3.73/1.83	YES
3.73/1.85	proof of /export/starexec/sandbox/benchmark/theBenchmark.itrs
3.73/1.85	# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty
3.73/1.85	
3.73/1.85	
3.73/1.85	Termination of the given ITRS could be proven:
3.73/1.85	
3.73/1.85	(0) ITRS
3.73/1.85	(1) ITRStoIDPProof [EQUIVALENT, 0 ms]
3.73/1.85	(2) IDP
3.73/1.85	(3) UsableRulesProof [EQUIVALENT, 0 ms]
3.73/1.85	(4) IDP
3.73/1.85	(5) IDPNonInfProof [SOUND, 128 ms]
3.73/1.85	(6) IDP
3.73/1.85	(7) PisEmptyProof [EQUIVALENT, 0 ms]
3.73/1.85	(8) YES
3.73/1.85	
3.73/1.85	
3.73/1.85	----------------------------------------
3.73/1.85	
3.73/1.85	(0)
3.73/1.85	Obligation:
3.73/1.85	ITRS problem:
3.73/1.85	
3.73/1.85	The following function symbols are pre-defined:
3.73/1.85	<<<
3.73/1.85	 & 	~ 	Bwand: (Integer, Integer) -> Integer
3.73/1.85	 >= 	~ 	Ge: (Integer, Integer) -> Boolean
3.73/1.85	 | 	~ 	Bwor: (Integer, Integer) -> Integer
3.73/1.85	 / 	~ 	Div: (Integer, Integer) -> Integer
3.73/1.85	 != 	~ 	Neq: (Integer, Integer) -> Boolean
3.73/1.85	 && 	~ 	Land: (Boolean, Boolean) -> Boolean
3.73/1.85	 ! 	~ 	Lnot: (Boolean) -> Boolean
3.73/1.85	 = 	~ 	Eq: (Integer, Integer) -> Boolean
3.73/1.85	 <= 	~ 	Le: (Integer, Integer) -> Boolean
3.73/1.85	 ^ 	~ 	Bwxor: (Integer, Integer) -> Integer
3.73/1.85	 % 	~ 	Mod: (Integer, Integer) -> Integer
3.73/1.85	 > 	~ 	Gt: (Integer, Integer) -> Boolean
3.73/1.85	 + 	~ 	Add: (Integer, Integer) -> Integer
3.73/1.85	 -1 	~ 	UnaryMinus: (Integer) -> Integer
3.73/1.85	 < 	~ 	Lt: (Integer, Integer) -> Boolean
3.73/1.85	 || 	~ 	Lor: (Boolean, Boolean) -> Boolean
3.73/1.85	 - 	~ 	Sub: (Integer, Integer) -> Integer
3.73/1.85	 ~ 	~ 	Bwnot: (Integer) -> Integer
3.73/1.85	 * 	~ 	Mul: (Integer, Integer) -> Integer
3.73/1.85	>>>
3.73/1.85	
3.73/1.85	The TRS R consists of the following rules:
3.73/1.85	eval(x) -> Cond_eval(x % 2 > 0 && x > 0, x)
3.73/1.85	Cond_eval(TRUE, x) -> eval(x - 3)
3.73/1.85	The set Q consists of the following terms:
3.73/1.85	eval(x0)
3.73/1.85	Cond_eval(TRUE, x0)
3.73/1.85	
3.73/1.85	----------------------------------------
3.73/1.85	
3.73/1.85	(1) ITRStoIDPProof (EQUIVALENT)
3.73/1.85	Added dependency pairs
3.73/1.85	----------------------------------------
3.73/1.85	
3.73/1.85	(2)
3.73/1.85	Obligation:
3.73/1.85	IDP problem:
3.73/1.85	The following function symbols are pre-defined:
3.73/1.85	<<<
3.73/1.85	 & 	~ 	Bwand: (Integer, Integer) -> Integer
3.73/1.85	 >= 	~ 	Ge: (Integer, Integer) -> Boolean
3.73/1.85	 | 	~ 	Bwor: (Integer, Integer) -> Integer
3.73/1.85	 / 	~ 	Div: (Integer, Integer) -> Integer
3.73/1.85	 != 	~ 	Neq: (Integer, Integer) -> Boolean
3.73/1.85	 && 	~ 	Land: (Boolean, Boolean) -> Boolean
3.73/1.85	 ! 	~ 	Lnot: (Boolean) -> Boolean
3.73/1.85	 = 	~ 	Eq: (Integer, Integer) -> Boolean
