/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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty Termination of the given ITRS could be proven: (0) ITRS (1) ITRStoIDPProof [EQUIVALENT, 0 ms] (2) IDP (3) UsableRulesProof [EQUIVALENT, 0 ms] (4) IDP (5)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) AND (7) IDP (8) UsableRulesProof [EQUIVALENT, 0 ms] (9) IDP (10) IDPNonInfProof [SOUND, 127 ms] (11) IDP (12) IDependencyGraphProof [EQUIVALENT, 0 ms] (13) TRUE (14) IDP (15) IDPNonInfProof [SOUND, 48 ms] (16) IDP (17) PisEmptyProof [EQUIVALENT, 0 ms] (18) 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 < exp(10, 2), x + 1) exp(x, y) -> if(y > 0, x, y) if(TRUE, x, y) -> x * exp(x, y - 1) if(FALSE, x, y) -> 1 The set Q consists of the following terms: cu(TRUE, x0) exp(x0, x1) if(TRUE, x0, x1) if(FALSE, 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 The ITRS R consists of the following rules: cu(TRUE, x) -> cu(x < exp(10, 2), x + 1) exp(x, y) -> if(y > 0, x, y) if(TRUE, x, y) -> x * exp(x, y - 1) if(FALSE, x, y) -> 1 The integer pair graph contains the following rules and edges: (0): CU(TRUE, x[0]) -> CU(x[0] < exp(10, 2), x[0] + 1) (1): CU(TRUE, x[1]) -> EXP(10, 2) (2): EXP(x[2], y[2]) -> IF(y[2] > 0, x[2], y[2]) (3): IF(TRUE, x[3], y[3]) -> EXP(x[3], y[3] - 1) (0) -> (0), if (x[0] < exp(10, 2) & x[0] + 1 ->^* x[0]') (0) -> (1), if (x[0] < exp(10, 2) & x[0] + 1 ->^* x[1]) (1) -> (2), if (10 ->^* x[2] & 2 ->^* y[2]) (2) -> (3), if (y[2] > 0 & x[2] ->^* x[3] & y[2] ->^* y[3]) (3) -> (2), if (x[3] ->^* x[2] & y[3] - 1 ->^* y[2]) The set Q consists of the following terms: cu(TRUE, x0) exp(x0, x1) if(TRUE, x0, x1) if(FALSE, 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 The ITRS R consists of the following rules: exp(x, y) -> if(y > 0, x, y) if(TRUE, x, y) -> x * exp(x, y - 1) if(FALSE, x, y) -> 1 The integer pair graph contains the following rules and edges: (0): CU(TRUE, x[0]) -> CU(x[0] < exp(10, 2), x[0] + 1) (1): CU(TRUE, x[1]) -> EXP(10, 2) (2): EXP(x[2], y[2]) -> IF(y[2] > 0, x[2], y[2]) (3): IF(TRUE, x[3], y[3]) -> EXP(x[3], y[3] - 1) (0) -> (0), if (x[0] < exp(10, 2) & x[0] + 1 ->^* x[0]') (0) -> (1), if (x[0] < exp(10, 2) & x[0] + 1 ->^* x[1]) (1) -> (2), if (10 ->^* x[2] & 2 ->^* y[2]) (2) -> (3), if (y[2] > 0 & x[2] ->^* x[3] & y[2] ->^* y[3]) (3) -> (2), if (x[3] ->^* x[2] & y[3] - 1 ->^* y[2]) The set Q consists of the following terms: cu(TRUE, x0) exp(x0, x1) if(TRUE, x0, x1) if(FALSE, x0, x1) ---------------------------------------- (5) IDependencyGraphProof (EQUIVALENT) The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 2 SCCs with 1 less node. ---------------------------------------- (6) Complex