3.63/1.77	YES
3.63/1.78	proof of /export/starexec/sandbox/benchmark/theBenchmark.itrs
3.63/1.78	# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty
3.63/1.78	
3.63/1.78	
3.63/1.78	Termination of the given ITRS could be proven:
3.63/1.78	
3.63/1.78	(0) ITRS
3.63/1.78	(1) ITRStoIDPProof [EQUIVALENT, 0 ms]
3.63/1.78	(2) IDP
3.63/1.78	(3) UsableRulesProof [EQUIVALENT, 0 ms]
3.63/1.78	(4) IDP
3.63/1.78	(5) IDPNonInfProof [SOUND, 100 ms]
3.63/1.78	(6) IDP
3.63/1.78	(7) PisEmptyProof [EQUIVALENT, 0 ms]
3.63/1.78	(8) YES
3.63/1.78	
3.63/1.78	
3.63/1.78	----------------------------------------
3.63/1.78	
3.63/1.78	(0)
3.63/1.78	Obligation:
3.63/1.78	ITRS problem:
3.63/1.78	
3.63/1.78	The following function symbols are pre-defined:
3.63/1.78	<<<
3.63/1.78	 & 	~ 	Bwand: (Integer, Integer) -> Integer
3.63/1.78	 >= 	~ 	Ge: (Integer, Integer) -> Boolean
3.63/1.78	 | 	~ 	Bwor: (Integer, Integer) -> Integer
3.63/1.78	 / 	~ 	Div: (Integer, Integer) -> Integer
3.63/1.78	 != 	~ 	Neq: (Integer, Integer) -> Boolean
3.63/1.78	 && 	~ 	Land: (Boolean, Boolean) -> Boolean
3.63/1.78	 ! 	~ 	Lnot: (Boolean) -> Boolean
3.63/1.78	 = 	~ 	Eq: (Integer, Integer) -> Boolean
3.63/1.78	 <= 	~ 	Le: (Integer, Integer) -> Boolean
3.63/1.78	 ^ 	~ 	Bwxor: (Integer, Integer) -> Integer
3.63/1.78	 % 	~ 	Mod: (Integer, Integer) -> Integer
3.63/1.78	 > 	~ 	Gt: (Integer, Integer) -> Boolean
3.63/1.78	 + 	~ 	Add: (Integer, Integer) -> Integer
3.63/1.78	 -1 	~ 	UnaryMinus: (Integer) -> Integer
3.63/1.78	 < 	~ 	Lt: (Integer, Integer) -> Boolean
3.63/1.78	 || 	~ 	Lor: (Boolean, Boolean) -> Boolean
3.63/1.78	 - 	~ 	Sub: (Integer, Integer) -> Integer
3.63/1.78	 ~ 	~ 	Bwnot: (Integer) -> Integer
3.63/1.78	 * 	~ 	Mul: (Integer, Integer) -> Integer
3.63/1.78	>>>
3.63/1.78	
3.63/1.78	The TRS R consists of the following rules:
3.63/1.78	f(x) -> Cond_f(x > x, x)
3.63/1.78	Cond_f(TRUE, x) -> f(x)
3.63/1.78	The set Q consists of the following terms:
3.63/1.78	f(x0)
3.63/1.78	Cond_f(TRUE, x0)
3.63/1.78	
3.63/1.78	----------------------------------------
3.63/1.78	
3.63/1.78	(1) ITRStoIDPProof (EQUIVALENT)
3.63/1.78	Added dependency pairs
3.63/1.78	----------------------------------------
3.63/1.78	
3.63/1.78	(2)
3.63/1.78	Obligation:
3.63/1.78	IDP problem:
3.63/1.78	The following function symbols are pre-defined:
3.63/1.78	<<<
3.63/1.78	 & 	~ 	Bwand: (Integer, Integer) -> Integer
3.63/1.78	 >= 	~ 	Ge: (Integer, Integer) -> Boolean
3.63/1.78	 | 	~ 	Bwor: (Integer, Integer) -> Integer
3.63/1.78	 / 	~ 	Div: (Integer, Integer) -> Integer
3.63/1.78	 != 	~ 	Neq: (Integer, Integer) -> Boolean
3.63/1.78	 && 	~ 	Land: (Boolean, Boolean) -> Boolean
3.63/1.78	 ! 	~ 	Lnot: (Boolean) -> Boolean
3.63/1.78	 = 	~ 	Eq: (Integer, Integer) -> Boolean
