3.74/1.86 YES 3.74/1.87 proof of /export/starexec/sandbox/benchmark/theBenchmark.itrs 3.74/1.87 # AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty 3.74/1.87 3.74/1.87 3.74/1.87 Termination of the given ITRS could be proven: 3.74/1.87 3.74/1.87 (0) ITRS 3.74/1.87 (1) ITRStoIDPProof [EQUIVALENT, 0 ms] 3.74/1.87 (2) IDP 3.74/1.87 (3) UsableRulesProof [EQUIVALENT, 0 ms] 3.74/1.87 (4) IDP 3.74/1.87 (5) IDPNonInfProof [SOUND, 182 ms] 3.74/1.87 (6) IDP 3.74/1.87 (7) PisEmptyProof [EQUIVALENT, 0 ms] 3.74/1.87 (8) YES 3.74/1.87 3.74/1.87 3.74/1.87 ---------------------------------------- 3.74/1.87 3.74/1.87 (0) 3.74/1.87 Obligation: 3.74/1.87 ITRS problem: 3.74/1.87 3.74/1.87 The following function symbols are pre-defined: 3.74/1.87 <<< 3.74/1.87 & ~ Bwand: (Integer, Integer) -> Integer 3.74/1.87 >= ~ Ge: (Integer, Integer) -> Boolean 3.74/1.87 | ~ Bwor: (Integer, Integer) -> Integer 3.74/1.87 / ~ Div: (Integer, Integer) -> Integer 3.74/1.87 != ~ Neq: (Integer, Integer) -> Boolean 3.74/1.87 && ~ Land: (Boolean, Boolean) -> Boolean 3.74/1.87 ! ~ Lnot: (Boolean) -> Boolean 3.74/1.87 = ~ Eq: (Integer, Integer) -> Boolean 3.74/1.87 <= ~ Le: (Integer, Integer) -> Boolean 3.74/1.87 ^ ~ Bwxor: (Integer, Integer) -> Integer 3.74/1.87 % ~ Mod: (Integer, Integer) -> Integer 3.74/1.87 > ~ Gt: (Integer, Integer) -> Boolean 3.74/1.87 + ~ Add: (Integer, Integer) -> Integer 3.74/1.87 -1 ~ UnaryMinus: (Integer) -> Integer 3.74/1.87 < ~ Lt: (Integer, Integer) -> Boolean 3.74/1.87 || ~ Lor: (Boolean, Boolean) -> Boolean 3.74/1.87 - ~ Sub: (Integer, Integer) -> Integer 3.74/1.87 ~ ~ Bwnot: (Integer) -> Integer 3.74/1.87 * ~ Mul: (Integer, Integer) -> Integer 3.74/1.87 >>> 3.74/1.87 3.74/1.87 The TRS R consists of the following rules: 3.74/1.87 f(TRUE, x, y) -> f(x > y, x + 1, y + 2) 3.74/1.87 The set Q consists of the following terms: 3.74/1.87 f(TRUE, x0, x1) 3.74/1.87 3.74/1.87 ---------------------------------------- 3.74/1.87 3.74/1.87 (1) ITRStoIDPProof (EQUIVALENT) 3.74/1.87 Added dependency pairs 3.74/1.87 ---------------------------------------- 3.74/1.87 3.74/1.87 (2) 3.74/1.87 Obligation: 3.74/1.87 IDP problem: 3.74/1.87 The following function symbols are pre-defined: 3.74/1.87 <<< 3.74/1.87 & ~ Bwand: (Integer, Integer) -> Integer 3.74/1.87 >= ~ Ge: (Integer, Integer) -> Boolean 3.74/1.87 | ~ Bwor: (Integer, Integer) -> Integer 3.74/1.87 / ~ Div: (Integer, Integer) -> Integer 3.74/1.87 != ~ Neq: (Integer, Integer) -> Boolean 3.74/1.87 && ~ Land: (Boolean, Boolean) -> Boolean 3.74/1.87 ! ~ Lnot: (Boolean) -> Boolean 