/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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty Termination of the given ITRS could be proven: (0) ITRS (1) ITRStoIDPProof [EQUIVALENT, 0 ms] (2) IDP (3) UsableRulesProof [EQUIVALENT, 0 ms] (4) IDP (5) IDependencyGraphProof [EQUIVALENT, 0 ms] (6) IDP (7) IDPNonInfProof [SOUND, 158 ms] (8) IDP (9) PisEmptyProof [EQUIVALENT, 0 ms] (10) 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: div(x, y) -> Cond_div(0 >= y, x, y) Cond_div(TRUE, x, y) -> 0 div(x, y) -> Cond_div1(y >= x, x, y) Cond_div1(TRUE, x, y) -> 0 div(x, y) -> Cond_div2(x > y && y > 0, x, y) Cond_div2(TRUE, x, y) -> div(x - y, y) + 1 The set Q consists of the following terms: div(x0, x1) Cond_div(TRUE, x0, x1) Cond_div1(TRUE, x0, x1) Cond_div2(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 + ~ 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, Boolean The ITRS R consists of the following rules: div(x, y) -> Cond_div(0 >= y, x, y) Cond_div(TRUE, x, y) -> 0 div(x, y) -> Cond_div1(y >= x, x, y) Cond_div1(TRUE, x, y) -> 0 div(x, y) -> Cond_div2(x > y && y > 0, x, y) Cond_div2(TRUE, x, y) -> div(x - y, y) + 1 The integer pair graph contains the following rules and edges: (0): DIV(x[0], y[0]) -> COND_DIV(0 >= y[0], x[0], y[0]) (1): DIV(x[1], y[1]) -> COND_DIV1(y[1] >= x[1], x[1], y[1]) (2): DIV(x[2], y[2]) -> COND_DIV2(x[2] > y[2] && y[2] > 0, x[2], y[2]) (3): COND_DIV2(TRUE, x[3], y[3]) -> DIV(x[3] - y[3], y[3]) (2) -> (3), if (x[2] > y[2] && y[2] > 0 & x[2] ->^* x[3] & y[2] ->^* y[3]) (3) -> (0), if (x[3] - y[3] ->^* x[0] & y[3] ->^* y[0]) (3) -> (1), if (x[3] - y[3] ->^* x[1] & y[3] ->^* y[1]) (3) -> (2), if (x[3] - y[3] ->^* x[2] & y[3] ->^* y[2]) The set Q consists of the following terms: div(x0, x1) Cond_div(TRUE, x0, x1) Cond_div1(TRUE, x0, x1) Cond_div2(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 + ~ 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, Boolean R is empty. The integer pair graph contains the following rules and edges: (0): DIV(x[0], y[0]) -> COND_DIV(0 >= y[0], x[0], y[0]) (1): DIV(x[1], y[1]) -> COND_DIV1(y[1] >= x[1], x[1], y[1]) (2): DIV(x[2], y[2]) -> COND_DIV2(x[2] > y[2] && y[2] > 0, x[2], y[2]) (3): COND_DIV2(TRUE, x[3], y[3]) -> DIV(x[3] - y[3], y[3]) (2) -> (3), if (x[2] > y[2] && y[2] > 0 & x[2] ->^* x[3] & y[2] ->^* y[3]) (3) -> (0), if (x[3] - y[3] ->^* x[0] & y[3] ->^* y[0]) (3) -> (1), if (x[3] - y[3] ->^* x[1] & y[3] ->^* y[1]) (3) -> (2), if (x[3] - y[3] ->^* x[2] & y[3] ->^* y[2]) The set Q consists of the following terms: div(x0, x1) Cond_div(TRUE, x0, x1) Cond_div1(TRUE, x0, x1) Cond_div2(TRUE, x0, x1) ---------------------------------------- (5) IDependencyGraphProof (EQUIVALENT) The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 2 less nodes. ---------------------------------------- (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: Integer, Boolean R is empty. The integer pair graph contains the following rules and edges: (3): COND_DIV2(TRUE, x[3], y[3]) -> DIV(x[3] - y[3], y[3]) (2): DIV(x[2], y[2]) -> COND_DIV2(x[2] > y[2] && y[2] > 0, x[2], y[2]) (3) -> (2), if (x[3] - y[3] ->^* x[2] & y[3] ->^* y[2]) (2) -> (3), if (x[2] > y[2] && y[2] > 0 & x[2] ->^* x[3] & y[2] ->^* y[3]) The set Q consists of the following terms: div(x0, x1) Cond_div(TRUE, x0, x1) Cond_div1(TRUE, x0, x1) Cond_div2(TRUE, x0, x1) ---------------------------------------- (7) 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@1294e867 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 COND_DIV2(TRUE, x[3], y[3]) -> DIV(-(x[3], y[3]), y[3]) the following chains were created: *We consider the chain DIV(x[2], y[2]) -> COND_DIV2(&&(>(x[2], y[2]), >(y[2], 0)), x[2], y[2]), COND_DIV2(TRUE, x[3], y[3]) -> DIV(-(x[3], y[3]), y[3]), DIV(x[2], y[2]) -> COND_DIV2(&&(>(x[2], y[2]), >(y[2], 0)), x[2], y[2]) which results in the following constraint: (1) (&&(>(x[2], y[2]), >(y[2], 0))=TRUE & x[2]=x[3] & y[2]=y[3] & -(x[3], y[3])=x[2]1 & y[3]=y[2]1 ==> COND_DIV2(TRUE, x[3], y[3])_>=_NonInfC & COND_DIV2(TRUE, x[3], y[3])_>=_DIV(-(x[3], y[3]), y[3]) & (U^Increasing(DIV(-(x[3], y[3]), y[3])), >=)) We simplified constraint (1) using rules (III), (IV), (IDP_BOOLEAN) which results in the following new constraint: (2) (>(x[2], y[2])=TRUE & >(y[2], 0)=TRUE ==> COND_DIV2(TRUE, x[2], y[2])_>=_NonInfC & COND_DIV2(TRUE, x[2], y[2])_>=_DIV(-(x[2], y[2]), y[2]) & (U^Increasing(DIV(-(x[3], y[3]), y[3])), >=)) We simplified constraint (2) using rule (POLY_CONSTRAINTS) which results in the following new constraint: (3) (x[2] + [-1] + [-1]y[2] >= 0 & y[2] + [-1] >= 0 ==> (U^Increasing(DIV(-(x[3], y[3]), y[3])), >=) & [(-1)bni_12 + (-1)Bound*bni_12] + [(-1)bni_12]y[2] + [(2)bni_12]x[2] >= 0 & [(-1)bso_13] + y[2] >= 0) We simplified constraint (3) using rule (IDP_POLY_SIMPLIFY) which results in the following new constraint: (4) (x[2] + [-1] + [-1]y[2] >= 0 & y[2] + [-1] >= 0 ==> (U^Increasing(DIV(-(x[3], y[3]), y[3])), >=) & [(-1)bni_12 + (-1)Bound*bni_12] + [(-1)bni_12]y[2] + [(2)bni_12]x[2] >= 0 & [(-1)bso_13] + y[2] >= 0) We simplified constraint (4) using rule (POLY_REMOVE_MIN_MAX) which results in the following new constraint: (5) (x[2] + [-1] + [-1]y[2] >= 0 & y[2] + [-1] >= 0 ==> (U^Increasing(DIV(-(x[3], y[3]), y[3])), >=) & [(-1)bni_12 + (-1)Bound*bni_12] + [(-1)bni_12]y[2] + [(2)bni_12]x[2] >= 0 & [(-1)bso_13] + y[2] >= 0) For Pair DIV(x[2], y[2]) -> COND_DIV2(&&(>(x[2], y[2]), >(y[2], 0)), x[2], y[2]) the following chains were created: *We consider the chain DIV(x[2], y[2]) -> COND_DIV2(&&(>(x[2], y[2]), >(y[2], 0)), x[2], y[2]), COND_DIV2(TRUE, x[3], y[3]) -> DIV(-(x[3], y[3]), y[3]) which results in the following constraint: (1) (&&(>(x[2], y[2]), >(y[2], 0))=TRUE & x[2]=x[3] & y[2]=y[3] ==> DIV(x[2], y[2])_>=_NonInfC & DIV(x[2], y[2])_>=_COND_DIV2(&&(>(x[2], y[2]), >(y[2], 0)), x[2], y[2]) & (U^Increasing(COND_DIV2(&&(>(x[2], y[2]), >(y[2], 0)), x[2], y[2])), >=)) We simplified constraint (1) using rules (IV), (IDP_BOOLEAN) which results in the following new constraint: (2) (>(x[2], y[2])=TRUE & >(y[2], 0)=TRUE ==> DIV(x[2], y[2])_>=_NonInfC & DIV(x[2], y[2])_>=_COND_DIV2(&&(>(x[2], y[2]), >(y[2], 0)), x[2], y[2]) & (U^Increasing(COND_DIV2(&&(>(x[2], y[2]), >(y[2], 0)), x[2], y[2])), >=)) We simplified constraint (2) using rule (POLY_CONSTRAINTS) which results in the following new constraint: (3) (x[2] + [-1] + [-1]y[2] >= 0 & y[2] + [-1] >= 0 ==> (U^Increasing(COND_DIV2(&&(>(x[2], y[2]), >(y[2], 0)), x[2], y[2])), >=) & [(-1)bni_14 + (-1)Bound*bni_14] + [(2)bni_14]x[2] >= 0 & [(-1)bso_15] + y[2] >= 0) We simplified constraint (3) using rule (IDP_POLY_SIMPLIFY) which results in the following new constraint: (4) (x[2] + [-1] + [-1]y[2] >= 0 & y[2] + [-1] >= 0 ==> (U^Increasing(COND_DIV2(&&(>(x[2], y[2]), >(y[2], 0)), x[2], y[2])), >=) & [(-1)bni_14 + (-1)Bound*bni_14] + [(2)bni_14]x[2] >= 0 & [(-1)bso_15] + y[2] >= 0) We simplified constraint (4) using rule (POLY_REMOVE_MIN_MAX) which results in the following new constraint: (5) (x[2] + [-1] + [-1]y[2] >= 0 & y[2] + [-1] >= 0 ==> (U^Increasing(COND_DIV2(&&(>(x[2], y[2]), >(y[2], 0)), x[2], y[2])), >=) & [(-1)bni_14 + (-1)Bound*bni_14] + [(2)bni_14]x[2] >= 0 & [(-1)bso_15] + y[2] >= 0) To summarize, we get the following constraints P__>=_ for the following pairs. *COND_DIV2(TRUE, x[3], y[3]) -> DIV(-(x[3], y[3]), y[3]) *(x[2] + [-1] + [-1]y[2] >= 0 & y[2] + [-1] >= 0 ==> (U^Increasing(DIV(-(x[3], y[3]), y[3])), >=) & [(-1)bni_12 + (-1)Bound*bni_12] + [(-1)bni_12]y[2] + [(2)bni_12]x[2] >= 0 & [(-1)bso_13] + y[2] >= 0) *DIV(x[2], y[2]) -> COND_DIV2(&&(>(x[2], y[2]), >(y[2], 0)), x[2], y[2]) *(x[2] + [-1] + [-1]y[2] >= 0 & y[2] + [-1] >= 0 ==> (U^Increasing(COND_DIV2(&&(>(x[2], y[2]), >(y[2], 0)), x[2], y[2])), >=) & [(-1)bni_14 + (-1)Bound*bni_14] + [(2)bni_14]x[2] >= 0 & [(-1)bso_15] + y[2] >= 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) = [1] POL(COND_DIV2(x_1, x_2, x_3)) = [-1] + [-1]x_3 + [2]x_2 POL(DIV(x_1, x_2)) = [-1] + [2]x_1 POL(-(x_1, x_2)) = x_1 + [-1]x_2 POL(&&(x_1, x_2)) = [-1] POL(>(x_1, x_2)) = [-1] POL(0) = 0 The following pairs are in P_>: COND_DIV2(TRUE, x[3], y[3]) -> DIV(-(x[3], y[3]), y[3]) DIV(x[2], y[2]) -> COND_DIV2(&&(>(x[2], y[2]), >(y[2], 0)), x[2], y[2]) The following pairs are in P_bound: COND_DIV2(TRUE, x[3], y[3]) -> DIV(-(x[3], y[3]), y[3]) DIV(x[2], y[2]) -> COND_DIV2(&&(>(x[2], y[2]), >(y[2], 0)), x[2], y[2]) The following pairs are in P_>=: none At least the following rules have been oriented under context sensitive arithmetic replacement: FALSE^1 -> &&(FALSE, TRUE)^1 FALSE^1 -> &&(FALSE, FALSE)^1 ---------------------------------------- (8) 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: div(x0, x1) Cond_div(TRUE, x0, x1) Cond_div1(TRUE, x0, x1) Cond_div2(TRUE, x0, x1) ---------------------------------------- (9) PisEmptyProof (EQUIVALENT) The TRS P is empty. Hence, there is no (P,Q,R) chain. ---------------------------------------- (10) YES