3.73/1.85	 <= 	~ 	Le: (Integer, Integer) -> Boolean
3.73/1.85	 ^ 	~ 	Bwxor: (Integer, Integer) -> Integer
3.73/1.85	 % 	~ 	Mod: (Integer, Integer) -> Integer
3.73/1.85	 > 	~ 	Gt: (Integer, Integer) -> Boolean
3.73/1.85	 + 	~ 	Add: (Integer, Integer) -> Integer
3.73/1.85	 -1 	~ 	UnaryMinus: (Integer) -> Integer
3.73/1.85	 < 	~ 	Lt: (Integer, Integer) -> Boolean
3.73/1.85	 || 	~ 	Lor: (Boolean, Boolean) -> Boolean
3.73/1.85	 - 	~ 	Sub: (Integer, Integer) -> Integer
3.73/1.85	 ~ 	~ 	Bwnot: (Integer) -> Integer
3.73/1.85	 * 	~ 	Mul: (Integer, Integer) -> Integer
3.73/1.85	>>>
3.73/1.85	
3.73/1.85	
3.73/1.85	The following domains are used:
3.73/1.85	   Boolean, Integer
3.73/1.85	
3.73/1.85	The ITRS R consists of the following rules:
3.73/1.85	eval(x) -> Cond_eval(x % 2 > 0 && x > 0, x)
3.73/1.85	Cond_eval(TRUE, x) -> eval(x - 3)
3.73/1.85	
3.73/1.85	The integer pair graph contains the following rules and edges:
3.73/1.85	(0): EVAL(x[0]) -> COND_EVAL(x[0] % 2 > 0 && x[0] > 0, x[0])
3.73/1.85	(1): COND_EVAL(TRUE, x[1]) -> EVAL(x[1] - 3)
3.73/1.85	
3.73/1.85	   (0) -> (1), if (x[0] % 2 > 0 && x[0] > 0  & x[0] ->^* x[1])
3.73/1.85	   (1) -> (0), if (x[1] - 3 ->^* x[0])
3.73/1.85	
3.73/1.85	The set Q consists of the following terms:
3.73/1.85	eval(x0)
3.73/1.85	Cond_eval(TRUE, x0)
3.73/1.85	
3.73/1.85	----------------------------------------
3.73/1.85	
3.73/1.85	(3) UsableRulesProof (EQUIVALENT)
3.73/1.85	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.
3.73/1.85	----------------------------------------
3.73/1.85	
3.73/1.85	(4)
3.73/1.85	Obligation:
3.73/1.85	IDP problem:
3.73/1.85	The following function symbols are pre-defined:
3.73/1.85	<<<
3.73/1.85	 & 	~ 	Bwand: (Integer, Integer) -> Integer
3.73/1.85	 >= 	~ 	Ge: (Integer, Integer) -> Boolean
3.73/1.85	 | 	~ 	Bwor: (Integer, Integer) -> Integer
3.73/1.85	 / 	~ 	Div: (Integer, Integer) -> Integer
3.73/1.85	 != 	~ 	Neq: (Integer, Integer) -> Boolean
3.73/1.85	 && 	~ 	Land: (Boolean, Boolean) -> Boolean
3.73/1.85	 ! 	~ 	Lnot: (Boolean) -> Boolean
3.73/1.85	 = 	~ 	Eq: (Integer, Integer) -> Boolean
3.73/1.85	 <= 	~ 	Le: (Integer, Integer) -> Boolean
3.73/1.85	 ^ 	~ 	Bwxor: (Integer, Integer) -> Integer
3.73/1.85	 % 	~ 	Mod: (Integer, Integer) -> Integer
3.73/1.85	 > 	~ 	Gt: (Integer, Integer) -> Boolean
3.73/1.85	 + 	~ 	Add: (Integer, Integer) -> Integer
3.73/1.85	 -1 	~ 	UnaryMinus: (Integer) -> Integer
3.73/1.85	 < 	~ 	Lt: (Integer, Integer) -> Boolean
3.73/1.85	 || 	~ 	Lor: (Boolean, Boolean) -> Boolean
3.73/1.85	 - 	~ 	Sub: (Integer, Integer) -> Integer
3.73/1.85	 ~ 	~ 	Bwnot: (Integer) -> Integer
3.73/1.85	 * 	~ 	Mul: (Integer, Integer) -> Integer
3.73/1.85	>>>
3.73/1.85	
3.73/1.85	
3.73/1.85	The following domains are used:
3.73/1.85	   Boolean, Integer
3.73/1.85	
3.73/1.85	R is empty.
