4.33/2.06 YES 4.33/2.08 proof of /export/starexec/sandbox/benchmark/theBenchmark.itrs 4.33/2.08 # AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty 4.33/2.08 4.33/2.08 4.33/2.08 Termination of the given ITRS could be proven: 4.33/2.08 4.33/2.08 (0) ITRS 4.33/2.08 (1) ITRStoIDPProof [EQUIVALENT, 0 ms] 4.33/2.08 (2) IDP 4.33/2.08 (3) UsableRulesProof [EQUIVALENT, 0 ms] 4.33/2.08 (4) IDP 4.33/2.08 (5) IDPNonInfProof [SOUND, 22 ms] 4.33/2.08 (6) IDP 4.33/2.08 (7) PisEmptyProof [EQUIVALENT, 0 ms] 4.33/2.08 (8) YES 4.33/2.08 4.33/2.08 4.33/2.08 ---------------------------------------- 4.33/2.08 4.33/2.08 (0) 4.33/2.08 Obligation: 4.33/2.08 ITRS problem: 4.33/2.08 4.33/2.08 The following function symbols are pre-defined: 4.33/2.08 <<< 4.33/2.08 & ~ Bwand: (Integer, Integer) -> Integer 4.33/2.08 >= ~ Ge: (Integer, Integer) -> Boolean 4.33/2.08 | ~ Bwor: (Integer, Integer) -> Integer 4.33/2.08 / ~ Div: (Integer, Integer) -> Integer 4.33/2.08 != ~ Neq: (Integer, Integer) -> Boolean 4.33/2.08 && ~ Land: (Boolean, Boolean) -> Boolean 4.33/2.08 ! ~ Lnot: (Boolean) -> Boolean 4.33/2.08 = ~ Eq: (Integer, Integer) -> Boolean 4.33/2.08 <= ~ Le: (Integer, Integer) -> Boolean 4.33/2.08 ^ ~ Bwxor: (Integer, Integer) -> Integer 4.33/2.08 % ~ Mod: (Integer, Integer) -> Integer 4.33/2.08 + ~ Add: (Integer, Integer) -> Integer 4.33/2.08 > ~ Gt: (Integer, Integer) -> Boolean 4.33/2.08 -1 ~ Sub: (Integer, Integer) -> Integer 4.33/2.08 < ~ Lt: (Integer, Integer) -> Boolean 4.33/2.08 || ~ Lor: (Boolean, Boolean) -> Boolean 4.33/2.08 - ~ UnaryMinus: (Integer) -> Integer 4.33/2.08 ~ ~ Bwnot: (Integer) -> Integer 4.33/2.08 * ~ Mul: (Integer, Integer) -> Integer 4.33/2.08 >>> 4.33/2.08 4.33/2.08 The TRS R consists of the following rules: 4.33/2.08 eval(x) -> Cond_eval(x >= 0, x) 4.33/2.08 Cond_eval(TRUE, x) -> eval(-(2) * x + 10) 4.33/2.08 The set Q consists of the following terms: 4.33/2.08 eval(x0) 4.33/2.08 Cond_eval(TRUE, x0) 4.33/2.08 4.33/2.08 ---------------------------------------- 4.33/2.08 4.33/2.08 (1) ITRStoIDPProof (EQUIVALENT) 4.33/2.08 Added dependency pairs 4.33/2.08 ---------------------------------------- 4.33/2.08 4.33/2.08 (2) 4.33/2.08 Obligation: 4.33/2.08 IDP problem: 4.33/2.08 The following function symbols are pre-defined: 4.33/2.08 <<< 4.33/2.08 & ~ Bwand: (Integer, Integer) -> Integer 4.33/2.08 >= ~ Ge: (Integer, Integer) -> Boolean 4.33/2.08 | ~ Bwor: (Integer, Integer) -> Integer 4.33/2.08 / ~ Div: (Integer, Integer) -> Integer 