/export/starexec/sandbox/solver/bin/starexec_run_standard /export/starexec/sandbox/benchmark/theBenchmark.itrs /export/starexec/sandbox/output/output_files -------------------------------------------------------------------------------- YES proof of /export/starexec/sandbox/benchmark/theBenchmark.itrs # AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty Termination of the given ITRS could be proven: (0) ITRS (1) ITRStoIDPProof [EQUIVALENT, 0 ms] (2) IDP (3) UsableRulesProof [EQUIVALENT, 0 ms] (4) IDP (5) IDPNonInfProof [SOUND, 180 ms] (6) IDP (7) PisEmptyProof [EQUIVALENT, 0 ms] (8) YES ---------------------------------------- (0) Obligation: ITRS problem: The following function symbols are pre-defined: <<< & ~ Bwand: (Integer, Integer) -> Integer >= ~ Ge: (Integer, Integer) -> Boolean | ~ Bwor: (Integer, Integer) -> Integer / ~ Div: (Integer, Integer) -> Integer != ~ Neq: (Integer, Integer) -> Boolean && ~ Land: (Boolean, Boolean) -> Boolean ! ~ Lnot: (Boolean) -> Boolean = ~ Eq: (Integer, Integer) -> Boolean <= ~ Le: (Integer, Integer) -> Boolean ^ ~ Bwxor: (Integer, Integer) -> Integer % ~ Mod: (Integer, Integer) -> Integer + ~ Add: (Integer, Integer) -> Integer > ~ Gt: (Integer, Integer) -> Boolean -1 ~ UnaryMinus: (Integer) -> Integer < ~ Lt: (Integer, Integer) -> Boolean || ~ Lor: (Boolean, Boolean) -> Boolean - ~ Sub: (Integer, Integer) -> Integer ~ ~ Bwnot: (Integer) -> Integer * ~ Mul: (Integer, Integer) -> Integer >>> The TRS R consists of the following rules: cu(TRUE, x) -> cu(x < 100000, x + 1) The set Q consists of the following terms: cu(TRUE, x0) ---------------------------------------- (1) ITRStoIDPProof (EQUIVALENT) Added dependency pairs ---------------------------------------- (2) Obligation: IDP problem: The following function symbols are pre-defined: <<< & ~ Bwand: (Integer, Integer) -> Integer >= ~ Ge: (Integer, Integer) -> Boolean | ~ Bwor: (Integer, Integer) -> Integer / ~ Div: (Integer, Integer) -> Integer != ~ Neq: (Integer, Integer) -> Boolean && ~ Land: (Boolean, Boolean) -> Boolean ! ~ Lnot: (Boolean) -> Boolean = ~ Eq: (Integer, Integer) -> Boolean <= ~ Le: (Integer, Integer) -> Boolean ^ ~ Bwxor: (Integer, Integer) -> Integer % ~ Mod: (Integer, Integer) -> Integer + ~ Add: (Integer, Integer) -> Integer > ~ Gt: (Integer, Integer) -> Boolean -1 ~ UnaryMinus: (Integer) -> Integer < ~ Lt: (Integer, Integer) -> Boolean || ~ Lor: (Boolean, Boolean) -> Boolean - ~ Sub: (Integer, Integer) -> Integer ~ ~ Bwnot: (Integer) -> Integer * ~ Mul: (Integer, Integer) -> Integer >>> The following domains are used: Integer The ITRS R consists of the following rules: cu(TRUE, x) -> cu(x < 100000, x + 1) The integer pair graph contains the following rules and edges: (0): CU(TRUE, x[0]) -> CU(x[0] < 100000, x[0] + 1) (0) -> (0), if (x[0] < 100000 & x[0] + 1 ->^* x[0]') The set Q consists of the following terms: cu(TRUE, x0) ---------------------------------------- (3) UsableRulesProof (EQUIVALENT) 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) Obligation: IDP problem: The following