/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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty Termination of the given ITRS could be proven: (0) ITRS (1) ITRStoIDPProof [EQUIVALENT, 0 ms] (2) IDP (3) UsableRulesProof [EQUIVALENT, 0 ms] (4) IDP (5) IDependencyGraphProof [EQUIVALENT, 0 ms] (6) IDP (7) IDPNonInfProof [SOUND, 242 ms] (8) IDP (9) IDependencyGraphProof [EQUIVALENT, 0 ms] (10) TRUE ---------------------------------------- (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, y) -> Cond_f(x >= 1 && y = x - 1, x, y) Cond_f(TRUE, x, y) -> f(x, round(x)) round(x) -> x round(x) -> x + 1 The set Q consists of the following terms: f(x0, x1) Cond_f(TRUE, x0, x1) round(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: Boolean, Integer The ITRS R consists of the following rules: f(x, y) -> Cond_f(x >= 1 && y = x - 1, x, y) Cond_f(TRUE, x, y) -> f(x, round(x)) round(x) -> x round(x) -> x + 1 The integer pair graph contains the following rules and edges: (0): F(x[0], y[0]) -> COND_F(x[0] >= 1 && y[0] = x[0] - 1, x[0], y[0]) (1): COND_F(TRUE, x[1], y[1]) -> F(x[1], round(x[1])) (2): COND_F(TRUE, x[2], y[2]) -> ROUND(x[2]) (0) -> (1), if (x[0] >= 1 && y[0] = x[0] - 1 & x[0] ->^* x[1] & y[0] ->^* y[1]) (0) -> (2), if (x[0] >= 1 && y[0] = x[0] - 1 & x[0] ->^* x[2] & y[0] ->^* y[2]) (1) -> (0), if (x[1] ->^* x[0] & round(x[1]) ->^* y[0]) The set Q consists of the following terms: f(x0, x1) Cond_f(TRUE, x0, x1) round(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, Boolean The ITRS R consists of the following rules: round(x) -> x round(x) -> x + 1 The integer pair graph contains the following rules and edges: (0): F(x[0], y[0]) -> COND_F(x[0] >= 1 && y[0] = x[0] - 1, x[0], y[0]) (1): COND_F(TRUE, x[1], y[1]) -> F(x[1], round(x[1])) (2): COND_F(TRUE, x[2], y[2]) -> ROUND(x[2]) (0) -> (1), if (x[0] >= 1 && y[0] = x[0] - 1 & x[0] ->^* x[1] & y[0] ->^* y[1]) (0) -> (2), if (x[0] >= 1 && y[0] = x[0] - 1 & x[0] ->^* x[2] & y[0] ->^* y[2]) (1) -> (0), if (x[1] ->^* x[0] & round(x[1]) ->^* y[0]) The set Q consists of the following terms: f(x0, x1) Cond_f(TRUE, x0, x1) round(x0) ---------------------------------------- (5) IDependencyGraphProof (EQUIVALENT) The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 1 less node. ---------------------------------------- (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 The ITRS R consists of the following rules: round(x) -> x round(x) -> x + 1 The integer pair graph contains the following rules and edges: (1): COND_F(TRUE, x[1], y[1]) -> F(x[1], round(x[1])) (0): F(x[0], y[0]) -> COND_F(x[0] >= 1 && y[0] = x[0] - 1, x[0], y[0]) (1) -> (0), if (x[1] ->^* x[0] & round(x[1]) ->^* y[0]) (0) -> (1), if (x[0] >= 1 && y[0] = x[0] - 1 & x[0] ->^* x[1] & y[0] ->^* y[1]) The set Q consists of the following terms: f(x0, x1) Cond_f(TRUE, x0, x1) round(x0) ---------------------------------------- (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@5d507df4 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_F(TRUE, x[1], y[1]) -> F(x[1], round(x[1])) the following chains were created: *We consider the chain F(x[0], y[0]) -> COND_F(&&(>=(x[0], 1), =(y[0], -(x[0], 1))), x[0], y[0]), COND_F(TRUE, x[1], y[1]) -> F(x[1], round(x[1])), F(x[0], y[0]) -> COND_F(&&(>=(x[0], 1), =(y[0], -(x[0], 1))), x[0], y[0]) which results in the following constraint: (1) (&&(>=(x[0], 1), =(y[0], -(x[0], 1)))=TRUE & x[0]=x[1] & y[0]=y[1] & x[1]=x[0]1 & round(x[1])=y[0]1 ==> COND_F(TRUE, x[1], y[1])_>=_NonInfC & COND_F(TRUE, x[1], y[1])_>=_F(x[1], round(x[1])) & (U^Increasing(F(x[1], round(x[1]))), >=)) We simplified constraint (1) using rules (III), (IV), (IDP_BOOLEAN) which results in the following new constraint: (2) (>=(x[0], 1)=TRUE & >=(y[0], -(x[0], 1))=TRUE & <=(y[0], -(x[0], 1))=TRUE ==> COND_F(TRUE, x[0], y[0])_>=_NonInfC & COND_F(TRUE, x[0], y[0])_>=_F(x[0], round(x[0])) & (U^Increasing(F(x[1], round(x[1]))), >=)) We simplified constraint (2) using rule (POLY_CONSTRAINTS) which results in the following new constraint: (3) (x[0] + [-1] >= 0 & y[0] + [1] + [-1]x[0] >= 0 & x[0] + [-1] + [-1]y[0] >= 0 ==> (U^Increasing(F(x[1], round(x[1]))), >=) & [(-1)bni_15 + (-1)Bound*bni_15] + [bni_15]x[0] >= 0 & [(-1)bso_16] >= 0) We simplified constraint (3) using rule (IDP_POLY_SIMPLIFY) which results in the following new constraint: (4) (x[0] + [-1] >= 0 & y[0] + [1] + [-1]x[0] >= 0 & x[0] + [-1] + [-1]y[0] >= 0 ==> (U^Increasing(F(x[1], round(x[1]))), >=) & [(-1)bni_15 + (-1)Bound*bni_15] + [bni_15]x[0] >= 0 & [(-1)bso_16] >= 0) We simplified constraint (4) using rule (POLY_REMOVE_MIN_MAX) which results in the following new constraint: (5) (x[0] + [-1] >= 0 & y[0] + [1] + [-1]x[0] >= 0 & x[0] + [-1] + [-1]y[0] >= 0 ==> (U^Increasing(F(x[1], round(x[1]))), >=) & [(-1)bni_15 + (-1)Bound*bni_15] + [bni_15]x[0] >= 0 & [(-1)bso_16] >= 0) We simplified constraint (5) using rule (IDP_SMT_SPLIT) which results in the following new constraints: (6) (x[0] + [-1] >= 0 & y[0] + [1] + [-1]x[0] >= 0 & x[0] + [-1] + [-1]y[0] >= 0 & y[0] >= 0 ==> (U^Increasing(F(x[1], round(x[1]))), >=) & [(-1)bni_15 + (-1)Bound*bni_15] + [bni_15]x[0] >= 0 & [(-1)bso_16] >= 0) (7) (x[0] + [-1] >= 0 & [-1]y[0] + [1] + [-1]x[0] >= 0 & x[0] + [-1] + y[0] >= 0 & y[0] >= 0 ==> (U^Increasing(F(x[1], round(x[1]))), >=) & [(-1)bni_15 + (-1)Bound*bni_15] + [bni_15]x[0] >= 0 & [(-1)bso_16] >= 0) For Pair F(x[0], y[0]) -> COND_F(&&(>=(x[0], 1), =(y[0], -(x[0], 1))), x[0], y[0]) the following chains were created: *We consider the chain F(x[0], y[0]) -> COND_F(&&(>=(x[0], 1), =(y[0], -(x[0], 1))), x[0], y[0]), COND_F(TRUE, x[1], y[1]) -> F(x[1], round(x[1])) which results in the following constraint: (1) (&&(>=(x[0], 1), =(y[0], -(x[0], 1)))=TRUE & x[0]=x[1] & y[0]=y[1] ==> F(x[0], y[0])_>=_NonInfC & F(x[0], y[0])_>=_COND_F(&&(>=(x[0], 1), =(y[0], -(x[0], 1))), x[0], y[0]) & (U^Increasing(COND_F(&&(>=(x[0], 1), =(y[0], -(x[0], 1))), x[0], y[0])), >=)) We simplified constraint (1) using rules (IV), (IDP_BOOLEAN) which results in the following new constraint: (2) (>=(x[0], 1)=TRUE & >=(y[0], -(x[0], 1))=TRUE & <=(y[0], -(x[0], 1))=TRUE ==> F(x[0], y[0])_>=_NonInfC & F(x[0], y[0])_>=_COND_F(&&(>=(x[0], 1), =(y[0], -(x[0], 1))), x[0], y[0]) & (U^Increasing(COND_F(&&(>=(x[0], 1), =(y[0], -(x[0], 1))), x[0], y[0])), >=)) We simplified constraint (2) using rule (POLY_CONSTRAINTS) which results in the following new constraint: (3) (x[0] + [-1] >= 0 & y[0] + [1] + [-1]x[0] >= 0 & x[0] + [-1] + [-1]y[0] >= 0 ==> (U^Increasing(COND_F(&&(>=(x[0], 1), =(y[0], -(x[0], 1))), x[0], y[0])), >=) & [(-1)bni_17 + (-1)Bound*bni_17] + [(-1)bni_17]y[0] + [(2)bni_17]x[0] >= 0 & [(-1)bso_18] + [-1]y[0] + x[0] >= 0) We simplified constraint (3) using rule (IDP_POLY_SIMPLIFY) which results in the following new constraint: (4) (x[0] + [-1] >= 0 & y[0] + [1] + [-1]x[0] >= 0 & x[0] + [-1] + [-1]y[0] >= 0 ==> (U^Increasing(COND_F(&&(>=(x[0], 1), =(y[0], -(x[0], 1))), x[0], y[0])), >=) & [(-1)bni_17 + (-1)Bound*bni_17] + [(-1)bni_17]y[0] + [(2)bni_17]x[0] >= 0 & [(-1)bso_18] + [-1]y[0] + x[0] >= 0) We simplified constraint (4) using rule (POLY_REMOVE_MIN_MAX) which results in the following new constraint: (5) (x[0] + [-1] >= 0 & y[0] + [1] + [-1]x[0] >= 0 & x[0] + [-1] + [-1]y[0] >= 0 ==> (U^Increasing(COND_F(&&(>=(x[0], 1), =(y[0], -(x[0], 1))), x[0], y[0])), >=) & [(-1)bni_17 + (-1)Bound*bni_17] + [(-1)bni_17]y[0] + [(2)bni_17]x[0] >= 0 & [(-1)bso_18] + [-1]y[0] + x[0] >= 0) We simplified constraint (5) using rule (IDP_SMT_SPLIT) which results in the following new constraints: (6) (x[0] + [-1] >= 0 & y[0] + [1] + [-1]x[0] >= 0 & x[0] + [-1] + [-1]y[0] >= 0 & y[0] >= 0 ==> (U^Increasing(COND_F(&&(>=(x[0], 1), =(y[0], -(x[0], 1))), x[0], y[0])), >=) & [(-1)bni_17 + (-1)Bound*bni_17] + [(-1)bni_17]y[0] + [(2)bni_17]x[0] >= 0 & [(-1)bso_18] + [-1]y[0] + x[0] >= 0) (7) (x[0] + [-1] >= 0 & [-1]y[0] + [1] + [-1]x[0] >= 0 & x[0] + [-1] + y[0] >= 0 & y[0] >= 0 ==> (U^Increasing(COND_F(&&(>=(x[0], 1), =(y[0], -(x[0], 1))), x[0], y[0])), >=) & [(-1)bni_17 + (-1)Bound*bni_17] + [bni_17]y[0] + [(2)bni_17]x[0] >= 0 & [(-1)bso_18] + y[0] + x[0] >= 0) To summarize, we get the following constraints P__>=_ for the following pairs. *COND_F(TRUE, x[1], y[1]) -> F(x[1], round(x[1])) *(x[0] + [-1] >= 0 & y[0] + [1] + [-1]x[0] >= 0 & x[0] + [-1] + [-1]y[0] >= 0 & y[0] >= 0 ==> (U^Increasing(F(x[1], round(x[1]))), >=) & [(-1)bni_15 + (-1)Bound*bni_15] + [bni_15]x[0] >= 0 & [(-1)bso_16] >= 0) *(x[0] + [-1] >= 0 & [-1]y[0] + [1] + [-1]x[0] >= 0 & x[0] + [-1] + y[0] >= 0 & y[0] >= 0 ==> (U^Increasing(F(x[1], round(x[1]))), >=) & [(-1)bni_15 + (-1)Bound*bni_15] + [bni_15]x[0] >= 0 & [(-1)bso_16] >= 0) *F(x[0], y[0]) -> COND_F(&&(>=(x[0], 1), =(y[0], -(x[0], 1))), x[0], y[0]) *(x[0] + [-1] >= 0 & y[0] + [1] + [-1]x[0] >= 0 & x[0] + [-1] + [-1]y[0] >= 0 & y[0] >= 0 ==> (U^Increasing(COND_F(&&(>=(x[0], 1), =(y[0], -(x[0], 1))), x[0], y[0])), >=) & [(-1)bni_17 + (-1)Bound*bni_17] + [(-1)bni_17]y[0] + [(2)bni_17]x[0] >= 0 & [(-1)bso_18] + [-1]y[0] + x[0] >= 0) *(x[0] + [-1] >= 0 & [-1]y[0] + [1] + [-1]x[0] >= 0 & x[0] + [-1] + y[0] >= 0 & y[0] >= 0 ==> (U^Increasing(COND_F(&&(>=(x[0], 1), =(y[0], -(x[0], 1))), x[0], y[0])), >=) & [(-1)bni_17 + (-1)Bound*bni_17] + [bni_17]y[0] + [(2)bni_17]x[0] >= 0 & [(-1)bso_18] + y[0] + x[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) = [1] POL(round(x_1)) = x_1 POL(+(x_1, x_2)) = x_1 + x_2 POL(1) = [1] POL(COND_F(x_1, x_2, x_3)) = [-1] + x_2 + [-1]x_1 POL(F(x_1, x_2)) = [-1] + [-1]x_2 + [2]x_1 POL(&&(x_1, x_2)) = 0 POL(>=(x_1, x_2)) = [-1] POL(=(x_1, x_2)) = [-1] POL(-(x_1, x_2)) = x_1 + [-1]x_2 The following pairs are in P_>: F(x[0], y[0]) -> COND_F(&&(>=(x[0], 1), =(y[0], -(x[0], 1))), x[0], y[0]) The following pairs are in P_bound: COND_F(TRUE, x[1], y[1]) -> F(x[1], round(x[1])) F(x[0], y[0]) -> COND_F(&&(>=(x[0], 1), =(y[0], -(x[0], 1))), x[0], y[0]) The following pairs are in P_>=: COND_F(TRUE, x[1], y[1]) -> F(x[1], round(x[1])) At least the following rules have been oriented under context sensitive arithmetic replacement: x^1 -> round(x)^1 +(x, 1)^1 -> round(x)^1 TRUE^1 -> &&(TRUE, TRUE)^1 FALSE^1 -> &&(TRUE, FALSE)^1 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: Integer The ITRS R consists of the following rules: round(x) -> x round(x) -> x + 1 The integer pair graph contains the following rules and edges: (1): COND_F(TRUE, x[1], y[1]) -> F(x[1], round(x[1])) The set Q consists of the following terms: f(x0, x1) Cond_f(TRUE, x0, x1) round(x0) ---------------------------------------- (9) IDependencyGraphProof (EQUIVALENT) The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 1 less node. ---------------------------------------- (10) TRUE