3.73/1.85	
3.73/1.85	The integer pair graph contains the following rules and edges:
3.73/1.85	(0): EVAL(x[0]) -> COND_EVAL(x[0] % 2 > 0 && x[0] > 0, x[0])
3.73/1.85	(1): COND_EVAL(TRUE, x[1]) -> EVAL(x[1] - 3)
3.73/1.85	
3.73/1.85	   (0) -> (1), if (x[0] % 2 > 0 && x[0] > 0  & x[0] ->^* x[1])
3.73/1.85	   (1) -> (0), if (x[1] - 3 ->^* x[0])
3.73/1.85	
3.73/1.85	The set Q consists of the following terms:
3.73/1.85	eval(x0)
3.73/1.85	Cond_eval(TRUE, x0)
3.73/1.85	
3.73/1.85	----------------------------------------
3.73/1.85	
3.73/1.85	(5) IDPNonInfProof (SOUND)
3.73/1.85	Used the following options for this NonInfProof:
3.73/1.85	
3.73/1.85	IDPGPoloSolver:
3.73/1.85	Range: [(-1,2)]
3.73/1.85	IsNat: false
3.73/1.85	Interpretation Shape Heuristic: aprove.DPFramework.IDPProblem.Processors.nonInf.poly.IdpDefaultShapeHeuristic@252654a4
3.73/1.85	Constraint Generator: NonInfConstraintGenerator:
3.73/1.85	PathGenerator: MetricPathGenerator:
3.73/1.85	Max Left Steps: 1
3.73/1.85	Max Right Steps: 1
3.73/1.85	
3.73/1.85	
3.73/1.85	
3.73/1.85	The constraints were generated the following way:
3.73/1.85	
3.73/1.85	The DP Problem is simplified using the Induction Calculus [NONINF] with the following steps:
3.73/1.85	
3.73/1.85	Note that final constraints are written in bold face.
3.73/1.85	
3.73/1.85	
3.73/1.85	
3.73/1.85	For Pair EVAL(x) -> COND_EVAL(&&(>(%(x, 2), 0), >(x, 0)), x) the following chains were created:
3.73/1.85	*We consider the chain EVAL(x[0]) -> COND_EVAL(&&(>(%(x[0], 2), 0), >(x[0], 0)), x[0]), COND_EVAL(TRUE, x[1]) -> EVAL(-(x[1], 3)) which results in the following constraint:
3.73/1.85	
3.73/1.85	(1)    (&&(>(%(x[0], 2), 0), >(x[0], 0))=TRUE & x[0]=x[1]  ==>  EVAL(x[0])_>=_NonInfC & EVAL(x[0])_>=_COND_EVAL(&&(>(%(x[0], 2), 0), >(x[0], 0)), x[0]) & (U^Increasing(COND_EVAL(&&(>(%(x[0], 2), 0), >(x[0], 0)), x[0])), >=))
3.73/1.85	
3.73/1.85	
3.73/1.85	
3.73/1.85	We simplified constraint (1) using rules (IV), (IDP_BOOLEAN) which results in the following new constraint:
3.73/1.85	
3.73/1.85	(2)    (>(%(x[0], 2), 0)=TRUE & >(x[0], 0)=TRUE  ==>  EVAL(x[0])_>=_NonInfC & EVAL(x[0])_>=_COND_EVAL(&&(>(%(x[0], 2), 0), >(x[0], 0)), x[0]) & (U^Increasing(COND_EVAL(&&(>(%(x[0], 2), 0), >(x[0], 0)), x[0])), >=))
3.73/1.85	
3.73/1.85	
3.73/1.85	