3.74/1.87 = ~ Eq: (Integer, Integer) -> Boolean 3.74/1.87 <= ~ Le: (Integer, Integer) -> Boolean 3.74/1.87 ^ ~ Bwxor: (Integer, Integer) -> Integer 3.74/1.87 % ~ Mod: (Integer, Integer) -> Integer 3.74/1.87 > ~ Gt: (Integer, Integer) -> Boolean 3.74/1.87 + ~ Add: (Integer, Integer) -> Integer 3.74/1.87 -1 ~ UnaryMinus: (Integer) -> Integer 3.74/1.87 < ~ Lt: (Integer, Integer) -> Boolean 3.74/1.87 || ~ Lor: (Boolean, Boolean) -> Boolean 3.74/1.87 - ~ Sub: (Integer, Integer) -> Integer 3.74/1.87 ~ ~ Bwnot: (Integer) -> Integer 3.74/1.87 * ~ Mul: (Integer, Integer) -> Integer 3.74/1.87 >>> 3.74/1.87 3.74/1.87 3.74/1.87 The following domains are used: 3.74/1.87 Integer 3.74/1.87 3.74/1.87 The ITRS R consists of the following rules: 3.74/1.87 f(TRUE, x, y) -> f(x > y, x + 1, y + 2) 3.74/1.87 3.74/1.87 The integer pair graph contains the following rules and edges: 3.74/1.87 (0): F(TRUE, x[0], y[0]) -> F(x[0] > y[0], x[0] + 1, y[0] + 2) 3.74/1.87 3.74/1.87 (0) -> (0), if (x[0] > y[0] & x[0] + 1 ->^* x[0]' & y[0] + 2 ->^* y[0]') 3.74/1.87 3.74/1.87 The set Q consists of the following terms: 3.74/1.87 f(TRUE, x0, x1) 3.74/1.87 3.74/1.87 ---------------------------------------- 3.74/1.87 3.74/1.87 (3) UsableRulesProof (EQUIVALENT) 3.74/1.87 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. 3.74/1.87 ---------------------------------------- 3.74/1.87 3.74/1.87 (4) 3.74/1.87 Obligation: 3.74/1.87 IDP problem: 3.74/1.87 The following function symbols are pre-defined: 3.74/1.87 <<< 3.74/1.87 & ~ Bwand: (Integer, Integer) -> Integer 3.74/1.87 >= ~ Ge: (Integer, Integer) -> Boolean 3.74/1.87 | ~ Bwor: (Integer, Integer) -> Integer 3.74/1.87 / ~ Div: (Integer, Integer) -> Integer 3.74/1.87 != ~ Neq: (Integer, Integer) -> Boolean 3.74/1.87 && ~ Land: (Boolean, Boolean) -> Boolean 3.74/1.87 ! ~ Lnot: (Boolean) -> Boolean 3.74/1.87 = ~ Eq: (Integer, Integer) -> Boolean 3.74/1.87 <= ~ Le: (Integer, Integer) -> Boolean 3.74/1.87 ^ ~ Bwxor: (Integer, Integer) -> Integer 3.74/1.87 % ~ Mod: (Integer, Integer) -> Integer 3.74/1.87 > ~ Gt: (Integer, Integer) -> Boolean 3.74/1.87 + ~ Add: (Integer, Integer) -> Integer 3.74/1.87 -1 ~ UnaryMinus: (Integer) -> Integer 3.74/1.87 < ~ Lt: (Integer, Integer) -> Boolean 3.74/1.87 || ~ Lor: (Boolean, Boolean) -> Boolean 3.74/1.87 - ~ Sub: (Integer, Integer) -> Integer 3.74/1.87 ~ ~ Bwnot: (Integer) -> Integer 3.74/1.87 * ~ Mul: (Integer, Integer) -> Integer 3.74/1.87 >>> 3.74/1.87 3.74/1.87 3.74/1.87 The following domains are used: 3.74/1.87 Integer 3.74/1.87 3.74/1.87 R is empty. 