We consider the chain COND_F(TRUE, x[1]) -> F(x[1]), F(x[0]) -> COND_F(<(*(x[0], x[0]), 0), x[0]), COND_F(TRUE, x[1]) -> F(x[1]) which results in the following constraint:
3.88/1.79	
3.88/1.79	(1)    (x[1]=x[0] & <(*(x[0], x[0]), 0)=TRUE & x[0]=x[1]1  ==>  F(x[0])_>=_NonInfC & F(x[0])_>=_COND_F(<(*(x[0], x[0]), 0), x[0]) & (U^Increasing(COND_F(<(*(x[0], x[0]), 0), x[0])), >=))
3.88/1.79	
3.88/1.79	
3.88/1.79	
3.88/1.79	We simplified constraint (1) using rules (III), (IV) which results in the following new constraint:
3.88/1.79	
3.88/1.79	(2)    (<(*(x[0], x[0]), 0)=TRUE  ==>  F(x[0])_>=_NonInfC & F(x[0])_>=_COND_F(<(*(x[0], x[0]), 0), x[0]) & (U^Increasing(COND_F(<(*(x[0], x[0]), 0), x[0])), >=))
3.88/1.79	
3.88/1.79	
3.88/1.79	
3.88/1.79	We simplified constraint (2) using rule (POLY_CONSTRAINTS) which results in the following new constraint:
3.88/1.79	
3.88/1.79	(3)    ([-1] + [-1]x[0]^2 >= 0  ==>  (U^Increasing(COND_F(<(*(x[0], x[0]), 0), x[0])), >=) & [(-1)bni_9 + (-1)Bound*bni_9] >= 0 & [(-1)bso_10] >= 0)
3.88/1.79	
3.88/1.79	
3.88/1.79	
3.88/1.79	We simplified constraint (3) using rule (IDP_POLY_SIMPLIFY) which results in the following new constraint:
3.88/1.79	
3.88/1.79	(4)    ([-1] + [-1]x[0]^2 >= 0  ==>  (U^Increasing(COND_F(<(*(x[0], x[0]), 0), x[0])), >=) & [(-1)bni_9 + (-1)Bound*bni_9] >= 0 & [(-1)bso_10] >= 0)
3.88/1.79	
3.88/1.79	
3.88/1.79	
3.88/1.79	We simplified constraint (4) using rule (POLY_REMOVE_MIN_MAX) which results in the following new constraint:
3.88/1.79	
3.88/1.79	(5)    ([-1] + [-1]x[0]^2 >= 0  ==>  (U^Increasing(COND_F(<(*(x[0], x[0]), 0), x[0])), >=) & [(-1)bni_9 + (-1)Bound*bni_9] >= 0 & [(-1)bso_10] >= 0)
3.88/1.79	
3.88/1.79	
3.88/1.79	
3.88/1.79	We simplified constraint (5) using rule (IDP_SMT_SPLIT) which results in the following new constraints:
3.88/1.79	
3.88/1.79	(6)    ([-1] + [-1]x[0]^2 >= 0 & x[0] >= 0  ==>  (U^Increasing(COND_F(<(*(x[0], x[0]), 0), x[0])), >=) & [(-1)bni_9 + (-1)Bound*bni_9] >= 0 & [(-1)bso_10] >= 0)
3.88/1.79	
3.88/1.79	(7)    ([-1] + [-1]x[0]^2 >= 0 & x[0] >= 0  ==>  (U^Increasing(COND_F(<(*(x[0], x[0]), 0), x[0])), >=) & [(-1)bni_9 + (-1)Bound*bni_9] >= 0 & [(-1)bso_10] >= 0)
3.88/1.79	
3.88/1.79	
3.88/1.79	
3.88/1.79	
3.88/1.79	
3.88/1.79	
3.88/1.79	
3.88/1.79	
3.88/1.79	For Pair COND_F(TRUE, x) -> F(x) the following chains were created:
3.88/1.79	*We consider the chain F(x[0]) -> COND_F(<(*(x[0], x[0]), 0), x[0]), COND_F(TRUE, x[1]) -> F(x[1]), F(x[0]) -> COND_F(<(*(x[0], x[0]), 0), x[0]) which results in the following constraint:
3.88/1.79	
3.88/1.79	(1)    (<(*(x[0], x[0]), 0)=TRUE & x[0]=x[1] & x[1]=x[0]1  ==>  COND_F(TRUE, x[1])_>=_NonInfC & COND_F(TRUE, x[1])_>=_F(x[1]) & (U^Increasing(F(x[1])), >=))
3.88/1.79	
3.88/1.79	
3.88/1.79	