Obligation (AND) ---------------------------------------- (7) 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: exp(x, y) -> if(y > 0, x, y) if(TRUE, x, y) -> x * exp(x, y - 1) if(FALSE, x, y) -> 1 The integer pair graph contains the following rules and edges: (3): IF(TRUE, x[3], y[3]) -> EXP(x[3], y[3] - 1) (2): EXP(x[2], y[2]) -> IF(y[2] > 0, x[2], y[2]) (3) -> (2), if (x[3] ->^* x[2] & y[3] - 1 ->^* y[2]) (2) -> (3), if (y[2] > 0 & x[2] ->^* x[3] & y[2] ->^* y[3]) The set Q consists of the following terms: cu(TRUE, x0) exp(x0, x1) if(TRUE, x0, x1) if(FALSE, x0, x1) ---------------------------------------- (8) 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. ---------------------------------------- (9) 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: (3): IF(TRUE, x[3], y[3]) -> EXP(x[3], y[3] - 1) (2): EXP(x[2], y[2]) -> IF(y[2] > 0, x[2], y[2]) (3) -> (2), if (x[3] ->^* x[2] & y[3] - 1 ->^* y[2]) (2) -> (3), if (y[2] > 0 & x[2] ->^* x[3] & y[2] ->^* y[3]) The set Q consists of the following terms: cu(TRUE, x0) exp(x0, x1) if(TRUE, x0, x1) if(FALSE, x0, x1) ---------------------------------------- (10) 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@35409643 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 IF(TRUE, x[3], y[3]) -> EXP(x[3], -(y[3], 1)) the following chains were created: *We consider the chain EXP(x[2], y[2]) -> IF(>(y[2], 0), x[2], y[2]), IF(TRUE, x[3], y[3]) -> EXP(x[3], -(y[3], 1)), EXP(x[2], y[2]) -> IF(>(y[2], 0), x[2], y[2]) which results in the following constraint: (1) (>(y[2], 0)=TRUE & x[2]=x[3] & y[2]=y[3] & x[3]=x[2]1 & -(y[3], 1)=y[2]1 ==> IF(TRUE, x[3], y[3])_>=_NonInfC & IF(TRUE, x[3], y[3])_>=_EXP(x[3], -(y[3], 1)) & (U^Increasing(EXP(x[3], -(y[3], 1))), >=)) We simplified constraint (1) using rules (III), (IV) which results in the following new constraint: (2) (>(y[2], 0)=TRUE ==> IF(TRUE, x[2], y[2])_>=_NonInfC & IF(TRUE, x[2], y[2])_>=_EXP(x[2], -(y[2], 1)) & (U^Increasing(EXP(x[3], -(y[3], 1))), >=)) We simplified constraint (2) using rule (POLY_CONSTRAINTS) which results in the following new constraint: (3) (y[2] + [-1] >= 0 ==> (U^Increasing(EXP(x[3], -(y[3], 1))), >=) & [(-1)bni_11 + (-1)Bound*bni_11] + [(2)bni_11]y[2] >= 0 & [(-1)bso_12] >= 0) We simplified constraint (3) using rule (IDP_POLY_SIMPLIFY) which results in the following new constraint: (4) (y[2] + [-1] >= 0 ==> (U^Increasing(EXP(x[3], -(y[3], 1))), >=) & [(-1)bni_11 + (-1)Bound*bni_11] + [(2)bni_11]y[2] >= 0 & [(-1)bso_12] >= 0) We simplified constraint (4) using rule (POLY_REMOVE_MIN_MAX) which results in the following new constraint: (5) (y[2] + [-1] >= 0 ==> (U^Increasing(EXP(x[3], -(y[3], 1))), >=) & [(-1)bni_11 + (-1)Bound*bni_11] + [(2)bni_11]y[2] >= 0 & [(-1)bso_12] >= 0) We simplified constraint (5) using rule (IDP_UNRESTRICTED_VARS) which results in the following new constraint: (6) (y[2] + [-1] >= 0 ==> (U^Increasing(EXP(x[3], -(y[3], 1))), >=) & 0 = 0 & [(-1)bni_11 + (-1)Bound*bni_11] + [(2)bni_11]y[2] >= 0 & [(-1)bso_12] >= 0) For Pair EXP(x[2], y[2]) -> IF(>(y[2], 0), x[2], y[2]) the following chains were created: *We consider the chain EXP(x[2], y[2]) -> IF(>(y[2], 0), x[2], y[2]), IF(TRUE, x[3], y[3]) -> EXP(x[3], -(y[3], 1)) which results in the following constraint: (1) (>(y[2], 0)=TRUE & x[2]=x[3] & y[2]=y[3] ==> EXP(x[2], y[2])_>=_NonInfC & EXP(x[2], y[2])_>=_IF(>(y[2], 0), x[2], y[2]) & (U^Increasing(IF(>(y[2], 0), x[2], y[2])), >=)) We simplified constraint (1) using rule (IV) which results in the following new constraint: (2) (>(y[2], 0)=TRUE ==> EXP(x[2], y[2])_>=_NonInfC & EXP(x[2], y[2])_>=_IF(>(y[2], 0), x[2], y[2]) & (U^Increasing(IF(>(y[2], 0), x[2], y[2])), >=)) We simplified constraint (2) using rule (POLY_CONSTRAINTS) which results in the following new constraint: (3) (y[2] + [-1] >= 0 ==> (U^Increasing(IF(>(y[2], 0), x[2], y[2])), >=) & [bni_13 + (-1)Bound*bni_13] + [(2)bni_13]y[2] >= 0 & [2 + (-1)bso_14] >= 0) We simplified constraint (3) using rule (IDP_POLY_SIMPLIFY) which results in the following new constraint: (4) (y[2] + [-1] >= 0 ==> (U^Increasing(IF(>(y[2], 0), x[2], y[2])), >=) & [bni_13 + (-1)Bound*bni_13] + [(2)bni_13]y[2] >= 0 & [2 + (-1)bso_14] >= 0) We simplified constraint (4) using rule (POLY_REMOVE_MIN_MAX) which results in the following new constraint: (5) (y[2] + [-1] >= 0 ==> (U^Increasing(IF(>(y[2], 0), x[2], y[2])), >=) & [bni_13 + (-1)Bound*bni_13] + [(2)bni_13]y[2] >= 0 & [2 + (-1)bso_14] >= 0) We simplified constraint (5) using rule (IDP_UNRESTRICTED_VARS) which results in the following new constraint: (6) (y[2] + [-1] >= 0 ==> (U^Increasing(IF(>(y[2], 0), x[2], y[2])), >=) & 0 = 0 & [bni_13 + (-1)Bound*bni_13] + [(2)bni_13]y[2] >= 0 & [2 + (-1)bso_14] >= 0) To summarize, we get the following constraints P__>=_ for the following pairs. *IF(TRUE, x[3], y[3]) -> EXP(x[3], -(y[3], 1)) *(y[2] + [-1] >= 0 ==> (U^Increasing(EXP(x[3], -(y[3], 1))), >=) & 0 = 0 & [(-1)bni_11 + (-1)Bound*bni_11] + [(2)bni_11]y[2] >= 0 & [(-1)bso_12] >= 0) *EXP(x[2], y[2]) -> IF(>(y[2], 0), x[2], y[2]) *(y[2] + [-1] >= 0 ==> (U^Increasing(IF(>(y[2], 0), x[2], y[2])), >=) & 0 = 0 & [bni_13 + (-1)Bound*bni_13] + [(2)bni_13]y[2] >= 0 & [2 + (-1)bso_14] >= 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(IF(x_1, x_2, x_3)) = [-1] + [2]x_3 POL(EXP(x_1, x_2)) = [1] + [2]x_2 POL(-(x_1, x_2)) = x_1 + [-1]x_2 POL(1) = [1] POL(>(x_1, x_2)) = [2] POL(0) = 