3.63/1.78	 <= 	~ 	Le: (Integer, Integer) -> Boolean
3.63/1.78	 ^ 	~ 	Bwxor: (Integer, Integer) -> Integer
3.63/1.78	 % 	~ 	Mod: (Integer, Integer) -> Integer
3.63/1.78	 > 	~ 	Gt: (Integer, Integer) -> Boolean
3.63/1.78	 + 	~ 	Add: (Integer, Integer) -> Integer
3.63/1.78	 -1 	~ 	UnaryMinus: (Integer) -> Integer
3.63/1.78	 < 	~ 	Lt: (Integer, Integer) -> Boolean
3.63/1.78	 || 	~ 	Lor: (Boolean, Boolean) -> Boolean
3.63/1.78	 - 	~ 	Sub: (Integer, Integer) -> Integer
3.63/1.78	 ~ 	~ 	Bwnot: (Integer) -> Integer
3.63/1.78	 * 	~ 	Mul: (Integer, Integer) -> Integer
3.63/1.78	>>>
3.63/1.78	
3.63/1.78	
3.63/1.78	The following domains are used:
3.63/1.78	   Integer
3.63/1.78	
3.63/1.78	The ITRS R consists of the following rules:
3.63/1.78	f(x) -> Cond_f(x > x, x)
3.63/1.78	Cond_f(TRUE, x) -> f(x)
3.63/1.78	
3.63/1.78	The integer pair graph contains the following rules and edges:
3.63/1.78	(0): F(x[0]) -> COND_F(x[0] > x[0], x[0])
3.63/1.78	(1): COND_F(TRUE, x[1]) -> F(x[1])
3.63/1.78	
3.63/1.78	   (0) -> (1), if (x[0] > x[0]  & x[0] ->^* x[1])
3.63/1.78	   (1) -> (0), if (x[1] ->^* x[0])
3.63/1.78	
3.63/1.78	The set Q consists of the following terms:
3.63/1.78	f(x0)
3.63/1.78	Cond_f(TRUE, x0)
3.63/1.78	
3.63/1.78	----------------------------------------
3.63/1.78	
3.63/1.78	(3) UsableRulesProof (EQUIVALENT)
3.63/1.78	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.
3.63/1.78	----------------------------------------
3.63/1.78	
3.63/1.78	(4)
3.63/1.78	Obligation:
3.63/1.78	IDP problem:
3.63/1.78	The following function symbols are pre-defined:
3.63/1.78	<<<
3.63/1.78	 & 	~ 	Bwand: (Integer, Integer) -> Integer
3.63/1.78	 >= 	~ 	Ge: (Integer, Integer) -> Boolean
3.63/1.78	 | 	~ 	Bwor: (Integer, Integer) -> Integer
3.63/1.78	 / 	~ 	Div: (Integer, Integer) -> Integer
3.63/1.78	 != 	~ 	Neq: (Integer, Integer) -> Boolean
3.63/1.78	 && 	~ 	Land: (Boolean, Boolean) -> Boolean
3.63/1.78	 ! 	~ 	Lnot: (Boolean) -> Boolean
3.63/1.78	 = 	~ 	Eq: (Integer, Integer) -> Boolean
3.63/1.78	 <= 	~ 	Le: (Integer, Integer) -> Boolean
3.63/1.78	 ^ 	~ 	Bwxor: (Integer, Integer) -> Integer
3.63/1.78	 % 	~ 	Mod: (Integer, Integer) -> Integer
3.63/1.78	 > 	~ 	Gt: (Integer, Integer) -> Boolean
3.63/1.78	 + 	~ 	Add: (Integer, Integer) -> Integer
3.63/1.78	 -1 	~ 	UnaryMinus: (Integer) -> Integer
3.63/1.78	 < 	~ 	Lt: (Integer, Integer) -> Boolean
3.63/1.78	 || 	~ 	Lor: (Boolean, Boolean) -> Boolean
3.63/1.78	 - 	~ 	Sub: (Integer, Integer) -> Integer
3.63/1.78	 ~ 	~ 	Bwnot: (Integer) -> Integer
3.63/1.78	 * 	~ 	Mul: (Integer, Integer) -> Integer
3.63/1.78	>>>
3.63/1.78	
3.63/1.78	
3.63/1.78	The following domains are used:
3.63/1.78	   Integer
3.63/1.78	
3.63/1.78	R is empty.
