/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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty Termination of the given ITRS could be proven: (0) ITRS (1) ITRStoIDPProof [EQUIVALENT, 0 ms] (2) IDP (3) UsableRulesProof [EQUIVALENT, 0 ms] (4) IDP (5)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, 163 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 > ~ Gt: (Integer, Integer) -> Boolean + ~ Add: (Integer, Integer) -> Integer -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, y) -> f(x > y && y > 2, 1 + x, 2 * y) The set Q consists of the following terms: f(TRUE, x0, x1) ---------------------------------------- (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 > ~ Gt: (Integer, Integer) -> Boolean + ~ Add: (Integer, Integer) -> Integer -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: Boolean, Integer The ITRS R consists of the following rules: f(TRUE, x, y) -> f(x > y && y > 2, 1 + x, 2 * y) The integer pair graph contains the following rules and edges: (0): F(TRUE, x[0], y[0]) -> F(x[0] > y[0] && y[0] > 2, 1 + x[0], 2 * y[0]) (0) -> (0), if (x[0] > y[0] && y[0] > 2 & 1 + x[0] ->^* x[0]' & 2 * y[0] ->^* y[0]') The set Q consists of the following terms: f(TRUE, x0, x1) ---------------------------------------- (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 > ~ Gt: (Integer, Integer) -> Boolean + ~ Add: (Integer, Integer) -> Integer -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: Boolean, Integer R is empty. The integer pair graph contains the following rules and edges: (0): F(TRUE, x[0], y[0]) -> F(x[0] > y[0] && y[0] > 2, 1 + x[0], 2 * y[0]) (0) -> (0), if (x[0] > y[0] && y[0] > 2 & 1 + x[0] ->^* x[0]' & 2 * y[0] ->^* y[0]') The set Q consists of the following terms: f(TRUE, x0, x1) ---------------------------------------- (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@62e12f08 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, y) -> F(&&(>(x, y), >(y, 2)), +(1, x), *(2, y)) the following chains were created: *We consider the chain F(TRUE, x[0], y[0]) -> F(&&(>(x[0], y[0]), >(y[0], 2)), +(1, x[0]), *(2, y[0])), F(TRUE, x[0], y[0]) -> F(&&(>(x[0], y[0]), >(y[0], 2)), +(1, x[0]), *(2, y[0])), F(TRUE, x[0], y[0]) -> F(&&(>(x[0], y[0]), >(y[0], 2)), +(1, x[0]), *(2, y[0])) which results in the following constraint: (1) (&&(>(x[0], y[0]), >(y[0], 2))=TRUE & +(1, x[0])=x[0]1 & *(2, y[0])=y[0]1 & &&(>(x[0]1, y[0]1), >(y[0]1, 2))=TRUE & +(1, x[0]1)=x[0]2 & *(2, y[0]1)=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), >(y[0]1, 2)), +(1, x[0]1), *(2, y[0]1)) & (U^Increasing(F(&&(>(x[0]1, y[0]1), >(y[0]1, 2)), +(1, x[0]1), *(2, y[0]1))), >=)) We simplified constraint (1) using rules (III), (IV), (IDP_BOOLEAN) which results in the following new constraint: (2) (>(x[0], y[0])=TRUE & >(y[0], 2)=TRUE & >(+(1, x[0]), *(2, y[0]))=TRUE & >(*(2, y[0]), 2)=TRUE ==> F(TRUE, +(1, x[0]), *(2, y[0]))_>=_NonInfC & F(TRUE, +(1, x[0]), *(2, y[0]))_>=_F(&&(>(+(1, x[0]), *(2, y[0])), >(*(2, y[0]), 2)), +(1, +(1, x[0])), *(2, *(2, y[0]))) & (U^Increasing(F(&&(>(x[0]1, y[0]1), >(y[0]1, 2)), +(1, x[0]1), *(2, y[0]1))), >=)) We simplified constraint (2) using rule (POLY_CONSTRAINTS) which results in the following new constraint: (3) (x[0] + [-1] + [-1]y[0] >= 0 & y[0] + [-3] >= 0 & x[0] + [-2]y[0] >= 0 & [2]y[0] + [-3] >= 0 ==> (U^Increasing(F(&&(>(x[0]1, y[0]1), >(y[0]1, 2)), +(1, x[0]1), *(2, y[0]1))), >=) & [(3)bni_9 + (-1)Bound*bni_9] + [(-2)bni_9]y[0] + [bni_9]x[0] >= 0 & [-2 + (-1)bso_10] + [2]y[0] >= 0) We simplified constraint (3) using rule (IDP_POLY_SIMPLIFY) which results in the following new constraint: (4) (x[0] + [-1] + [-1]y[0] >= 0 & y[0] + [-3] >= 0 & x[0] + [-2]y[0] >= 0 & [2]y[0] + [-3] >= 0 ==> (U^Increasing(F(&&(>(x[0]1, y[0]1), >(y[0]1, 2)), +(1, x[0]1), *(2, y[0]1))), >=) & [(3)bni_9 + (-1)Bound*bni_9] + [(-2)bni_9]y[0] + [bni_9]x[0] >= 0 & [-2 + (-1)bso_10] + [2]y[0] >= 0) We simplified constraint (4) using rule (POLY_REMOVE_MIN_MAX) which results in the following new constraint: (5) (x[0] + [-1] + [-1]y[0] >= 0 & y[0] + [-3] >= 0 & x[0] + [-2]y[0] >= 0 & [2]y[0] + [-3] >= 0 ==> (U^Increasing(F(&&(>(x[0]1, y[0]1), >(y[0]1, 2)), +(1, x[0]1), *(2, y[0]1))), >=) & [(3)bni_9 + (-1)Bound*bni_9] + [(-2)bni_9]y[0] + [bni_9]x[0] >= 0 & [-2 + (-1)bso_10] + [2]y[0] >= 0) To summarize, we get the following constraints P__>=_ for the following pairs. *F(TRUE, x, y) -> F(&&(>(x, y), >(y, 2)), +(1, x), *(2, y)) *(x[0] + [-1] + [-1]y[0] >= 0 & y[0] + [-3] >= 0 & x[0] + [-2]y[0] >= 0 & [2]y[0] + [-3] >= 0 ==> (U^Increasing(F(&&(>(x[0]1, y[0]1), >(y[0]1, 2)), +(1, x[0]1), *(2, y[0]1))), >=) & [(3)bni_9 + (-1)Bound*bni_9] + [(-2)bni_9]y[0] + [bni_9]x[0] >= 0 & [-2 + (-1)bso_10] + [2]y[0] >= 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) = [3] POL(F(x_1, x_2, x_3)) = [2] + [-1]x_3 + x_2 + [-1]x_1 POL(&&(x_1, x_2)) = [-1] POL(>(x_1, x_2)) = [-1] POL(2) = [2] POL(+(x_1, x_2)) = x_1 + x_2 POL(1) = [1] POL(*(x_1, x_2)) = x_1*x_2 The following pairs are in P_>: F(TRUE, x[0], y[0]) -> F(&&(>(x[0], y[0]), >(y[0], 2)), +(1, x[0]), *(2, y[0])) The following pairs are in P_bound: F(TRUE, x[0], y[0]) -> F(&&(>(x[0], y[0]), >(y[0], 2)), +(1, x[0]), *(2, y[0])) The following pairs are in P_>=: none At least the following rules have been oriented under context sensitive arithmetic replacement: TRUE^1 -> &&(TRUE, TRUE)^1 FALSE^1 -> &&(TRUE, FALSE)^1 FALSE^1 -> &&(FALSE, TRUE)^1 FALSE^1 -> &&(FALSE, FALSE)^1 ---------------------------------------- (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 > ~ Gt: (Integer, Integer) -> Boolean + ~ Add: (Integer, Integer) -> Integer -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, x1) ---------------------------------------- (7) PisEmptyProof (EQUIVALENT) The TRS P is empty. Hence, there is no (P,Q,R) chain. ---------------------------------------- (8) YES