0 The following pairs are in P_>: EXP(x[2], y[2]) -> IF(>(y[2], 0), x[2], y[2]) The following pairs are in P_bound: IF(TRUE, x[3], y[3]) -> EXP(x[3], -(y[3], 1)) EXP(x[2], y[2]) -> IF(>(y[2], 0), x[2], y[2]) The following pairs are in P_>=: IF(TRUE, x[3], y[3]) -> EXP(x[3], -(y[3], 1)) There are no usable rules. ---------------------------------------- (11) 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: (3): IF(TRUE, x[3], y[3]) -> EXP(x[3], y[3] - 1) The set Q consists of the following terms: cu(TRUE, x0) exp(x0, x1) if(TRUE, x0, x1) if(FALSE, x0, x1) ---------------------------------------- (12) IDependencyGraphProof (EQUIVALENT) The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 1 less node. ---------------------------------------- (13) TRUE ---------------------------------------- (14) 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: exp(x, y) -> if(y > 0, x, y) if(TRUE, x, y) -> x * exp(x, y - 1) if(FALSE, x, y) -> 1 The integer pair graph contains the following rules and edges: (0): CU(TRUE, x[0]) -> CU(x[0] < exp(10, 2), x[0] + 1) (0) -> (0), if (x[0] < exp(10, 2) & x[0] + 1 ->^* x[0]') The set Q consists of the following terms: cu(TRUE, x0) exp(x0, x1) if(TRUE, x0, x1) if(FALSE, x0, x1) ---------------------------------------- (15) 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@35409643 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[0]) -> CU(<(x[0], exp(10, 2)), +(x[0], 1)) the following chains were created: *We consider the chain CU(TRUE, x[0]) -> CU(<(x[0], exp(10, 2)), +(x[0], 1)), CU(TRUE, x[0]) -> CU(<(x[0], exp(10, 2)), +(x[0], 1)), CU(TRUE, x[0]) -> CU(<(x[0], exp(10, 2)), +(x[0], 1)) which results in the following constraint: (1) (<(x[0], exp(10, 2))=TRUE & +(x[0], 1)=x[0]1 & <(x[0]1, exp(10, 2))=TRUE & +(x[0]1, 1)=x[0]2 ==> CU(TRUE, x[0]1)_>=_NonInfC & CU(TRUE, x[0]1)_>=_CU(<(x[0]1, exp(10, 2)), +(x[0]1, 1)) & (U^Increasing(CU(<(x[0]1, exp(10, 2)), +(x[0]1, 1))), >=)) We simplified constraint (1) using rules (III), (IV), (IDP_CONSTANT_FOLD), (REWRITING) which results in the following new constraint: (2) (<(x[0], 100)=TRUE & <(+(x[0], 1), 100)=TRUE ==> CU(TRUE, +(x[0], 1))_>=_NonInfC & CU(TRUE, +(x[0], 1))_>=_CU(<(+(x[0], 1), exp(10, 2)), +(+(x[0], 1), 1)) & (U^Increasing(CU(<(x[0]1, exp(10, 2)), +(x[0]1, 1))), >=)) We simplified constraint (2) using rule (POLY_CONSTRAINTS) which results in the following new constraint: (3) ([99] + [-1]x[0] >= 0 & [98] + [-1]x[0] >= 0 ==> (U^Increasing(CU(<(x[0]1, exp(10, 2)), +(x[0]1, 1))), >=) & [bni_15 + (-1)Bound*bni_15] + [(-1)bni_15]x[0] >= 0 & [1 + (-1)bso_16] >= 0) We simplified constraint (3) using rule (IDP_POLY_SIMPLIFY) which results