3.63/1.78	
3.63/1.78	The integer pair graph contains the following rules and edges:
3.63/1.78	(0): F(x[0]) -> COND_F(x[0] > x[0], x[0])
3.63/1.78	(1): COND_F(TRUE, x[1]) -> F(x[1])
3.63/1.78	
3.63/1.78	   (0) -> (1), if (x[0] > x[0]  & x[0] ->^* x[1])
3.63/1.78	   (1) -> (0), if (x[1] ->^* x[0])
3.63/1.78	
3.63/1.78	The set Q consists of the following terms:
3.63/1.78	f(x0)
3.63/1.78	Cond_f(TRUE, x0)
3.63/1.78	
3.63/1.78	----------------------------------------
3.63/1.78	
3.63/1.78	(5) IDPNonInfProof (SOUND)
3.63/1.78	Used the following options for this NonInfProof:
3.63/1.78	
3.63/1.78	IDPGPoloSolver:
3.63/1.78	Range: [(-1,2)]
3.63/1.78	IsNat: false
3.63/1.78	Interpretation Shape Heuristic: aprove.DPFramework.IDPProblem.Processors.nonInf.poly.IdpDefaultShapeHeuristic@11353166
3.63/1.78	Constraint Generator: NonInfConstraintGenerator:
3.63/1.78	PathGenerator: MetricPathGenerator:
3.63/1.78	Max Left Steps: 1
3.63/1.78	Max Right Steps: 1
3.63/1.78	
3.63/1.78	
3.63/1.78	
3.63/1.78	The constraints were generated the following way:
3.63/1.78	
3.63/1.78	The DP Problem is simplified using the Induction Calculus [NONINF] with the following steps:
3.63/1.78	
3.63/1.78	Note that final constraints are written in bold face.
3.63/1.78	
3.63/1.78	
3.63/1.78	
3.63/1.78	For Pair F(x) -> COND_F(>(x, x), x) the following chains were created:
3.63/1.78	*We consider the chain COND_F(TRUE, x[1]) -> F(x[1]), F(x[0]) -> COND_F(>(x[0], x[0]), x[0]), COND_F(TRUE, x[1]) -> F(x[1]) which results in the following constraint:
3.63/1.78	
3.63/1.78	(1)    (x[1]=x[0] & >(x[0], x[0])=TRUE & x[0]=x[1]1  ==>  F(x[0])_>=_NonInfC & F(x[0])_>=_COND_F(>(x[0], x[0]), x[0]) & (U^Increasing(COND_F(>(x[0], x[0]), x[0])), >=))
3.63/1.78	
3.63/1.78	
3.63/1.78	
3.63/1.78	We simplified constraint (1) using rules (III), (IV) which results in the following new constraint:
3.63/1.78	
3.63/1.78	(2)    (>(x[0], x[0])=TRUE  ==>  F(x[0])_>=_NonInfC & F(x[0])_>=_COND_F(>(x[0], x[0]), x[0]) & (U^Increasing(COND_F(>(x[0], x[0]), x[0])), >=))
3.63/1.78	
3.63/1.78	
3.63/1.78	
3.63/1.78	We simplified constraint (2) using rule (POLY_CONSTRAINTS) which results in the following new constraint:
3.63/1.78	
3.63/1.78	(3)    ([-1] >= 0  ==>  (U^Increasing(COND_F(>(x[0], x[0]), x[0])), >=) & [(-1)bni_9 + (-1)Bound*bni_9] + [(-1)bni_9]x[0] >= 0 & [-1 + (-1)bso_10] >= 0)
3.63/1.78	
3.63/1.78	
3.63/1.78	
3.63/1.78	We simplified constraint (3) using rule (IDP_POLY_SIMPLIFY) which results in the following new constraint:
3.63/1.78	
3.63/1.78	(4)    ([-1] >= 0  ==>  (U^Increasing(COND_F(>(x[0], x[0]), x[0])), >=) & [(-1)bni_9 + (-1)Bound*bni_9] + [(-1)bni_9]x[0] >= 0 & [-1 + (-1)bso_10] >= 0)
3.63/1.78	
3.63/1.78	
3.63/1.78	
3.63/1.78	We simplified constraint (4) using rule (POLY_REMOVE_MIN_MAX) which results in the following new constraint:
3.63/1.78	
3.63/1.78	(5)    ([-1] >= 0  ==>  (U^Increasing(COND_F(>(x[0], x[0]), x[0])), >=) & [(-1)bni_9 + (-1)Bound*bni_9] + [(-1)bni_9]x[0] >= 0 & [-1 + (-1)bso_10] >= 0)
3.63/1.78	
3.63/1.78	
3.63/1.78	
3.63/1.78	We simplified constraint (5) using rule (IDP_UNRESTRICTED_VARS) which results in the following new constraint:
3.63/1.78	
3.63/1.78	(6)    ([-1] >= 0  ==>  (U^Increasing(COND_F(>(x[0], x[0]), x[0])), >=) & [(-1)bni_9] = 0 & [(-1)bni_9 + (-1)Bound*bni_9] >= 0 & [-1 + (-1)bso_10] >= 0)
3.63/1.78	
3.63/1.78	
3.63/1.78	
3.63/1.78	We solved constraint (6) using rule (IDP_SMT_SPLIT).