3.74/1.87 3.74/1.87 The integer pair graph contains the following rules and edges: 3.74/1.87 (0): F(TRUE, x[0], y[0]) -> F(x[0] > y[0], x[0] + 1, y[0] + 2) 3.74/1.87 3.74/1.87 (0) -> (0), if (x[0] > y[0] & x[0] + 1 ->^* x[0]' & y[0] + 2 ->^* y[0]') 3.74/1.87 3.74/1.87 The set Q consists of the following terms: 3.74/1.87 f(TRUE, x0, x1) 3.74/1.87 3.74/1.87 ---------------------------------------- 3.74/1.87 3.74/1.87 (5) IDPNonInfProof (SOUND) 3.74/1.87 Used the following options for this NonInfProof: 3.74/1.87 3.74/1.87 IDPGPoloSolver: 3.74/1.87 Range: [(-1,2)] 3.74/1.87 IsNat: false 3.74/1.87 Interpretation Shape Heuristic: aprove.DPFramework.IDPProblem.Processors.nonInf.poly.IdpDefaultShapeHeuristic@199a9c6d 3.74/1.87 Constraint Generator: NonInfConstraintGenerator: 3.74/1.87 PathGenerator: MetricPathGenerator: 3.74/1.87 Max Left Steps: 1 3.74/1.87 Max Right Steps: 1 3.74/1.87 3.74/1.87 3.74/1.87 3.74/1.87 The constraints were generated the following way: 3.74/1.87 3.74/1.87 The DP Problem is simplified using the Induction Calculus [NONINF] with the following steps: 3.74/1.87 3.74/1.87 Note that final constraints are written in bold face. 3.74/1.87 3.74/1.87 3.74/1.87 3.74/1.87 For Pair F(TRUE, x, y) -> F(>(x, y), +(x, 1), +(y, 2)) the following chains were created: 3.74/1.87 *We consider the chain F(TRUE, x[0], y[0]) -> F(>(x[0], y[0]), +(x[0], 1), +(y[0], 2)), F(TRUE, x[0], y[0]) -> F(>(x[0], y[0]), +(x[0], 1), +(y[0], 2)), F(TRUE, x[0], y[0]) -> F(>(x[0], y[0]), +(x[0], 1), +(y[0], 2)) which results in the following constraint: 3.74/1.87 3.74/1.87 (1) (>(x[0], y[0])=TRUE & +(x[0], 1)=x[0]1 & +(y[0], 2)=y[0]1 & >(x[0]1, y[0]1)=TRUE & +(x[0]1, 1)=x[0]2 & +(y[0]1, 2)=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), +(x[0]1, 1), +(y[0]1, 2)) & (U^Increasing(F(>(x[0]1, y[0]1), +(x[0]1, 1), +(y[0]1, 2))), >=)) 3.74/1.87 3.74/1.87 3.74/1.87 3.74/1.87 We simplified constraint (1) using rules (III), (IV) which results in the following new constraint: 3.74/1.87 3.74/1.87 (2) (>(x[0], y[0])=TRUE & >(+(x[0], 1), +(y[0], 2))=TRUE ==> F(TRUE, +(x[0], 1), +(y[0], 2))_>=_NonInfC & F(TRUE, +(x[0], 1), +(y[0], 2))_>=_F(>(+(x[0], 1), +(y[0], 2)), +(+(x[0], 1), 1), +(+(y[0], 2), 2)) & (U^Increasing(F(>(x[0]1, y[0]1), +(x[0]1, 1), +(y[0]1, 2))), >=)) 3.74/1.87 3.74/1.87 3.74/1.87 3.74/1.87 We simplified constraint (2) using rule (POLY_CONSTRAINTS) which results in the following new constraint: 3.74/1.87 3.74/1.87 (3) (x[0] + [-1] + [-1]y[0] >= 0 & x[0] + [-2] + [-1]y[0] >= 0 ==> (U^Increasing(F(>(x[0]1, y[0]1), +(x[0]1, 1), +(y[0]1, 