3.73/1.85	We simplified constraint (2) using rule (POLY_CONSTRAINTS) which results in the following new constraint:
3.73/1.85	
3.73/1.85	(3)    (max{[2], [-2]} + [-1] >= 0 & x[0] + [-1] >= 0  ==>  (U^Increasing(COND_EVAL(&&(>(%(x[0], 2), 0), >(x[0], 0)), x[0])), >=) & [bni_10 + (-1)Bound*bni_10] + [bni_10]x[0] >= 0 & [1 + (-1)bso_11] >= 0)
3.73/1.85	
3.73/1.85	
3.73/1.85	
3.73/1.85	We simplified constraint (3) using rule (IDP_POLY_SIMPLIFY) which results in the following new constraint:
3.73/1.85	
3.73/1.85	(4)    (max{[2], [-2]} + [-1] >= 0 & x[0] + [-1] >= 0  ==>  (U^Increasing(COND_EVAL(&&(>(%(x[0], 2), 0), >(x[0], 0)), x[0])), >=) & [bni_10 + (-1)Bound*bni_10] + [bni_10]x[0] >= 0 & [1 + (-1)bso_11] >= 0)
3.73/1.85	
3.73/1.85	
3.73/1.85	
3.73/1.85	We simplified constraint (4) using rule (POLY_REMOVE_MIN_MAX) which results in the following new constraint:
3.73/1.85	
3.73/1.85	(5)    (x[0] + [-1] >= 0 & [4] >= 0 & [1] >= 0  ==>  (U^Increasing(COND_EVAL(&&(>(%(x[0], 2), 0), >(x[0], 0)), x[0])), >=) & [bni_10 + (-1)Bound*bni_10] + [bni_10]x[0] >= 0 & [1 + (-1)bso_11] >= 0)
3.73/1.85	
3.73/1.85	
3.73/1.85	
3.73/1.85	We simplified constraint (5) using rule (IDP_POLY_GCD) which results in the following new constraint:
3.73/1.85	
3.73/1.85	(6)    (x[0] + [-1] >= 0 & [1] >= 0 & [1] >= 0  ==>  (U^Increasing(COND_EVAL(&&(>(%(x[0], 2), 0), >(x[0], 0)), x[0])), >=) & [bni_10 + (-1)Bound*bni_10] + [bni_10]x[0] >= 0 & [1 + (-1)bso_11] >= 0)
3.73/1.85	
3.73/1.85	
3.73/1.85	
3.73/1.85	
3.73/1.85	
3.73/1.85	
3.73/1.85	
3.73/1.85	
3.73/1.85	For Pair COND_EVAL(TRUE, x) -> EVAL(-(x, 3)) the following chains were created:
3.73/1.85	*We consider the chain EVAL(x[0]) -> COND_EVAL(&&(>(%(x[0], 2), 0), >(x[0], 0)), x[0]), COND_EVAL(TRUE, x[1]) -> EVAL(-(x[1], 3)), EVAL(x[0]) -> COND_EVAL(&&(>(%(x[0], 2), 0), >(x[0], 0)), x[0]) which results in the following constraint:
3.73/1.85	
3.73/1.85	(1)    (&&(>(%(x[0], 2), 0), >(x[0], 0))=TRUE & x[0]=x[1] & -(x[1], 3)=x[0]1  ==>  COND_EVAL(TRUE, x[1])_>=_NonInfC & COND_EVAL(TRUE, x[1])_>=_EVAL(-(x[1], 3)) & (U^Increasing(EVAL(-(x[1], 3))), >=))
3.73/1.85	
3.73/1.85	
3.73/1.85	
3.73/1.85	We simplified constraint (1) using rules (III), (IV), (IDP_BOOLEAN) which results in the following new constraint:
3.73/1.85	