3.88/1.79	We simplified constraint (1) using rules (III), (IV) which results in the following new constraint:
3.88/1.79	
3.88/1.79	(2)    (<(*(x[0], x[0]), 0)=TRUE  ==>  COND_F(TRUE, x[0])_>=_NonInfC & COND_F(TRUE, x[0])_>=_F(x[0]) & (U^Increasing(F(x[1])), >=))
3.88/1.79	
3.88/1.79	
3.88/1.79	
3.88/1.79	We simplified constraint (2) using rule (POLY_CONSTRAINTS) which results in the following new constraint:
3.88/1.79	
3.88/1.79	(3)    ([-1] + [-1]x[0]^2 >= 0  ==>  (U^Increasing(F(x[1])), >=) & [(-1)bni_11 + (-1)Bound*bni_11] >= 0 & [(-1)bso_12] >= 0)
3.88/1.79	
3.88/1.79	
3.88/1.79	
3.88/1.79	We simplified constraint (3) using rule (IDP_POLY_SIMPLIFY) which results in the following new constraint:
3.88/1.79	
3.88/1.79	(4)    ([-1] + [-1]x[0]^2 >= 0  ==>  (U^Increasing(F(x[1])), >=) & [(-1)bni_11 + (-1)Bound*bni_11] >= 0 & [(-1)bso_12] >= 0)
3.88/1.79	
3.88/1.79	
3.88/1.79	
3.88/1.79	We simplified constraint (4) using rule (POLY_REMOVE_MIN_MAX) which results in the following new constraint:
3.88/1.79	
3.88/1.79	(5)    ([-1] + [-1]x[0]^2 >= 0  ==>  (U^Increasing(F(x[1])), >=) & [(-1)bni_11 + (-1)Bound*bni_11] >= 0 & [(-1)bso_12] >= 0)
3.88/1.79	
3.88/1.79	
3.88/1.79	
3.88/1.79	We simplified constraint (5) using rule (IDP_SMT_SPLIT) which results in the following new constraints:
3.88/1.79	
3.88/1.79	(6)    ([-1] + [-1]x[0]^2 >= 0 & x[0] >= 0  ==>  (U^Increasing(F(x[1])), >=) & [(-1)bni_11 + (-1)Bound*bni_11] >= 0 & [(-1)bso_12] >= 0)
3.88/1.79	
3.88/1.79	(7)    ([-1] + [-1]x[0]^2 >= 0 & x[0] >= 0  ==>  (U^Increasing(F(x[1])), >=) & [(-1)bni_11 + (-1)Bound*bni_11] >= 0 & [(-1)bso_12] >= 0)
3.88/1.79	
3.88/1.79	
3.88/1.79	
3.88/1.79	
3.88/1.79	
3.88/1.79	
3.88/1.79	
3.88/1.79	
3.88/1.79	To summarize, we get the following constraints P__>=_ for the following pairs.
3.88/1.79	
3.88/1.79	*F(x) -> COND_F(<(*(x, x), 0), x)
3.88/1.79	
3.88/1.79	*([-1] + [-1]x[0]^2 >= 0 & x[0] >= 0  ==>  (U^Increasing(COND_F(<(*(x[0], x[0]), 0), x[0])), >=) & [(-1)bni_9 + (-1)Bound*bni_9] >= 0 & [(-1)bso_10] >= 0)
3.88/1.79	
3.88/1.79	
3.88/1.79	*([-1] + [-1]x[0]^2 >= 0 & x[0] >= 0  ==>  (U^Increasing(COND_F(<(*(x[0], x[0]), 0), x[0])), >=) & [(-1)bni_9 + (-1)Bound*bni_9] >= 0 & [(-1)bso_10] >= 0)
3.88/1.79	
3.88/1.79	
3.88/1.79	
3.88/1.79	
3.88/1.79	*COND_F(TRUE, x) -> F(x)
3.88/1.79	
3.88/1.79	*([-1] + [-1]x[0]^2 >= 0 & x[0] >= 0  ==>  (U^Increasing(F(x[1])), >=) & [(-1)bni_11 + (-1)Bound*bni_11] >= 0 & [(-1)bso_12] >= 0)
3.88/1.79	
3.88/1.79	
3.88/1.79	*([-1] + [-1]x[0]^2 >= 0 & x[0] >= 0  ==>  (U^Increasing(F(x[1])), >=) & [(-1)bni_11 + (-1)Bound*bni_11] >= 0 & [(-1)bso_12] >= 0)
3.88/1.79	
3.88/1.79	
3.88/1.79	
3.88/1.79	
3.88/1.79	
3.88/1.79	
3.88/1.79	
3.88/1.79	
3.88/1.79	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. 