2))), >=) & [bni_8 + (-1)Bound*bni_8] + [(-1)bni_8]y[0] + [bni_8]x[0] >= 0 & [1 + (-1)bso_9] >= 0) 3.74/1.87 3.74/1.87 3.74/1.87 3.74/1.87 We simplified constraint (3) using rule (IDP_POLY_SIMPLIFY) which results in the following new constraint: 3.74/1.87 3.74/1.87 (4) (x[0] + [-1] + [-1]y[0] >= 0 & x[0] + [-2] + [-1]y[0] >= 0 ==> (U^Increasing(F(>(x[0]1, y[0]1), +(x[0]1, 1), +(y[0]1, 2))), >=) & [bni_8 + (-1)Bound*bni_8] + [(-1)bni_8]y[0] + [bni_8]x[0] >= 0 & [1 + (-1)bso_9] >= 0) 3.74/1.87 3.74/1.87 3.74/1.87 3.74/1.87 We simplified constraint (4) using rule (POLY_REMOVE_MIN_MAX) which results in the following new constraint: 3.74/1.87 3.74/1.87 (5) (x[0] + [-1] + [-1]y[0] >= 0 & x[0] + [-2] + [-1]y[0] >= 0 ==> (U^Increasing(F(>(x[0]1, y[0]1), +(x[0]1, 1), +(y[0]1, 2))), >=) & [bni_8 + (-1)Bound*bni_8] + [(-1)bni_8]y[0] + [bni_8]x[0] >= 0 & [1 + (-1)bso_9] >= 0) 3.74/1.87 3.74/1.87 3.74/1.87 3.74/1.87 We simplified constraint (5) using rule (IDP_SMT_SPLIT) which results in the following new constraint: 3.74/1.87 3.74/1.87 (6) (x[0] >= 0 & [-1] + x[0] >= 0 ==> (U^Increasing(F(>(x[0]1, y[0]1), +(x[0]1, 1), +(y[0]1, 2))), >=) & [(2)bni_8 + (-1)Bound*bni_8] + [bni_8]x[0] >= 0 & [1 + (-1)bso_9] >= 0) 3.74/1.87 3.74/1.87 3.74/1.87 3.74/1.87 We simplified constraint (6) using rule (IDP_SMT_SPLIT) which results in the following new constraints: 3.74/1.87 3.74/1.87 (7) (x[0] >= 0 & [-1] + x[0] >= 0 & y[0] >= 0 ==> (U^Increasing(F(>(x[0]1, y[0]1), +(x[0]1, 1), +(y[0]1, 2))), >=) & [(2)bni_8 + (-1)Bound*bni_8] + [bni_8]x[0] >= 0 & [1 + (-1)bso_9] >= 0) 3.74/1.87 3.74/1.87 (8) (x[0] >= 0 & [-1] + x[0] >= 0 & y[0] >= 0 ==> (U^Increasing(F(>(x[0]1, y[0]1), +(x[0]1, 1), +(y[0]1, 2))), >=) & [(2)bni_8 + (-1)Bound*bni_8] + [bni_8]x[0] >= 0 & [1 + (-1)bso_9] >= 0) 3.74/1.87 3.74/1.87 3.74/1.87 3.74/1.87 3.74/1.87 3.74/1.87 3.74/1.87 3.74/1.87 3.74/1.87 To summarize, we get the following constraints P__>=_ for the following pairs. 3.74/1.87 3.74/1.87 *F(TRUE, x, y) -> F(>(x, y), +(x, 1), +(y, 2)) 3.74/1.87 3.74/1.87 *(x[0] >= 0 & [-1] + x[0] >= 0 & y[0] >= 0 ==> (U^Increasing(F(>(x[0]1, y[0]1), +(x[0]1, 1), +(y[0]1, 2))), >=) & [(2)bni_8 + (-1)Bound*bni_8] + [bni_8]x[0] >= 0 & [1 + (-1)bso_9] >= 0) 3.74/1.87 3.74/1.87 3.74/1.87 *(x[0] >= 0 & [-1] + x[0] >= 0 & y[0] >= 0 ==> (U^Increasing(F(>(x[0]1, y[0]1), +(x[0]1, 1), +(y[0]1, 2))), >=) & [(2)bni_8 + (-1)Bound*bni_8] + [bni_8]x[0] >= 0 & [1 + (-1)bso_9] >= 0) 3.74/1.87 3.74/1.87 3.74/1.87 3.74/1.87 3.74/1.87 3.74/1.87 3.74/1.87 3.74/1.87 3.74/1.87 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. 