/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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty Termination of the given ITRS could be proven: (0) ITRS (1) ITRStoIDPProof [EQUIVALENT, 0 ms] (2) IDP (3) UsableRulesProof [EQUIVALENT, 0 ms] (4) IDP (5)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, 151 ms] (6) IDP (7) IDependencyGraphProof [EQUIVALENT, 0 ms] (8) 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: minus(x, y) -> cond(x >= y + 1, x, y) cond(FALSE, x, y) -> 0 cond(TRUE, x, y) -> 1 + minus(x, y + 1) The set Q consists of the following terms: minus(x0, x1) cond(FALSE, x0, x1) cond(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 The ITRS R consists of the following rules: minus(x, y) -> cond(x >= y + 1, x, y) cond(FALSE, x, y) -> 0 cond(TRUE, x, y) -> 1 + minus(x, y + 1) The integer pair graph contains the following rules and edges: (0): MINUS(x[0], y[0]) -> COND(x[0] >= y[0] + 1, x[0], y[0]) (1): COND(TRUE, x[1], y[1]) -> MINUS(x[1], y[1] + 1) (0) -> (1), if (x[0] >= y[0] + 1 & x[0] ->^* x[1] & y[0] ->^* y[1]) (1) -> (0), if (x[1] ->^* x[0] & y[1] + 1 ->^* y[0]) The set Q consists of the following terms: minus(x0, x1) cond(FALSE, x0, x1) cond(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 R is empty. The integer pair graph contains the following rules and edges: (0): MINUS(x[0], y[0]) -> COND(x[0] >= y[0] + 1, x[0], y[0]) (1): COND(TRUE, x[1], y[1]) -> MINUS(x[1], y[1] + 1) (0) -> (1), if (x[0] >= y[0] + 1 & x[0] ->^* x[1] & y[0] ->^* y[1]) (1) -> (0), if (x[1] ->^* x[0] & y[1] + 1 ->^* y[0]) The set Q consists of the following terms: minus(x0, x1) cond(FALSE, x0, x1) cond(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@7fad6e16 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 MINUS(x, y) -> COND(>=(x, +(y, 1)), x, y) the following chains were created: *We consider the chain MINUS(x[0], y[0]) -> COND(>=(x[0], +(y[0], 1)), x[0], y[0]), COND(TRUE, x[1], y[1]) -> MINUS(x[1], +(y[1], 1)) which results in the following constraint: (1) (>=(x[0], +(y[0], 1))=TRUE & x[0]=x[1] & y[0]=y[1] ==> MINUS(x[0], y[0])_>=_NonInfC & MINUS(x[0], y[0])_>=_COND(>=(x[0], +(y[0], 1)), x[0], y[0]) & (U^Increasing(COND(>=(x[0], +(y[0], 1)), x[0], y[0])), >=)) We simplified constraint (1) using rule (IV) which results in the following new constraint: (2) (>=(x[0], +(y[0], 1))=TRUE ==> MINUS(x[0], y[0])_>=_NonInfC & MINUS(x[0], y[0])_>=_COND(>=(x[0], +(y[0], 1)), x[0], y[0]) & (U^Increasing(COND(>=(x[0], +(y[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] + [-1]y[0] >= 0 ==> (U^Increasing(COND(>=(x[0], +(y[0], 1)), x[0], y[0])), >=) & [(-1)bni_11 + (-1)Bound*bni_11] + [(-1)bni_11]y[0] + [bni_11]x[0] >= 0 & [(-1)bso_12] >= 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 ==> (U^Increasing(COND(>=(x[0], +(y[0], 1)), x[0], y[0])), >=) & [(-1)bni_11 + (-1)Bound*bni_11] + [(-1)bni_11]y[0] + [bni_11]x[0] >= 0 & [(-1)bso_12] >= 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 ==> (U^Increasing(COND(>=(x[0], +(y[0], 1)), x[0], y[0])), >=) & [(-1)bni_11 + (-1)Bound*bni_11] + [(-1)bni_11]y[0] + [bni_11]x[0] >= 0 & [(-1)bso_12] >= 0) We simplified constraint (5) using rule (IDP_SMT_SPLIT) which results in the following new constraint: (6) (x[0] >= 0 ==> (U^Increasing(COND(>=(x[0], +(y[0], 1)), x[0], y[0])), >=) & [(-1)Bound*bni_11] + [bni_11]x[0] >= 0 & [(-1)bso_12] >= 0) We simplified constraint (6) using rule (IDP_SMT_SPLIT) which results in the following new constraints: (7) (x[0] >= 0 & y[0] >= 0 ==> (U^Increasing(COND(>=(x[0], +(y[0], 1)), x[0], y[0])), >=) & [(-1)Bound*bni_11] + [bni_11]x[0] >= 0 & [(-1)bso_12] >= 0) (8) (x[0] >= 0 & y[0] >= 0 ==> (U^Increasing(COND(>=(x[0], +(y[0], 1)), x[0], y[0])), >=) & [(-1)Bound*bni_11] + [bni_11]x[0] >= 0 & [(-1)bso_12] >= 0) For Pair COND(TRUE, x, y) -> MINUS(x, +(y, 