/export/starexec/sandbox2/solver/bin/starexec_run_standard /export/starexec/sandbox2/benchmark/theBenchmark.itrs /export/starexec/sandbox2/output/output_files -------------------------------------------------------------------------------- YES proof of /export/starexec/sandbox2/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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty Termination of the given ITRS could be proven: (0) ITRS (1) ITRStoIDPProof [EQUIVALENT, 0 ms] (2) IDP (3) UsableRulesProof [EQUIVALENT, 0 ms] (4) IDP (5)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, 141 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(x) -> Cond_f(x * x < 0, x) Cond_f(TRUE, x) -> f(x) The set Q consists of the following terms: f(x0) Cond_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(x) -> Cond_f(x * x < 0, x) Cond_f(TRUE, x) -> f(x) The integer pair graph contains the following rules and edges: (0): F(x[0]) -> COND_F(x[0] * x[0] < 0, x[0]) (1): COND_F(TRUE, x[1]) -> F(x[1]) (0) -> (1), if (x[0] * x[0] < 0 & x[0] ->^* x[1]) (1) -> (0), if (x[1] ->^* x[0]) The set Q consists of the following terms: f(x0) Cond_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(x[0]) -> COND_F(x[0] * x[0] < 0, x[0]) (1): COND_F(TRUE, x[1]) -> F(x[1]) (0) -> (1), if (x[0] * x[0] < 0 & x[0] ->^* x[1]) (1) -> (0), if (x[1] ->^* x[0]) The set Q consists of the following terms: f(x0) Cond_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@7bb9edcb 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(x) -> COND_F(<(*(x, x), 0), x) the following chains were created: *We consider the chain COND_F(TRUE, x[1]) -> F(x[1]), F(x[0]) -> COND_F(<(*(x[0], x[0]), 0), x[0]), COND_F(TRUE, x[1]) -> F(x[1]) which results in the following constraint: (1) (x[1]=x[0] & <(*(x[0], x[0]), 0)=TRUE & x[0]=x[1]1 ==> F(x[0])_>=_NonInfC & F(x[0])_>=_COND_F(<(*(x[0], x[0]), 0), x[0]) & (U^Increasing(COND_F(<(*(x[0], x[0]), 0), x[0])), >=)) We simplified constraint (1) using rules (III), (IV) which results in the following new constraint: (2) (<(*(x[0], x[0]), 0)=TRUE ==> F(x[0])_>=_NonInfC & F(x[0])_>=_COND_F(<(*(x[0], x[0]), 0), x[0]) & (U^Increasing(COND_F(<(*(x[0], x[0]), 0), x[0])), >=)) We simplified constraint (2) using rule (POLY_CONSTRAINTS) which results in the following new constraint: (3) ([-1] + [-1]x[0]^2 >= 0 ==> (U^Increasing(COND_F(<(*(x[0], x[0]), 0), x[0])), >=) & [(-1)bni_9 + (-1)Bound*bni_9] >= 0 & [(-1)bso_10] >= 0) We simplified constraint (3) using rule (IDP_POLY_SIMPLIFY) which results in the following new constraint: (4) ([-1] + [-1]x[0]^2 >= 0 ==> (U^Increasing(COND_F(<(*(x[0], x[0]), 0), x[0])), >=) & [(-1)bni_9 + (-1)Bound*bni_9] >= 0 & [(-1)bso_10] >= 0) We simplified constraint (4) using rule (POLY_REMOVE_MIN_MAX) which results in the following new constraint: (5) ([-1] + [-1]x[0]^2 >= 0 ==> (U^Increasing(COND_F(<(*(x[0], x[0]), 0), x[0])), >=) & [(-1)bni_9 + (-1)Bound*bni_9] >= 0 & [(-1)bso_10] >= 0) We simplified constraint (5) using rule (IDP_SMT_SPLIT) which results in the following new constraints: (6) ([-1] + [-1]x[0]^2 >= 0 & x[0] >= 0 ==> (U^Increasing(COND_F(<(*(x[0], x[0]), 0), x[0])), >=) & [(-1)bni_9 + (-1)Bound*bni_9] >= 