3.63/1.78	
3.63/1.78	
3.63/1.78	
3.63/1.78	
3.63/1.78	For Pair COND_F(TRUE, x) -> F(x) the following chains were created:
3.63/1.78	*We consider the chain F(x[0]) -> COND_F(>(x[0], x[0]), x[0]), COND_F(TRUE, x[1]) -> F(x[1]), F(x[0]) -> COND_F(>(x[0], x[0]), x[0]) which results in the following constraint:
3.63/1.78	
3.63/1.78	(1)    (>(x[0], x[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])), >=))
3.63/1.78	
3.63/1.78	
3.63/1.78	
3.63/1.78	We simplified constraint (1) using rules (III), (IV) which results in the following new constraint:
3.63/1.78	
3.63/1.78	(2)    (>(x[0], x[0])=TRUE  ==>  COND_F(TRUE, x[0])_>=_NonInfC & COND_F(TRUE, x[0])_>=_F(x[0]) & (U^Increasing(F(x[1])), >=))
3.63/1.78	
3.63/1.78	
3.63/1.78	
3.63/1.78	We simplified constraint (2) using rule (POLY_CONSTRAINTS) which results in the following new constraint:
3.63/1.78	
3.63/1.78	(3)    ([-1] >= 0  ==>  (U^Increasing(F(x[1])), >=) & [(-1)bni_11 + (-1)Bound*bni_11] + [(-1)bni_11]x[0] >= 0 & [(-1)bso_12] >= 0)
3.63/1.78	
3.63/1.78	
3.63/1.78	
3.63/1.78	We simplified constraint (3) using rule (IDP_POLY_SIMPLIFY) which results in the following new constraint:
3.63/1.78	
3.63/1.78	(4)    ([-1] >= 0  ==>  (U^Increasing(F(x[1])), >=) & [(-1)bni_11 + (-1)Bound*bni_11] + [(-1)bni_11]x[0] >= 0 & [(-1)bso_12] >= 0)
3.63/1.78	
3.63/1.78	
3.63/1.78	
3.63/1.78	We simplified constraint (4) using rule (POLY_REMOVE_MIN_MAX) which results in the following new constraint:
3.63/1.78	
3.63/1.78	(5)    ([-1] >= 0  ==>  (U^Increasing(F(x[1])), >=) & [(-1)bni_11 + (-1)Bound*bni_11] + [(-1)bni_11]x[0] >= 0 & [(-1)bso_12] >= 0)
3.63/1.78	
3.63/1.78	
3.63/1.78	
3.63/1.78	We simplified constraint (5) using rule (IDP_UNRESTRICTED_VARS) which results in the following new constraint:
3.63/1.78	
3.63/1.78	(6)    ([-1] >= 0  ==>  (U^Increasing(F(x[1])), >=) & [(-1)bni_11] = 0 & [(-1)bni_11 + (-1)Bound*bni_11] >= 0 & [(-1)bso_12] >= 0)
3.63/1.78	
3.63/1.78	
3.63/1.78	
3.63/1.78	We solved constraint (6) using rule (IDP_SMT_SPLIT).
3.63/1.78	
3.63/1.78	
3.63/1.78	
3.63/1.78	
3.63/1.78	To summarize, we get the following constraints P__>=_ for the following pairs.
3.63/1.78	
3.63/1.78	*F(x) -> COND_F(>(x, x), x)
3.63/1.78	
3.63/1.78	
3.63/1.78	
3.63/1.78	
3.63/1.78	*COND_F(TRUE, x) -> F(x)
3.63/1.78	
3.63/1.78	
3.63/1.78	
3.63/1.78	
3.63/1.78	
3.63/1.78	
3.63/1.78	
3.63/1.78	
3.63/1.78	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. 