4.33/2.08 != ~ Neq: (Integer, Integer) -> Boolean 4.33/2.08 && ~ Land: (Boolean, Boolean) -> Boolean 4.33/2.08 ! ~ Lnot: (Boolean) -> Boolean 4.33/2.08 = ~ Eq: (Integer, Integer) -> Boolean 4.33/2.08 <= ~ Le: (Integer, Integer) -> Boolean 4.33/2.08 ^ ~ Bwxor: (Integer, Integer) -> Integer 4.33/2.08 % ~ Mod: (Integer, Integer) -> Integer 4.33/2.08 + ~ Add: (Integer, Integer) -> Integer 4.33/2.08 > ~ Gt: (Integer, Integer) -> Boolean 4.33/2.08 -1 ~ Sub: (Integer, Integer) -> Integer 4.33/2.08 < ~ Lt: (Integer, Integer) -> Boolean 4.33/2.08 || ~ Lor: (Boolean, Boolean) -> Boolean 4.33/2.08 - ~ UnaryMinus: (Integer) -> Integer 4.33/2.08 ~ ~ Bwnot: (Integer) -> Integer 4.33/2.08 * ~ Mul: (Integer, Integer) -> Integer 4.33/2.08 >>> 4.33/2.08 4.33/2.08 4.33/2.08 The following domains are used: 4.33/2.08 Integer 4.33/2.08 4.33/2.08 The ITRS R consists of the following rules: 4.33/2.08 eval(x) -> Cond_eval(x >= 0, x) 4.33/2.08 Cond_eval(TRUE, x) -> eval(-(2) * x + 10) 4.33/2.08 4.33/2.08 The integer pair graph contains the following rules and edges: 4.33/2.08 (0): EVAL(x[0]) -> COND_EVAL(x[0] >= 0, x[0]) 4.33/2.08 (1): COND_EVAL(TRUE, x[1]) -> EVAL(-(2) * x[1] + 10) 4.33/2.08 4.33/2.08 (0) -> (1), if (x[0] >= 0 & x[0] ->^* x[1]) 4.33/2.08 (1) -> (0), if (-(2) * x[1] + 10 ->^* x[0]) 4.33/2.08 4.33/2.08 The set Q consists of the following terms: 4.33/2.08 eval(x0) 4.33/2.08 Cond_eval(TRUE, x0) 4.33/2.08 4.33/2.08 ---------------------------------------- 4.33/2.08 4.33/2.08 (3) UsableRulesProof (EQUIVALENT) 4.33/2.08 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. 4.33/2.08 ---------------------------------------- 4.33/2.08 4.33/2.08 (4) 4.33/2.08 Obligation: 4.33/2.08 IDP problem: 4.33/2.08 The following function symbols are pre-defined: 4.33/2.08 <<< 4.33/2.08 & ~ Bwand: (Integer, Integer) -> Integer 4.33/2.08 >= ~ Ge: (Integer, Integer) -> Boolean 4.33/2.08 | ~ Bwor: (Integer, Integer) -> Integer 4.33/2.08 / ~ Div: (Integer, Integer) -> Integer 4.33/2.08 != ~ Neq: (Integer, Integer) -> Boolean 4.33/2.08 && ~ Land: (Boolean, Boolean) -> Boolean 4.33/2.08 ! ~ Lnot: (Boolean) -> Boolean 4.33/2.08 = ~ Eq: (Integer, Integer) -> Boolean 4.33/2.08 <= ~ Le: (Integer, Integer) -> Boolean 4.33/2.08 ^ ~ Bwxor: (Integer, Integer) -> Integer 4.33/2.08 % ~ Mod: (Integer, Integer) -> Integer 4.33/2.08 + ~ Add: (Integer, Integer) -> Integer 4.33/2.08 > ~ Gt: (Integer, Integer) -> Boolean 4.33/2.08 -1 ~ Sub: (Integer, Integer) -> Integer 4.33/2.08 < ~ Lt: (Integer, Integer) -> Boolean 4.33/2.08 || ~ Lor: (Boolean, Boolean) -> Boolean 4.33/2.08 - ~ UnaryMinus: (Integer) -> Integer 4.33/2.08 ~ ~ Bwnot: (Integer) -> Integer 4.33/2.08 * ~ Mul: (Integer, Integer) -> Integer 4.33/2.08 >>> 4.33/2.08 4.33/2.08 4.33/2.08 The following domains are used: 4.33/2.08 Integer 4.33/2.08 4.33/2.08 R is empty. 