/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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty Termination of the given ITRS could be proven: (0) ITRS (1) ITRStoIDPProof [EQUIVALENT, 0 ms] (2) IDP (3) UsableRulesProof [EQUIVALENT, 0 ms] (4) IDP (5)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, 130 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: f(TRUE, x) -> f(1000 >= x, x + 1) The set Q consists of the following terms: f(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: f(TRUE, x) -> f(1000 >= x, x + 1) The integer pair graph contains the following rules and edges: (0): F(TRUE, x[0]) -> F(1000 >= x[0], x[0] + 1) (0) -> (0), if (1000 >= x[0] & x[0] + 1 ->^* x[0]') The set Q consists of the following terms: f(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): F(TRUE, x[0]) -> F(1000 >= x[0], x[0] + 1) (0) -> (0), if (1000 >= x[0] & x[0] + 1 ->^* x[0]') The set Q consists of the following terms: f(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@1b53ff85 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 F(TRUE, x) -> F(>=(1000, x), +(x, 1)) the following chains were created: *We consider the chain F(TRUE, x[0]) -> F(>=(1000, x[0]), +(x[0], 1)), F(TRUE, x[0]) -> F(>=(1000, x[0]), +(x[0], 1)), F(TRUE, x[0]) -> F(>=(1000, x[0]), +(x[0], 1)) which results in the following constraint: (1) (>=(1000, x[0])=TRUE & +(x[0], 1)=x[0]1 & >=(1000, x[0]1)=TRUE & +(x[0]1, 1)=x[0]2 ==> F(TRUE, x[0]1)_>=_NonInfC & F(TRUE, x[0]1)_>=_F(>=(1000, x[0]1), +(x[0]1, 1)) & (U^Increasing(F(>=(1000, x[0]1), +(x[0]1, 1))), >=)) We simplified constraint (1) using rules (III), (IV) which results in the following new constraint: (2) (>=(1000, x[0])=TRUE & >=(1000, +(x[0], 1))=TRUE ==> F(TRUE, +(x[0], 1))_>=_NonInfC & F(TRUE, +(x[0], 1))_>=_F(>=(1000, +(x[0], 1)), +(+(x[0], 1), 1)) & (U^Increasing(F(>=(1000, x[0]1), +(x[0]1, 1))), >=)) We simplified constraint (2) using rule (POLY_CONSTRAINTS) which results in the following new constraint: (3) ([1000] + [-1]x[0] >= 0 & [999] + [-1]x[0] >= 0 ==> (U^Increasing(F(>=(1000, x[0]1), +(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) ([1000] + [-1]x[0] >= 0 & [999] + [-1]x[0] >= 0 ==> (U^Increasing(F(>=(1000, x[0]1), +(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) ([1000] + [-1]x[0] >= 0 & [999] + [-1]x[0] >= 0 ==> (U^Increasing(F(>=(1000, x[0]1), +(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) ([1000] + x[0] >= 0 & [999] + x[0] >= 0 & x[0] >= 0 ==> (U^Increasing(F(>=(1000, x[0]1), +(x[0]1, 1))), >=) & [bni_7 + (-1)Bound*bni_7] + [bni_7]x[0] >= 0 & [1 + (-1)bso_8] >= 0) (7) ([1000] + [-1]x[0] >= 0 & [999] + [-1]x[0] >= 0 & x[0] >= 0 ==> (U^Increasing(F(>=(1000, x[0]1), +(x[0]1, 1))), >=) & [bni_7 + (-1)Bound*bni_7] + [(-1)bni_7]x[0] >= 0 & [1 + (-1)bso_8] >= 0) To summarize, we get the following constraints P__>=_ for the following pairs. *F(TRUE, x) -> F(>=(1000, x), +(x, 1)) *([1000] + x[0] >= 0 & [999] + x[0] >= 0 & x[0] >= 0 ==> (U^Increasing(F(>=(1000, x[0]1), +(x[0]1, 1))), >=) & [bni_7 + (-1)Bound*bni_7] + [bni_7]x[0] >= 0 & [1 + (-1)bso_8] >= 0) *([1000] + [-1]x[0] >= 0 & [999] + [-1]x[0] >= 0 & x[0] >= 0 ==> (U^Increasing(F(>=(1000, x[0]1), +(x[0]1, 1))), >=) & [bni_7 + (-1)Bound*bni_7] + [(-1)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(F(x_1, x_2)) = [2] + [-1]x_2 POL(>=(x_1, x_2)) = [-1] POL(1000) = [1000] POL(+(x_1, x_2)) = x_1 + x_2 POL(1) = [1] The following pairs are in P_>: F(TRUE, x[0]) -> F(>=(1000, x[0]), +(x[0], 1)) The following pairs are in P_bound: F(TRUE, x[0]) -> F(>=(1000, x[0]), +(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: f(TRUE, x0) ---------------------------------------- (7) PisEmptyProof (EQUIVALENT) The TRS P is empty. Hence, there is no (P,Q,R) chain. ---------------------------------------- (8) YES