3.63/1.78	
3.63/1.78	Using the following integer polynomial ordering the  resulting constraints can be solved 
3.63/1.78	
3.63/1.78	Polynomial interpretation over integers[POLO]:
3.63/1.78	
3.63/1.78	   POL(TRUE) = 0
3.63/1.78	   POL(FALSE) = 0
3.63/1.78	   POL(F(x_1)) = [-1] + [-1]x_1
3.63/1.78	   POL(COND_F(x_1, x_2)) = [-1] + [-1]x_2 + [-1]x_1
3.63/1.78	   POL(>(x_1, x_2)) = [-1]
3.63/1.78	
3.63/1.78	
3.63/1.78	The following pairs are in P_>:
3.63/1.78	
3.63/1.78	
3.63/1.78	   F(x[0]) -> COND_F(>(x[0], x[0]), x[0])
3.63/1.78	   COND_F(TRUE, x[1]) -> F(x[1])
3.63/1.78	
3.63/1.78	
3.63/1.78	The following pairs are in P_bound:
3.63/1.78	
3.63/1.78	
3.63/1.78	   F(x[0]) -> COND_F(>(x[0], x[0]), x[0])
3.63/1.78	   COND_F(TRUE, x[1]) -> F(x[1])
3.63/1.78	
3.63/1.78	
3.63/1.78	The following pairs are in P_>=:
3.63/1.78	
3.63/1.78	none
3.63/1.78	
3.63/1.78	
3.63/1.78	There are no usable rules.
3.63/1.78	----------------------------------------
3.63/1.78	
3.63/1.78	(6)
3.63/1.78	Obligation:
3.63/1.78	IDP problem:
3.63/1.78	The following function symbols are pre-defined:
3.63/1.78	<<<
3.63/1.78	 & 	~ 	Bwand: (Integer, Integer) -> Integer
3.63/1.78	 >= 	~ 	Ge: (Integer, Integer) -> Boolean
3.63/1.78	 | 	~ 	Bwor: (Integer, Integer) -> Integer
3.63/1.78	 / 	~ 	Div: (Integer, Integer) -> Integer
3.63/1.78	 != 	~ 	Neq: (Integer, Integer) -> Boolean
3.63/1.78	 && 	~ 	Land: (Boolean, Boolean) -> Boolean
3.63/1.78	 ! 	~ 	Lnot: (Boolean) -> Boolean
3.63/1.78	 = 	~ 	Eq: (Integer, Integer) -> Boolean
3.63/1.78	 <= 	~ 	Le: (Integer, Integer) -> Boolean
3.63/1.78	 ^ 	~ 	Bwxor: (Integer, Integer) -> Integer
3.63/1.78	 % 	~ 	Mod: (Integer, Integer) -> Integer
3.63/1.78	 > 	~ 	Gt: (Integer, Integer) -> Boolean
3.63/1.78	 + 	~ 	Add: (Integer, Integer) -> Integer
3.63/1.78	 -1 	~ 	UnaryMinus: (Integer) -> Integer
3.63/1.78	 < 	~ 	Lt: (Integer, Integer) -> Boolean
3.63/1.78	 || 	~ 	Lor: (Boolean, Boolean) -> Boolean
3.63/1.78	 - 	~ 	Sub: (Integer, Integer) -> Integer
3.63/1.78	 ~ 	~ 	Bwnot: (Integer) -> Integer
3.63/1.78	 * 	~ 	Mul: (Integer, Integer) -> Integer
3.63/1.78	>>>
3.63/1.78	
3.63/1.78	
3.63/1.78	The following domains are used:
3.63/1.78	none
3.63/1.78	
3.63/1.78	R is empty.
3.63/1.78	
3.63/1.78	The integer pair graph is empty.
3.63/1.78	
3.63/1.78	The set Q consists of the following terms:
3.63/1.78	f(x0)
3.63/1.78	Cond_f(TRUE, x0)
3.63/1.78	
3.63/1.78	----------------------------------------
3.63/1.78	
3.63/1.78	(7) PisEmptyProof (EQUIVALENT)
3.63/1.78	The TRS P is empty. Hence, there is no (P,Q,R) chain.
3.63/1.78	----------------------------------------
3.63/1.78	
3.63/1.78	(8)
3.63/1.78	YES
3.67/1.80	EOF