4.33/2.08 4.33/2.08 The integer pair graph contains the following rules and edges: 4.33/2.08 (0): EVAL(x[0]) -> COND_EVAL(x[0] >= 0, x[0]) 4.33/2.08 (1): COND_EVAL(TRUE, x[1]) -> EVAL(-(2) * x[1] + 10) 4.33/2.08 4.33/2.08 (0) -> (1), if (x[0] >= 0 & x[0] ->^* x[1]) 4.33/2.08 (1) -> (0), if (-(2) * x[1] + 10 ->^* x[0]) 4.33/2.08 4.33/2.08 The set Q consists of the following terms: 4.33/2.08 eval(x0) 4.33/2.08 Cond_eval(TRUE, x0) 4.33/2.08 4.33/2.08 ---------------------------------------- 4.33/2.08 4.33/2.08 (5) IDPNonInfProof (SOUND) 4.33/2.08 Used the following options for this NonInfProof: 4.33/2.08 4.33/2.08 IDPGPoloSolver: 4.33/2.08 Range: [(-1,2)] 4.33/2.08 IsNat: false 4.33/2.08 Interpretation Shape Heuristic: aprove.DPFramework.IDPProblem.Processors.nonInf.poly.IdpDefaultShapeHeuristic@2aaab1f9 4.33/2.08 Constraint Generator: NonInfConstraintGenerator: 4.33/2.08 PathGenerator: MetricPathGenerator: 4.33/2.08 Max Left Steps: 8 4.33/2.08 Max Right Steps: 1 4.33/2.08 4.33/2.08 4.33/2.08 4.33/2.08 The constraints were generated the following way: 4.33/2.08 4.33/2.08 The DP Problem is simplified using the Induction Calculus [NONINF] with the following steps: 4.33/2.08 4.33/2.08 Note that final constraints are written in bold face. 4.33/2.08 4.33/2.08 4.33/2.08 4.33/2.08 For Pair EVAL(x) -> COND_EVAL(>=(x, 0), x) the following chains were created: 4.33/2.08 *We consider the chain EVAL(x[0]) -> COND_EVAL(>=(x[0], 0), x[0]), COND_EVAL(TRUE, x[1]) -> EVAL(+(*(-(2), x[1]), 10)) which results in the following constraint: 4.33/2.08 4.33/2.08 (1) (>=(x[0], 0)=TRUE & x[0]=x[1] ==> EVAL(x[0])_>=_NonInfC & EVAL(x[0])_>=_COND_EVAL(>=(x[0], 0), x[0]) & (U^Increasing(COND_EVAL(>=(x[0], 0), x[0])), >=)) 4.33/2.08 4.33/2.08 4.33/2.08 4.33/2.08 We simplified constraint (1) using rule (IV) which results in the following new constraint: 4.33/2.08 4.33/2.08 (2) (>=(x[0], 0)=TRUE ==> EVAL(x[0])_>=_NonInfC & EVAL(x[0])_>=_COND_EVAL(>=(x[0], 0), x[0]) & (U^Increasing(COND_EVAL(>=(x[0], 0), x[0])), >=)) 4.33/2.08 4.33/2.08 4.33/2.08 4.33/2.08 We simplified constraint (2) using rule (POLY_CONSTRAINTS) which