function symbols are pre-defined: <<< & ~ Bwand: (Integer, Integer) -> Integer >= ~ Ge: (Integer, Integer) -> Boolean | ~ Bwor: (Integer, Integer) -> Integer / ~ Div: (Integer, Integer) -> Integer != ~ Neq: (Integer, Integer) -> Boolean && ~ Land: (Boolean, Boolean) -> Boolean ! ~ Lnot: (Boolean) -> Boolean = ~ Eq: (Integer, Integer) -> Boolean <= ~ Le: (Integer, Integer) -> Boolean ^ ~ Bwxor: (Integer, Integer) -> Integer % ~ Mod: (Integer, Integer) -> Integer + ~ Add: (Integer, Integer) -> Integer > ~ Gt: (Integer, Integer) -> Boolean -1 ~ UnaryMinus: (Integer) -> Integer < ~ Lt: (Integer, Integer) -> Boolean || ~ Lor: (Boolean, Boolean) -> Boolean - ~ Sub: (Integer, Integer) -> Integer ~ ~ Bwnot: (Integer) -> Integer * ~ Mul: (Integer, Integer) -> Integer >>> The following domains are used: Integer R is empty. The integer pair graph contains the following rules and edges: (0): CU(TRUE, x[0]) -> CU(x[0] < 100000, x[0] + 1) (0) -> (0), if (x[0] < 100000 & x[0] + 1 ->^* x[0]') The set Q consists of the following terms: cu(TRUE, x0) ---------------------------------------- (5) IDPNonInfProof (SOUND) Used the following options for this NonInfProof: IDPGPoloSolver: Range: [(-1,2)] IsNat: false Interpretation Shape Heuristic: aprove.DPFramework.IDPProblem.Processors.nonInf.poly.IdpDefaultShapeHeuristic@6012b267 Constraint Generator: NonInfConstraintGenerator: PathGenerator: MetricPathGenerator: Max Left Steps: 1 Max Right Steps: 1 The constraints were generated the following way: The DP Problem is simplified using the Induction Calculus [NONINF] with the following steps: Note that final constraints are written in bold face. For Pair CU(TRUE, x) -> CU(<(x, 100000), +(x, 1)) the following chains were created: *We consider the chain CU(TRUE, x[0]) -> CU(<(x[0], 100000), +(x[0], 1)), CU(TRUE, x[0]) -> CU(<(x[0], 100000), +(x[0], 1)), CU(TRUE, x[0]) -> CU(<(x[0], 100000), +(x[0], 1)) which results in the following constraint: (1) (<(x[0], 100000)=TRUE & +(x[0], 1)=x[0]1 & <(x[0]1, 100000)=TRUE & +(x[0]1, 1)=x[0]2 ==> CU(TRUE, x[0]1)_>=_NonInfC & CU(TRUE, x[0]1)_>=_CU(<(x[0]1, 100000), +(x[0]1, 1)) & (U^Increasing(CU(<(x[0]1, 100000), +(x[0]1, 1))), >=)) We simplified constraint (1) using rules (III), (IV) which results in the following new constraint: (2) (<(x[0], 100000)=TRUE & <(+(x[0], 1), 100000)=TRUE ==> CU(TRUE, +(x[0], 1))_>=_NonInfC & CU(TRUE, +(x[0], 1))_>=_CU(<(+(x[0], 1), 100000), +(+(x[0], 1), 1)) & (U^Increasing(CU(<(x[0]1, 100000), +(x[0]1, 1))), >=)) We simplified constraint (2) using rule (POLY_CONSTRAINTS) which results in the following new constraint: (3) ([99999] + [-1]x[0] >= 0 & [99998] + [-1]x[0] >= 0 ==> (U^Increasing(CU(<(x[0]1, 100000), +(x[0]1, 1))), >=) & [bni_7 + (-1)Bound*bni_7] + [(-1)bni_7]x[0] >= 0 & [1 + (-1)bso_8] >= 0) We simplified constraint (3) using rule (IDP_POLY_SIMPLIFY) which results in the following new constraint: (4) ([99999] + [-1]x[0] >= 0 & [99998] + [-1]x[0] >= 0 ==> (U^Increasing(CU(<(x[0]1, 100000), +(x[0]1, 1))), >=) & [bni_7 + (-1)Bound*bni_7] + [(-1)bni_7]x[0] >= 