3.74/1.87 3.74/1.87 Using the following integer polynomial ordering the resulting constraints can be solved 3.74/1.87 3.74/1.87 Polynomial interpretation over integers[POLO]: 3.74/1.87 3.74/1.87 POL(TRUE) = 0 3.74/1.87 POL(FALSE) = 0 3.74/1.87 POL(F(x_1, x_2, x_3)) = [2] + [-1]x_3 + x_2 3.74/1.87 POL(>(x_1, x_2)) = [-1] 3.74/1.87 POL(+(x_1, x_2)) = x_1 + x_2 3.74/1.87 POL(1) = [1] 3.74/1.87 POL(2) = [2] 3.74/1.87 3.74/1.87 3.74/1.87 The following pairs are in P_>: 3.74/1.87 3.74/1.87 3.74/1.87 F(TRUE, x[0], y[0]) -> F(>(x[0], y[0]), +(x[0], 1), +(y[0], 2)) 3.74/1.87 3.74/1.87 3.74/1.87 The following pairs are in P_bound: 3.74/1.87 3.74/1.87 3.74/1.87 F(TRUE, x[0], y[0]) -> F(>(x[0], y[0]), +(x[0], 1), +(y[0], 2)) 3.74/1.87 3.74/1.87 3.74/1.87 The following pairs are in P_>=: 3.74/1.87 3.74/1.87 none 3.74/1.87 3.74/1.87 3.74/1.87 There are no usable rules. 3.74/1.87 ---------------------------------------- 3.74/1.87 3.74/1.87 (6) 3.74/1.87 Obligation: 3.74/1.87 IDP problem: 3.74/1.87 The following function symbols are pre-defined: 3.74/1.87 <<< 3.74/1.87 & ~ Bwand: (Integer, Integer) -> Integer 3.74/1.87 >= ~ Ge: (Integer, Integer) -> Boolean 3.74/1.87 | ~ Bwor: (Integer, Integer) -> Integer 3.74/1.87 / ~ Div: (Integer, Integer) -> Integer 3.74/1.87 != ~ Neq: (Integer, Integer) -> Boolean 3.74/1.87 && ~ Land: (Boolean, Boolean) -> Boolean 3.74/1.87 ! ~ Lnot: (Boolean) -> Boolean 3.74/1.87 = ~ Eq: (Integer, Integer) -> Boolean 3.74/1.87 <= ~ Le: (Integer, Integer) -> Boolean 3.74/1.87 ^ ~ Bwxor: (Integer, Integer) -> Integer 3.74/1.87 % ~ Mod: (Integer, Integer) -> Integer 3.74/1.87 > ~ Gt: (Integer, Integer) -> Boolean 3.74/1.87 + ~ Add: (Integer, Integer) -> Integer 3.74/1.87 -1 ~ UnaryMinus: (Integer) -> Integer 3.74/1.87 < ~ Lt: (Integer, Integer) -> Boolean 3.74/1.87 || ~ Lor: (Boolean, Boolean) -> Boolean 3.74/1.87 - ~ Sub: (Integer, Integer) -> Integer 3.74/1.87 ~ ~ Bwnot: (Integer) -> Integer 3.74/1.87 * ~ Mul: (Integer, Integer) -> Integer 3.74/1.87 >>> 3.74/1.87 3.74/1.87 3.74/1.87 The following domains are used: 3.74/1.87 none 3.74/1.87 3.74/1.87 R is empty. 3.74/1.87 3.74/1.87 The integer pair graph is empty. 3.74/1.87 3.74/1.87 The set Q consists of the following terms: 3.74/1.87 f(TRUE, x0, x1) 3.74/1.87 3.74/1.87 ---------------------------------------- 3.74/1.87 3.74/1.87 (7) PisEmptyProof (EQUIVALENT) 3.74/1.87 The TRS P is empty. Hence, there is no (P,Q,R) chain. 3.74/1.87 ---------------------------------------- 3.74/1.87 3.74/1.87 (8) 3.74/1.87 YES 3.74/1.90 EOF