3.73/1.85	(2)    (>(%(x[0], 2), 0)=TRUE & >(x[0], 0)=TRUE  ==>  COND_EVAL(TRUE, x[0])_>=_NonInfC & COND_EVAL(TRUE, x[0])_>=_EVAL(-(x[0], 3)) & (U^Increasing(EVAL(-(x[1], 3))), >=))
3.73/1.85	
3.73/1.85	
3.73/1.85	
3.73/1.85	We simplified constraint (2) using rule (POLY_CONSTRAINTS) which results in the following new constraint:
3.73/1.85	
3.73/1.85	(3)    (max{[2], [-2]} + [-1] >= 0 & x[0] + [-1] >= 0  ==>  (U^Increasing(EVAL(-(x[1], 3))), >=) & [(-1)Bound*bni_12] + [bni_12]x[0] >= 0 & [2 + (-1)bso_13] >= 0)
3.73/1.85	
3.73/1.85	
3.73/1.85	
3.73/1.85	We simplified constraint (3) using rule (IDP_POLY_SIMPLIFY) which results in the following new constraint:
3.73/1.85	
3.73/1.85	(4)    (max{[2], [-2]} + [-1] >= 0 & x[0] + [-1] >= 0  ==>  (U^Increasing(EVAL(-(x[1], 3))), >=) & [(-1)Bound*bni_12] + [bni_12]x[0] >= 0 & [2 + (-1)bso_13] >= 0)
3.73/1.85	
3.73/1.85	
3.73/1.85	
3.73/1.85	We simplified constraint (4) using rule (POLY_REMOVE_MIN_MAX) which results in the following new constraint:
3.73/1.85	
3.73/1.85	(5)    (x[0] + [-1] >= 0 & [4] >= 0 & [1] >= 0  ==>  (U^Increasing(EVAL(-(x[1], 3))), >=) & [(-1)Bound*bni_12] + [bni_12]x[0] >= 0 & [2 + (-1)bso_13] >= 0)
3.73/1.85	
3.73/1.85	
3.73/1.85	
3.73/1.85	We simplified constraint (5) using rule (IDP_POLY_GCD) which results in the following new constraint:
3.73/1.85	
3.73/1.85	(6)    (x[0] + [-1] >= 0 & [1] >= 0 & [1] >= 0  ==>  (U^Increasing(EVAL(-(x[1], 3))), >=) & [(-1)Bound*bni_12] + [bni_12]x[0] >= 0 & [2 + (-1)bso_13] >= 0)
3.73/1.85	
3.73/1.85	
3.73/1.85	
3.73/1.85	
3.73/1.85	
3.73/1.85	
3.73/1.85	
3.73/1.85	
3.73/1.85	To summarize, we get the following constraints P__>=_ for the following pairs.
3.73/1.85	
3.73/1.85	*EVAL(x) -> COND_EVAL(&&(>(%(x, 2), 0), >(x, 0)), x)
3.73/1.85	
3.73/1.85	*(x[0] + [-1] >= 0 & [1] >= 0 & [1] >= 0  ==>  (U^Increasing(COND_EVAL(&&(>(%(x[0], 2), 0), >(x[0], 0)), x[0])), >=) & [bni_10 + (-1)Bound*bni_10] + [bni_10]x[0] >= 0 & [1 + (-1)bso_11] >= 0)
3.73/1.85	
3.73/1.85	
3.73/1.85	
3.73/1.85	
3.73/1.85	*COND_EVAL(TRUE, x) -> EVAL(-(x, 3))
3.73/1.85	
3.73/1.85	*(x[0] + [-1] >= 0 & [1] >= 0 & [1] >= 0  ==>  (U^Increasing(EVAL(-(x[1], 3))), >=) & [(-1)Bound*bni_12] + [bni_12]x[0] >= 0 & [2 + (-1)bso_13] >= 0)
3.73/1.85	
3.73/1.85	
3.73/1.85	
3.73/1.85	
3.73/1.85	
3.73/1.85	
3.73/1.85	
3.73/1.85	
3.73/1.85	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. 