0 & [(-1)bso_10] >= 0) (7) ([-1] + [-1]x[0]^2 >= 0 & x[0] >= 0 ==> (U^Increasing(COND_F(<(*(x[0], x[0]), 0), x[0])), >=) & [(-1)bni_9 + (-1)Bound*bni_9] >= 0 & [(-1)bso_10] >= 0) For Pair COND_F(TRUE, x) -> F(x) the following chains were created: *We consider the chain F(x[0]) -> COND_F(<(*(x[0], x[0]), 0), x[0]), COND_F(TRUE, x[1]) -> F(x[1]), F(x[0]) -> COND_F(<(*(x[0], x[0]), 0), x[0]) which results in the following constraint: (1) (<(*(x[0], x[0]), 0)=TRUE & x[0]=x[1] & x[1]=x[0]1 ==> COND_F(TRUE, x[1])_>=_NonInfC & COND_F(TRUE, x[1])_>=_F(x[1]) & (U^Increasing(F(x[1])), >=)) We simplified constraint (1) using rules (III), (IV) which results in the following new constraint: (2) (<(*(x[0], x[0]), 0)=TRUE ==> COND_F(TRUE, x[0])_>=_NonInfC & COND_F(TRUE, x[0])_>=_F(x[0]) & (U^Increasing(F(x[1])), >=)) We simplified constraint (2) using rule (POLY_CONSTRAINTS) which results in the following new constraint: (3) ([-1] + [-1]x[0]^2 >= 0 ==> (U^Increasing(F(x[1])), >=) & [(-1)bni_11 + (-1)Bound*bni_11] >= 0 & [(-1)bso_12] >= 0) We simplified constraint (3) using rule (IDP_POLY_SIMPLIFY) which results in the following new constraint: (4) ([-1] + [-1]x[0]^2 >= 0 ==> (U^Increasing(F(x[1])), >=) & [(-1)bni_11 + (-1)Bound*bni_11] >= 0 & [(-1)bso_12] >= 0) We simplified constraint (4) using rule (POLY_REMOVE_MIN_MAX) which results in the following new constraint: (5) ([-1] + [-1]x[0]^2 >= 0 ==> (U^Increasing(F(x[1])), >=) & [(-1)bni_11 + (-1)Bound*bni_11] >= 0 & [(-1)bso_12] >= 0) We simplified constraint (5) using rule (IDP_SMT_SPLIT) which results in the following new constraints: (6) ([-1] + [-1]x[0]^2 >= 0 & x[0] >= 0 ==> (U^Increasing(F(x[1])), >=) & [(-1)bni_11 + (-1)Bound*bni_11] >= 0 & [(-1)bso_12] >= 0) (7) ([-1] + [-1]x[0]^2 >= 0 & x[0] >= 0 ==> (U^Increasing(F(x[1])), >=) & [(-1)bni_11 + (-1)Bound*bni_11] >= 0 & [(-1)bso_12] >= 0) To summarize, we get the following constraints P__>=_ for the following pairs. *F(x) -> COND_F(<(*(x, x), 0), x) *([-1] + [-1]x[0]^2 >= 0 & x[0] >= 0 ==> (U^Increasing(COND_F(<(*(x[0], x[0]), 0), x[0])), >=) & [(-1)bni_9 + (-1)Bound*bni_9] >= 0 & [(-1)bso_10] >= 0) *([-1] + [-1]x[0]^2 >= 0 & x[0] >= 0 ==> (U^Increasing(COND_F(<(*(x[0], x[0]), 0), x[0])), >=) & [(-1)bni_9 + (-1)Bound*bni_9] >= 0 & [(-1)bso_10] >= 0) *COND_F(TRUE, x) -> F(x) *([-1] + [-1]x[0]^2 >= 0 & x[0] >= 0 ==> (U^Increasing(F(x[1])), >=) & [(-1)bni_11 + (-1)Bound*bni_11] >= 0 & [(-1)bso_12] >= 0) *([-1] + [-1]x[0]^2 >= 0 & x[0] >= 0 ==> (U^Increasing(F(x[1])), >=) & [(-1)bni_11 + (-1)Bound*bni_11] >= 0 & [(-1)bso_12] >= 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)) = [-1] POL(COND_F(x_1, x_2)) = [-1] POL(<(x_1, x_2)) = [-1] POL(*(x_1, x_2)) = x_1*x_2 POL(0) = 0 The following pairs are in P_>: F(x[0]) -> COND_F(<(*(x[0], x[0]), 0), x[0]) COND_F(TRUE, x[1]) -> F(x[1]) The following pairs are in P_bound: F(x[0]) -> COND_F(<(*(x[0], x[0]), 0), x[0]) COND_F(TRUE, x[1]) -> F(x[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(x0) Cond_f(TRUE, x0) ---------------------------------------- (7) PisEmptyProof (EQUIVALENT) The TRS P is empty. Hence, there is no (P,Q,R) chain. ---------------------------------------- (8) YES