0 & [1 + (-1)bso_8] >= 0) We simplified constraint (4) using rule (POLY_REMOVE_MIN_MAX) which results in the following new constraint: (5) ([99999] + [-1]x[0] >= 0 & [99998] + [-1]x[0] >= 0 ==> (U^Increasing(CU(<(x[0]1, 100000), +(x[0]1, 1))), >=) & [bni_7 + (-1)Bound*bni_7] + [(-1)bni_7]x[0] >= 0 & [1 + (-1)bso_8] >= 0) We simplified constraint (5) using rule (IDP_SMT_SPLIT) which results in the following new constraints: (6) ([99999] + [-1]x[0] >= 0 & [99998] + [-1]x[0] >= 0 & x[0] >= 0 ==> (U^Increasing(CU(<(x[0]1, 100000), +(x[0]1, 1))), >=) & [bni_7 + (-1)Bound*bni_7] + [(-1)bni_7]x[0] >= 0 & [1 + (-1)bso_8] >= 0) (7) ([99999] + x[0] >= 0 & [99998] + x[0] >= 0 & x[0] >= 0 ==> (U^Increasing(CU(<(x[0]1, 100000), +(x[0]1, 1))), >=) & [bni_7 + (-1)Bound*bni_7] + [bni_7]x[0] >= 0 & [1 + (-1)bso_8] >= 0) To summarize, we get the following constraints P__>=_ for the following pairs. *CU(TRUE, x) -> CU(<(x, 100000), +(x, 1)) *([99999] + [-1]x[0] >= 0 & [99998] + [-1]x[0] >= 0 & x[0] >= 0 ==> (U^Increasing(CU(<(x[0]1, 100000), +(x[0]1, 1))), >=) & [bni_7 + (-1)Bound*bni_7] + [(-1)bni_7]x[0] >= 0 & [1 + (-1)bso_8] >= 0) *([99999] + x[0] >= 0 & [99998] + x[0] >= 0 & x[0] >= 0 ==> (U^Increasing(CU(<(x[0]1, 100000), +(x[0]1, 1))), >=) & [bni_7 + (-1)Bound*bni_7] + [bni_7]x[0] >= 0 & [1 + (-1)bso_8] >= 0) 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. Using the following integer polynomial ordering the resulting constraints can be solved Polynomial interpretation over integers[POLO]: POL(TRUE) = 0 POL(FALSE) = 0 POL(CU(x_1, x_2)) = [2] + [-1]x_2 POL(<(x_1, x_2)) = [-1] POL(100000) = [100000] POL(+(x_1, x_2)) = x_1 + x_2 POL(1) = [1] The following pairs are in P_>: CU(TRUE, x[0]) -> CU(<(x[0], 100000), +(x[0], 1)) The following pairs are in P_bound: CU(TRUE, x[0]) -> CU(<(x[0], 100000), +(x[0], 1)) The following pairs are in P_>=: none There are no usable rules. ---------------------------------------- (6) Obligation: IDP problem: The following function symbols are pre-defined: <<< & ~ Bwand: (Integer, Integer) -> Integer >= ~ Ge: (Integer, Integer) -> Boolean | ~ Bwor: (Integer, Integer) -> Integer / ~ Div: (Integer, Integer) -> Integer != ~ Neq: (Integer, Integer) -> Boolean && ~ Land: (Boolean, Boolean) -> Boolean ! ~ Lnot: (Boolean) -> Boolean = ~ Eq: (Integer, Integer) -> Boolean <= ~ Le: (Integer, Integer) -> Boolean ^ ~ Bwxor: (Integer, Integer) -> Integer % ~ Mod: (Integer, Integer) -> Integer + ~ Add: (Integer, Integer) -> Integer > ~ Gt: (Integer, Integer) -> Boolean -1 ~ UnaryMinus: (Integer) -> Integer < ~ Lt: (Integer, Integer) -> Boolean || ~ Lor: (Boolean, Boolean) -> Boolean - ~ Sub: (Integer, Integer) -> Integer ~ ~ Bwnot: (Integer) -> Integer * ~ Mul: (Integer, Integer) -> Integer >>> The following domains are used: none R is empty. The integer pair graph is empty. The set Q consists of the following terms: cu(TRUE, x0) ---------------------------------------- (7) PisEmptyProof (EQUIVALENT) The TRS P is empty. Hence, there is no (P,Q,R) chain. ---------------------------------------- (8) YES