in the following new constraint: (4) ([99] + [-1]x[0] >= 0 & [98] + [-1]x[0] >= 0 ==> (U^Increasing(CU(<(x[0]1, exp(10, 2)), +(x[0]1, 1))), >=) & [bni_15 + (-1)Bound*bni_15] + [(-1)bni_15]x[0] >= 0 & [1 + (-1)bso_16] >= 0) We simplified constraint (4) using rule (POLY_REMOVE_MIN_MAX) which results in the following new constraint: (5) ([99] + [-1]x[0] >= 0 & [98] + [-1]x[0] >= 0 ==> (U^Increasing(CU(<(x[0]1, exp(10, 2)), +(x[0]1, 1))), >=) & [bni_15 + (-1)Bound*bni_15] + [(-1)bni_15]x[0] >= 0 & [1 + (-1)bso_16] >= 0) We simplified constraint (5) using rule (IDP_SMT_SPLIT) which results in the following new constraints: (6) ([99] + x[0] >= 0 & [98] + x[0] >= 0 & x[0] >= 0 ==> (U^Increasing(CU(<(x[0]1, exp(10, 2)), +(x[0]1, 1))), >=) & [bni_15 + (-1)Bound*bni_15] + [bni_15]x[0] >= 0 & [1 + (-1)bso_16] >= 0) (7) ([99] + [-1]x[0] >= 0 & [98] + [-1]x[0] >= 0 & x[0] >= 0 ==> (U^Increasing(CU(<(x[0]1, exp(10, 2)), +(x[0]1, 1))), >=) & [bni_15 + (-1)Bound*bni_15] + [(-1)bni_15]x[0] >= 0 & [1 + (-1)bso_16] >= 0) To summarize, we get the following constraints P__>=_ for the following pairs. *CU(TRUE, x[0]) -> CU(<(x[0], exp(10, 2)), +(x[0], 1)) *([99] + x[0] >= 0 & [98] + x[0] >= 0 & x[0] >= 0 ==> (U^Increasing(CU(<(x[0]1, exp(10, 2)), +(x[0]1, 1))), >=) & [bni_15 + (-1)Bound*bni_15] + [bni_15]x[0] >= 0 & [1 + (-1)bso_16] >= 0) *([99] + [-1]x[0] >= 0 & [98] + [-1]x[0] >= 0 & x[0] >= 0 ==> (U^Increasing(CU(<(x[0]1, exp(10, 2)), +(x[0]1, 1))), >=) & [bni_15 + (-1)Bound*bni_15] + [(-1)bni_15]x[0] >= 0 & [1 + (-1)bso_16] >= 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) = [1] POL(FALSE) = 0 POL(exp(x_1, x_2)) = [-1] + [-1]x_2 + [-1]x_1 POL(if(x_1, x_2, x_3)) = [-1] + [-1]x_3 + [-1]x_2 + [-1]x_1 POL(>(x_1, x_2)) = [-1] POL(0) = 0 POL(*(x_1, x_2)) = x_1*x_2 POL(-(x_1, x_2)) = x_1 + [-1]x_2 POL(1) = [1] POL(CU(x_1, x_2)) = [2] + [-1]x_2 POL(<(x_1, x_2)) = [-1] POL(10) = [10] POL(2) = [2] POL(+(x_1, x_2)) = x_1 + x_2 POL(100) = [100] The following pairs are in P_>: CU(TRUE, x[0]) -> CU(<(x[0], exp(10, 2)), +(x[0], 1)) The following pairs are in P_bound: CU(TRUE, x[0]) -> CU(<(x[0], exp(10, 2)), +(x[0], 1)) The following pairs are in P_>=: none There are no usable rules. ---------------------------------------- (16) 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: exp(x, y) -> if(y > 0, x, y) if(TRUE, x, y) -> x * exp(x, y - 1) if(FALSE, x, y) -> 1 The integer pair graph is empty. The set Q consists of the following terms: cu(TRUE, x0) exp(x0, x1) if(TRUE, x0, x1) if(FALSE, x0, x1) ---------------------------------------- (17) PisEmptyProof (EQUIVALENT) The TRS P is empty. Hence, there is no (P,Q,R) chain. ---------------------------------------- (18) YES