3.73/1.85	
3.73/1.85	Using the following integer polynomial ordering the  resulting constraints can be solved 
3.73/1.85	
3.73/1.85	Polynomial interpretation over integers[POLO]:
3.73/1.85	
3.73/1.85	   POL(TRUE) = [1]
3.73/1.85	   POL(FALSE) = [3]
3.73/1.85	   POL(EVAL(x_1)) = [1] + x_1
3.73/1.85	   POL(COND_EVAL(x_1, x_2)) = [1] + x_2 + [-1]x_1
3.73/1.85	   POL(&&(x_1, x_2)) = [1]
3.73/1.85	   POL(>(x_1, x_2)) = [-1]
3.73/1.85	   POL(2) = [2]
3.73/1.85	   POL(0) = 0
3.73/1.85	   POL(-(x_1, x_2)) = x_1 + [-1]x_2
3.73/1.85	   POL(3) = [3]
3.73/1.85	   
3.73/1.85	   Polynomial Interpretations with Context Sensitive Arithemetic Replacement
3.73/1.85	   POL(Term^CSAR-Mode @ Context)
3.73/1.85	   
3.73/1.85	   POL(%(x_1, 2)^1 @ {}) = max{x_2, [-1]x_2}
3.73/1.85	
3.73/1.85	
3.73/1.85	The following pairs are in P_>:
3.73/1.85	
3.73/1.85	
3.73/1.85	   EVAL(x[0]) -> COND_EVAL(&&(>(%(x[0], 2), 0), >(x[0], 0)), x[0])
3.73/1.85	   COND_EVAL(TRUE, x[1]) -> EVAL(-(x[1], 3))
3.73/1.85	
3.73/1.85	
3.73/1.85	The following pairs are in P_bound:
3.73/1.85	
3.73/1.85	
3.73/1.85	   EVAL(x[0]) -> COND_EVAL(&&(>(%(x[0], 2), 0), >(x[0], 0)), x[0])
3.73/1.85	   COND_EVAL(TRUE, x[1]) -> EVAL(-(x[1], 3))
3.73/1.85	
3.73/1.85	
3.73/1.85	The following pairs are in P_>=:
3.73/1.85	
3.73/1.85	none
3.73/1.85	
3.73/1.85	
3.73/1.85	At least the following rules have been oriented under context sensitive arithmetic replacement:
3.73/1.85	
3.73/1.85	   TRUE^1 -> &&(TRUE, TRUE)^1
3.73/1.85	   FALSE^1 -> &&(TRUE, FALSE)^1
3.73/1.85	   FALSE^1 -> &&(FALSE, TRUE)^1
3.73/1.85	   FALSE^1 -> &&(FALSE, FALSE)^1
3.73/1.85	
3.73/1.85	----------------------------------------
3.73/1.85	
3.73/1.85	(6)
3.73/1.85	Obligation:
3.73/1.85	IDP problem:
3.73/1.85	The following function symbols are pre-defined:
3.73/1.85	<<<
3.73/1.85	 & 	~ 	Bwand: (Integer, Integer) -> Integer
3.73/1.85	 >= 	~ 	Ge: (Integer, Integer) -> Boolean
3.73/1.85	 | 	~ 	Bwor: (Integer, Integer) -> Integer
3.73/1.85	 / 	~ 	Div: (Integer, Integer) -> Integer
3.73/1.85	 != 	~ 	Neq: (Integer, Integer) -> Boolean
3.73/1.85	 && 	~ 	Land: (Boolean, Boolean) -> Boolean
3.73/1.85	 ! 	~ 	Lnot: (Boolean) -> Boolean
3.73/1.85	 = 	~ 	Eq: (Integer, Integer) -> Boolean
3.73/1.85	 <= 	~ 	Le: (Integer, Integer) -> Boolean
3.73/1.85	 ^ 	~ 	Bwxor: (Integer, Integer) -> Integer
3.73/1.85	 % 	~ 	Mod: (Integer, Integer) -> Integer
3.73/1.85	 > 	~ 	Gt: (Integer, Integer) -> Boolean
3.73/1.85	 + 	~ 	Add: (Integer, Integer) -> Integer
3.73/1.85	 -1 	~ 	UnaryMinus: (Integer) -> Integer
3.73/1.85	 < 	~ 	Lt: (Integer, Integer) -> Boolean
3.73/1.85	 || 	~ 	Lor: (Boolean, Boolean) -> Boolean
3.73/1.85	 - 	~ 	Sub: (Integer, Integer) -> Integer
3.73/1.85	 ~ 	~ 	Bwnot: (Integer) -> Integer
3.73/1.85	 * 	~ 	Mul: (Integer, Integer) -> Integer
3.73/1.85	>>>
3.73/1.85	
3.73/1.85	
3.73/1.85	The following domains are used:
3.73/1.85	none
3.73/1.85	
3.73/1.85	R is empty.
3.73/1.85	
3.73/1.85	The integer pair graph is empty.
3.73/1.85	
3.73/1.85	The set Q consists of the following terms:
3.73/1.85	eval(x0)
3.73/1.85	Cond_eval(TRUE, x0)
3.73/1.85	
3.73/1.85	----------------------------------------
3.73/1.85	
3.73/1.85	(7) PisEmptyProof (EQUIVALENT)
3.73/1.85	The TRS P is empty. Hence, there is no (P,Q,R) chain.
3.73/1.85	----------------------------------------
3.73/1.85	
3.73/1.85	(8)
3.73/1.85	YES
3.73/1.86	EOF