3.88/1.79	
3.88/1.79	Using the following integer polynomial ordering the  resulting constraints can be solved 
3.88/1.79	
3.88/1.79	Polynomial interpretation over integers[POLO]:
3.88/1.79	
3.88/1.79	   POL(TRUE) = 0
3.88/1.79	   POL(FALSE) = 0
3.88/1.79	   POL(F(x_1)) = [-1]
3.88/1.79	   POL(COND_F(x_1, x_2)) = [-1]
3.88/1.79	   POL(<(x_1, x_2)) = [-1]
3.88/1.79	   POL(*(x_1, x_2)) = x_1*x_2
3.88/1.79	   POL(0) = 0
3.88/1.79	
3.88/1.79	
3.88/1.79	The following pairs are in P_>:
3.88/1.79	
3.88/1.79	
3.88/1.79	   F(x[0]) -> COND_F(<(*(x[0], x[0]), 0), x[0])
3.88/1.79	   COND_F(TRUE, x[1]) -> F(x[1])
3.88/1.79	
3.88/1.79	
3.88/1.79	The following pairs are in P_bound:
3.88/1.79	
3.88/1.79	
3.88/1.79	   F(x[0]) -> COND_F(<(*(x[0], x[0]), 0), x[0])
3.88/1.79	   COND_F(TRUE, x[1]) -> F(x[1])
3.88/1.79	
3.88/1.79	
3.88/1.79	The following pairs are in P_>=:
3.88/1.79	
3.88/1.79	none
3.88/1.79	
3.88/1.79	
3.88/1.79	There are no usable rules.
3.88/1.79	----------------------------------------
3.88/1.79	
3.88/1.79	(6)
3.88/1.79	Obligation:
3.88/1.79	IDP problem:
3.88/1.79	The following function symbols are pre-defined:
3.88/1.79	<<<
3.88/1.79	 & 	~ 	Bwand: (Integer, Integer) -> Integer
3.88/1.79	 >= 	~ 	Ge: (Integer, Integer) -> Boolean
3.88/1.79	 | 	~ 	Bwor: (Integer, Integer) -> Integer
3.88/1.79	 / 	~ 	Div: (Integer, Integer) -> Integer
3.88/1.79	 != 	~ 	Neq: (Integer, Integer) -> Boolean
3.88/1.79	 && 	~ 	Land: (Boolean, Boolean) -> Boolean
3.88/1.79	 ! 	~ 	Lnot: (Boolean) -> Boolean
3.88/1.79	 = 	~ 	Eq: (Integer, Integer) -> Boolean
3.88/1.79	 <= 	~ 	Le: (Integer, Integer) -> Boolean
3.88/1.79	 ^ 	~ 	Bwxor: (Integer, Integer) -> Integer
3.88/1.79	 % 	~ 	Mod: (Integer, Integer) -> Integer
3.88/1.79	 + 	~ 	Add: (Integer, Integer) -> Integer
3.88/1.79	 > 	~ 	Gt: (Integer, Integer) -> Boolean
3.88/1.79	 -1 	~ 	UnaryMinus: (Integer) -> Integer
3.88/1.79	 < 	~ 	Lt: (Integer, Integer) -> Boolean
3.88/1.79	 || 	~ 	Lor: (Boolean, Boolean) -> Boolean
3.88/1.79	 - 	~ 	Sub: (Integer, Integer) -> Integer
3.88/1.79	 ~ 	~ 	Bwnot: (Integer) -> Integer
3.88/1.79	 * 	~ 	Mul: (Integer, Integer) -> Integer
3.88/1.79	>>>
3.88/1.79	
3.88/1.79	
3.88/1.79	The following domains are used:
3.88/1.79	none
3.88/1.79	
3.88/1.79	R is empty.
3.88/1.79	
3.88/1.79	The integer pair graph is empty.
3.88/1.79	
3.88/1.79	The set Q consists of the following terms:
3.88/1.79	f(x0)
3.88/1.79	Cond_f(TRUE, x0)
3.88/1.79	
3.88/1.79	----------------------------------------
3.88/1.79	
3.88/1.79	(7) PisEmptyProof (EQUIVALENT)
3.88/1.79	The TRS P is empty. Hence, there is no (P,Q,R) chain.
3.88/1.79	----------------------------------------
3.88/1.79	
3.88/1.79	(8)
3.88/1.79	YES
3.96/1.81	EOF