results in the following new constraint: 4.33/2.08 4.33/2.08 (3) (x[0] >= 0 ==> (U^Increasing(COND_EVAL(>=(x[0], 0), x[0])), >=) & [bni_9 + (-1)Bound*bni_9] + [(2)bni_9]x[0] >= 0 & [2 + (-1)bso_10] + [2]x[0] >= 0) 4.33/2.08 4.33/2.08 4.33/2.08 4.33/2.08 We simplified constraint (3) using rule (IDP_POLY_SIMPLIFY) which results in the following new constraint: 4.33/2.08 4.33/2.08 (4) (x[0] >= 0 ==> (U^Increasing(COND_EVAL(>=(x[0], 0), x[0])), >=) & [bni_9 + (-1)Bound*bni_9] + [(2)bni_9]x[0] >= 0 & [2 + (-1)bso_10] + [2]x[0] >= 0) 4.33/2.08 4.33/2.08 4.33/2.08 4.33/2.08 We simplified constraint (4) using rule (POLY_REMOVE_MIN_MAX) which results in the following new constraint: 4.33/2.08 4.33/2.08 (5) (x[0] >= 0 ==> (U^Increasing(COND_EVAL(>=(x[0], 0), x[0])), >=) & [bni_9 + (-1)Bound*bni_9] + [(2)bni_9]x[0] >= 0 & [2 + (-1)bso_10] + [2]x[0] >= 0) 4.33/2.08 4.33/2.08 4.33/2.08 4.33/2.08 4.33/2.08 4.33/2.08 4.33/2.08 4.33/2.08 4.33/2.08 For Pair COND_EVAL(TRUE, x) -> EVAL(+(*(-(2), x), 10)) the following chains were created: 4.33/2.08 *We consider the chain COND_EVAL(TRUE, x[1]) -> EVAL(+(*(-(2), x[1]), 10)), EVAL(x[0]) -> COND_EVAL(>=(x[0], 0), x[0]), COND_EVAL(TRUE, x[1]) -> EVAL(+(*(-(2), x[1]), 10)), EVAL(x[0]) -> COND_EVAL(>=(x[0], 0), x[0]), COND_EVAL(TRUE, x[1]) -> EVAL(+(*(-(2), x[1]), 10)), EVAL(x[0]) -> COND_EVAL(>=(x[0], 0), x[0]), COND_EVAL(TRUE, x[1]) -> EVAL(+(*(-(2), x[1]), 10)), EVAL(x[0]) -> COND_EVAL(>=(x[0], 0), x[0]), COND_EVAL(TRUE, x[1]) -> EVAL(+(*(-(2), x[1]), 10)), EVAL(x[0]) -> COND_EVAL(>=(x[0], 0), x[0]) which results in the following constraint: 4.33/2.08 4.33/2.08 (1) (+(*(-(2), x[1]), 10)=x[0] & >=(x[0], 0)=TRUE & x[0]=x[1]1 & +(*(-(2), x[1]1), 10)=x[0]1 & >=(x[0]1, 0)=TRUE & x[0]1=x[1]2 & +(*(-(2), x[1]2), 10)=x[0]2 & >=(x[0]2, 0)=TRUE & x[0]2=x[1]3 & +(*(-(2), x[1]3), 10)=x[0]3 & >=(x[0]3, 0)=TRUE & x[0]3=x[1]4 & +(*(-(2), x[1]4), 10)=x[0]4 ==> COND_EVAL(TRUE, x[1]4)_>=_NonInfC & COND_EVAL(TRUE, x[1]4)_>=_EVAL(+(*(-(2), x[1]4), 10)) & (U^Increasing(EVAL(+(*(-(2), x[1]4), 10))), >=)) 4.33/2.08 4.33/2.08 4.33/2.08 4.33/2.08 We simplified constraint (1) using rules (III), (IV), (IDP_CONSTANT_FOLD) which results in the following new constraint: 4.33/2.08 4.33/2.08 (2) (>=(+(*(-2, x[1]), 10), 0)=TRUE & >=(+(*(-2, +(*(-2, x[1]), 10)), 10), 0)=TRUE & >=(+(*(-2, +(*(-2, +(*(-2, x[1]), 10)), 10)), 10), 0)=TRUE & >=(+(*(-2, +(*(-2, +(*(-2, +(*(-2, x[1]), 10)), 10)), 10)), 10), 0)=TRUE ==> COND_EVAL(TRUE, +(*(-2, +(*(-2, +(*(-2, +(*(-2, x[1]), 10)), 10)), 10)), 10))_>=_NonInfC & COND_EVAL(TRUE, +(*(-2, +(*(-2, +(*(-2, +(*(-2, x[1]), 10)), 10)), 10)), 10))_>=_EVAL(+(*(-(2), +(*(-2, +(*(-2, +(*(-2, +(*(-2, x[1]), 10)), 10)), 10)), 10)), 10)) & (U^Increasing(EVAL(+(*(-(2), x[1]4), 10))), >=)) 4.33/2.08 4.33/2.08 4.33/2.08 4.33/2.08 We simplified constraint (2) using rule (POLY_CONSTRAINTS) which results in the following new constraint: 4.33/2.08 4.33/2.08 (3) ([-2]x[1] + [10] >= 0 & [4]x[1] + [-10] >= 0 & [-8]x[1] + [30] >= 0 & [16]x[1] + [-50] >= 0 ==> (U^Increasing(EVAL(+(*(-(2), x[1]4), 10))), >=) & [(-1)bni_11 + (-1)Bound*bni_11] >= 0 & [-222 + (-1)bso_12] + [64]x[1] >= 0) 4.33/2.08 4.33/2.08 4.33/2.08 4.33/2.08 We simplified constraint (3) using rule (IDP_POLY_SIMPLIFY) which results in the following new constraint: 4.33/2.08 4.33/2.08 (4) ([-2]x[1] + [10] >= 0 & [4]x[1] + [-10] >= 0 & [-8]x[1] + [30] >= 0 & [16]x[1] + [-50] >= 0 ==> (U^Increasing(EVAL(+(*(-(2), x[1]4), 10))), >=) & [(-1)bni_11 + (-1)Bound*bni_11] >= 0 & [-222 + (-1)bso_12] + [64]x[1] >= 0) 4.33/2.08 4.33/2.08 4.33/2.08 4.33/2.08 We simplified constraint (4) using rule (POLY_REMOVE_MIN_MAX) which results in the following new constraint: 4.33/2.08 4.33/2.08 (5) ([-2]x[1] + [10] >= 0 & [4]x[1] + [-10] >= 0 & [-8]x[1] + [30] >= 0 & [16]x[1] + [-50] >= 0 ==> (U^Increasing(EVAL(+(*(-(2), x[1]4), 10))), >=) & [(-1)bni_11 + (-1)Bound*bni_11] >= 0 & [-222 + (-1)bso_12] + [64]x[1] >= 0) 4.33/2.08 4.33/2.08 4.33/2.08 4.33/2.08 We solved constraint (5) using rule (IDP_SMT_SPLIT). 4.33/2.08 4.33/2.08 4.33/2.08 4.33/2.08 4.33/2.08 To summarize, we get the following constraints P__>=_ for the following pairs. 4.33/2.08 4.33/2.08 *EVAL(x) -> COND_EVAL(>=(x, 0), x) 4.33/2.08 4.33/2.08 *(x[0] >= 0 ==> (U^Increasing(COND_EVAL(>=(x[0], 0), x[0])), >=) & [bni_9 + (-1)Bound*bni_9] + [(2)bni_9]x[0] >= 0 & [2 + (-1)bso_10] + [2]x[0] >= 0) 4.33/2.08 4.33/2.08 4.33/2.08 4.33/2.08 4.33/2.08 *COND_EVAL(TRUE, x) -> EVAL(+(*(-(2), x), 10)) 4.33/2.08 4.33/2.08 4.33/2.08 4.33/2.08 4.33/2.08 4.33/2.08 4.33/2.08 4.33/2.08 4.33/2.08 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. 