1)) the following chains were created: *We consider the chain MINUS(x[0], y[0]) -> COND(>=(x[0], +(y[0], 1)), x[0], y[0]), COND(TRUE, x[1], y[1]) -> MINUS(x[1], +(y[1], 1)), MINUS(x[0], y[0]) -> COND(>=(x[0], +(y[0], 1)), x[0], y[0]) which results in the following constraint: (1) (>=(x[0], +(y[0], 1))=TRUE & x[0]=x[1] & y[0]=y[1] & x[1]=x[0]1 & +(y[1], 1)=y[0]1 ==> COND(TRUE, x[1], y[1])_>=_NonInfC & COND(TRUE, x[1], y[1])_>=_MINUS(x[1], +(y[1], 1)) & (U^Increasing(MINUS(x[1], +(y[1], 1))), >=)) We simplified constraint (1) using rules (III), (IV) which results in the following new constraint: (2) (>=(x[0], +(y[0], 1))=TRUE ==> COND(TRUE, x[0], y[0])_>=_NonInfC & COND(TRUE, x[0], y[0])_>=_MINUS(x[0], +(y[0], 1)) & (U^Increasing(MINUS(x[1], +(y[1], 1))), >=)) We simplified constraint (2) using rule (POLY_CONSTRAINTS) which results in the following new constraint: (3) (x[0] + [-1] + [-1]y[0] >= 0 ==> (U^Increasing(MINUS(x[1], +(y[1], 1))), >=) & [(-1)bni_13 + (-1)Bound*bni_13] + [(-1)bni_13]y[0] + [bni_13]x[0] >= 0 & [1 + (-1)bso_14] >= 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 ==> (U^Increasing(MINUS(x[1], +(y[1], 1))), >=) & [(-1)bni_13 + (-1)Bound*bni_13] + [(-1)bni_13]y[0] + [bni_13]x[0] >= 0 & [1 + (-1)bso_14] >= 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 ==> (U^Increasing(MINUS(x[1], +(y[1], 1))), >=) & [(-1)bni_13 + (-1)Bound*bni_13] + [(-1)bni_13]y[0] + [bni_13]x[0] >= 0 & [1 + (-1)bso_14] >= 0) We simplified constraint (5) using rule (IDP_SMT_SPLIT) which results in the following new constraint: (6) (x[0] >= 0 ==> (U^Increasing(MINUS(x[1], +(y[1], 1))), >=) & [(-1)Bound*bni_13] + [bni_13]x[0] >= 0 & [1 + (-1)bso_14] >= 0) We simplified constraint (6) using rule (IDP_SMT_SPLIT) which results in the following new constraints: (7) (x[0] >= 0 & y[0] >= 0 ==> (U^Increasing(MINUS(x[1], +(y[1], 1))), >=) & [(-1)Bound*bni_13] + [bni_13]x[0] >= 0 & [1 + (-1)bso_14] >= 0) (8) (x[0] >= 0 & y[0] >= 0 ==> (U^Increasing(MINUS(x[1], +(y[1], 1))), >=) & [(-1)Bound*bni_13] + [bni_13]x[0] >= 0 & [1 + (-1)bso_14] >= 0) To summarize, we get the following constraints P__>=_ for the following pairs. *MINUS(x, y) -> COND(>=(x, +(y, 1)), x, y) *(x[0] >= 0 & y[0] >= 0 ==> (U^Increasing(COND(>=(x[0], +(y[0], 1)), x[0], y[0])), >=) & [(-1)Bound*bni_11] + [bni_11]x[0] >= 0 & [(-1)bso_12] >= 0) *(x[0] >= 0 & y[0] >= 0 ==> (U^Increasing(COND(>=(x[0], +(y[0], 1)), x[0], y[0])), >=) & [(-1)Bound*bni_11] + [bni_11]x[0] >= 0 & [(-1)bso_12] >= 0) *COND(TRUE, x, y) -> MINUS(x, +(y, 1)) *(x[0] >= 0 & y[0] >= 0 ==> (U^Increasing(MINUS(x[1], +(y[1], 1))), >=) & [(-1)Bound*bni_13] + [bni_13]x[0] >= 0 & [1 + (-1)bso_14] >= 0) *(x[0] >= 0 & y[0] >= 0 ==> (U^Increasing(MINUS(x[1], +(y[1], 1))), >=) & [(-1)Bound*bni_13] + [bni_13]x[0] >= 0 & [1 + (-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(MINUS(x_1, x_2)) = [-1] + [-1]x_2 + x_1 POL(COND(x_1, x_2, x_3)) = [-1] + [-1]x_3 + x_2 POL(>=(x_1, x_2)) = [-1] POL(+(x_1, x_2)) = x_1 + x_2 POL(1) = [1] The following pairs are in P_>: COND(TRUE, x[1], y[1]) -> MINUS(x[1], +(y[1], 1)) The following pairs are in P_bound: MINUS(x[0], y[0]) -> COND(>=(x[0], +(y[0], 1)), x[0], y[0]) COND(TRUE, x[1], y[1]) -> MINUS(x[1], +(y[1], 1)) The following pairs are in P_>=: MINUS(x[0], y[0]) -> COND(>=(x[0], +(y[0], 1)), x[0], y[0]) 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: Integer R is empty. The integer pair graph contains the following rules and edges: (0): MINUS(x[0], y[0]) -> COND(x[0] >= y[0] + 1, x[0], y[0]) The set Q consists of the following terms: minus(x0, x1) cond(FALSE, x0, x1) cond(TRUE, x0, x1) ---------------------------------------- (7) IDependencyGraphProof (EQUIVALENT) The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 1 less node. ---------------------------------------- (8) TRUE