4.33/2.08 4.33/2.08 Using the following integer polynomial ordering the resulting constraints can be solved 4.33/2.08 4.33/2.08 Polynomial interpretation over integers[POLO]: 4.33/2.08 4.33/2.08 POL(TRUE) = 0 4.33/2.08 POL(FALSE) = 0 4.33/2.08 POL(EVAL(x_1)) = [1] + [2]x_1 4.33/2.08 POL(COND_EVAL(x_1, x_2)) = [-1] 4.33/2.08 POL(>=(x_1, x_2)) = [2] 4.33/2.08 POL(0) = 0 4.33/2.08 POL(+(x_1, x_2)) = x_1 + x_2 4.33/2.08 POL(*(x_1, x_2)) = x_1*x_2 4.33/2.08 POL(-(x_1)) = [-1]x_1 4.33/2.08 POL(2) = [2] 4.33/2.08 POL(10) = [10] 4.33/2.08 POL(-2) = [-2] 4.33/2.08 4.33/2.08 4.33/2.08 The following pairs are in P_>: 4.33/2.08 4.33/2.08 4.33/2.08 EVAL(x[0]) -> COND_EVAL(>=(x[0], 0), x[0]) 4.33/2.08 COND_EVAL(TRUE, x[1]) -> EVAL(+(*(-(2), x[1]), 10)) 4.33/2.08 4.33/2.08 4.33/2.08 The following pairs are in P_bound: 4.33/2.08 4.33/2.08 4.33/2.08 EVAL(x[0]) -> COND_EVAL(>=(x[0], 0), x[0]) 4.33/2.08 COND_EVAL(TRUE, x[1]) -> EVAL(+(*(-(2), x[1]), 10)) 4.33/2.08 4.33/2.08 4.33/2.08 The following pairs are in P_>=: 4.33/2.08 4.33/2.08 none 4.33/2.08 4.33/2.08 4.33/2.08 There are no usable rules. 4.33/2.08 ---------------------------------------- 4.33/2.08 4.33/2.08 (6) 4.33/2.08 Obligation: 4.33/2.08 IDP problem: 4.33/2.08 The following function symbols are pre-defined: 4.33/2.08 <<< 4.33/2.08 & ~ Bwand: (Integer, Integer) -> Integer 4.33/2.08 >= ~ Ge: (Integer, Integer) -> Boolean 4.33/2.08 | ~ Bwor: (Integer, Integer) -> Integer 4.33/2.08 / ~ Div: (Integer, Integer) -> Integer 4.33/2.08 != ~ Neq: (Integer, Integer) -> Boolean 4.33/2.08 && ~ Land: (Boolean, Boolean) -> Boolean 4.33/2.08 ! ~ Lnot: (Boolean) -> Boolean 4.33/2.08 = ~ Eq: (Integer, Integer) -> Boolean 4.33/2.08 <= ~ Le: (Integer, Integer) -> Boolean 4.33/2.08 ^ ~ Bwxor: (Integer, Integer) -> Integer 4.33/2.08 % ~ Mod: (Integer, Integer) -> Integer 4.33/2.08 + ~ Add: (Integer, Integer) -> Integer 4.33/2.08 > ~ Gt: (Integer, Integer) -> Boolean 4.33/2.08 -1 ~ Sub: (Integer, Integer) -> Integer 4.33/2.08 < ~ Lt: (Integer, Integer) -> Boolean 4.33/2.08 || ~ Lor: (Boolean, Boolean) -> Boolean 4.33/2.08 - ~ UnaryMinus: (Integer) -> Integer 4.33/2.08 ~ ~ Bwnot: (Integer) -> Integer 4.33/2.08 * ~ Mul: (Integer, Integer) -> Integer 4.33/2.08 >>> 4.33/2.08 4.33/2.08 4.33/2.08 The following domains are used: 4.33/2.08 none 4.33/2.08 4.33/2.08 R is empty. 4.33/2.08 4.33/2.08 The integer pair graph is empty. 4.33/2.08 4.33/2.08 The set Q consists of the following terms: 4.33/2.08 eval(x0) 4.33/2.08 Cond_eval(TRUE, x0) 4.33/2.08 4.33/2.08 ---------------------------------------- 4.33/2.08 4.33/2.08 (7) PisEmptyProof (EQUIVALENT) 4.33/2.08 The TRS P is empty. Hence, there is no (P,Q,R) chain. 4.33/2.08 ---------------------------------------- 4.33/2.08 4.33/2.08 (8) 4.33/2.08 YES 4.55/2.12 EOF