3.61/1.87 YES 3.61/1.88 proof of /export/starexec/sandbox/benchmark/theBenchmark.itrs 3.61/1.88 # AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty 3.61/1.88 3.61/1.88 3.61/1.88 Termination of the given ITRS could be proven: 3.61/1.88 3.61/1.88 (0) ITRS 3.61/1.88 (1) ITRStoIDPProof [EQUIVALENT, 0 ms] 3.61/1.88 (2) IDP 3.61/1.88 (3) UsableRulesProof [EQUIVALENT, 0 ms] 3.61/1.88 (4) IDP 3.61/1.88 (5) IDPNonInfProof [SOUND, 123 ms] 3.61/1.88 (6) IDP 3.61/1.88 (7) IDependencyGraphProof [EQUIVALENT, 0 ms] 3.61/1.88 (8) TRUE 3.61/1.88 3.61/1.88 3.61/1.88 ---------------------------------------- 3.61/1.88 3.61/1.88 (0) 3.61/1.88 Obligation: 3.61/1.88 ITRS problem: 3.61/1.88 3.61/1.88 The following function symbols are pre-defined: 3.61/1.88 <<< 3.61/1.88 & ~ Bwand: (Integer, Integer) -> Integer 3.61/1.88 >= ~ Ge: (Integer, Integer) -> Boolean 3.61/1.88 | ~ Bwor: (Integer, Integer) -> Integer 3.61/1.88 / ~ Div: (Integer, Integer) -> Integer 3.61/1.88 != ~ Neq: (Integer, Integer) -> Boolean 3.61/1.88 && ~ Land: (Boolean, Boolean) -> Boolean 3.61/1.88 ! ~ Lnot: (Boolean) -> Boolean 3.61/1.88 = ~ Eq: (Integer, Integer) -> Boolean 3.61/1.88 <= ~ Le: (Integer, Integer) -> Boolean 3.61/1.88 ^ ~ Bwxor: (Integer, Integer) -> Integer 3.61/1.88 % ~ Mod: (Integer, Integer) -> Integer 3.61/1.88 > ~ Gt: (Integer, Integer) -> Boolean 3.61/1.88 + ~ Add: (Integer, Integer) -> Integer 3.61/1.88 -1 ~ UnaryMinus: (Integer) -> Integer 3.61/1.88 < ~ Lt: (Integer, Integer) -> Boolean 3.61/1.88 || ~ Lor: (Boolean, Boolean) -> Boolean 3.61/1.88 - ~ Sub: (Integer, Integer) -> Integer 3.61/1.88 ~ ~ Bwnot: (Integer) -> Integer 3.61/1.88 * ~ Mul: (Integer, Integer) -> Integer 3.61/1.88 >>> 3.61/1.88 3.61/1.88 The TRS R consists of the following rules: 3.61/1.88 eval(x) -> Cond_eval(x % 2 > 0 && x > 0, x) 3.61/1.88 Cond_eval(TRUE, x) -> eval(x - 1) 3.61/1.88 The set Q consists of the following terms: 3.61/1.88 eval(x0) 3.61/1.88 Cond_eval(TRUE, x0) 3.61/1.88 3.61/1.88 ---------------------------------------- 3.61/1.88 3.61/1.88 (1) ITRStoIDPProof (EQUIVALENT) 3.61/1.88 Added dependency pairs 3.61/1.88 ---------------------------------------- 3.61/1.88 3.61/1.88 (2) 3.61/1.88 Obligation: 3.61/1.88 IDP problem: 3.61/1.88 The following function symbols are pre-defined: 3.61/1.88 <<< 3.61/1.88 & ~ Bwand: (Integer, Integer) -> Integer 3.61/1.88 >= ~ Ge: (Integer, Integer) -> Boolean 3.61/1.88 | ~ Bwor: (Integer, Integer) -> Integer 3.61/1.88 / ~ Div: (Integer, Integer) -> Integer 3.61/1.88 != ~ Neq: (Integer, Integer) -> Boolean 3.61/1.88 && ~ Land: (Boolean, Boolean) -> Boolean 3.61/1.88 ! ~ Lnot: (Boolean) -> Boolean 3.61/1.88 = ~ Eq: (Integer, Integer) -> Boolean 3.61/1.88 <= ~ Le: (Integer, Integer) -> Boolean 3.61/1.88 ^ ~ Bwxor: (Integer, Integer) -> Integer 3.61/1.88 % ~ Mod: (Integer, Integer) -> Integer 3.61/1.88 > ~ Gt: (Integer, Integer) -> Boolean 3.61/1.88 + ~ Add: (Integer, Integer) -> Integer 3.61/1.88 -1 ~ UnaryMinus: (Integer) -> Integer 3.61/1.88 < ~ Lt: (Integer, Integer) -> Boolean 3.61/1.88 || ~ Lor: (Boolean, Boolean) -> Boolean 3.61/1.88 - ~ Sub: (Integer, Integer) -> Integer 3.61/1.88 ~ ~ Bwnot: (Integer) -> Integer 3.61/1.88 * ~ Mul: (Integer, Integer) -> Integer 3.61/1.88 >>> 3.61/1.88 3.61/1.88 3.61/1.88 The following domains are used: 3.61/1.88 Boolean, Integer 3.61/1.88 3.61/1.88 The ITRS R consists of the following rules: 3.61/1.88 eval(x) -> Cond_eval(x % 2 > 0 && x > 0, x) 3.61/1.88 Cond_eval(TRUE, x) -> eval(x - 1) 3.61/1.88 3.61/1.88 The integer pair graph contains the following rules and edges: 3.61/1.88 (0): EVAL(x[0]) -> COND_EVAL(x[0] % 2 > 0 && x[0] > 0, x[0]) 3.61/1.88 (1): COND_EVAL(TRUE, x[1]) -> EVAL(x[1] - 1) 3.61/1.88 3.61/1.88 (0) -> (1), if (x[0] % 2 > 0 && x[0] > 0 & x[0] ->^* x[1]) 3.61/1.88 (1) -> (0), if (x[1] - 1 ->^* x[0]) 3.61/1.88 3.61/1.88 The set Q consists of the following terms: 3.61/1.88 eval(x0) 3.61/1.88 Cond_eval(TRUE, x0) 3.61/1.88 3.61/1.88 ---------------------------------------- 3.61/1.88 3.61/1.88 (3) UsableRulesProof (EQUIVALENT) 3.61/1.88 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. 3.61/1.88 ---------------------------------------- 3.61/1.88 3.61/1.88 (4) 3.61/1.88 Obligation: 3.61/1.88 IDP problem: 3.61/1.88 The following function symbols are pre-defined: 3.61/1.88 <<< 3.61/1.88 & ~ Bwand: (Integer, Integer) -> Integer 3.61/1.88 >= ~ Ge: (Integer, Integer) -> Boolean 3.61/1.88 | ~ Bwor: (Integer, Integer) -> Integer 3.61/1.88 / ~ Div: (Integer, Integer) -> Integer 3.61/1.88 != ~ Neq: (Integer, Integer) -> Boolean 3.61/1.88 && ~ Land: (Boolean, Boolean) -> Boolean 3.61/1.88 ! ~ Lnot: (Boolean) -> Boolean 3.61/1.88 = ~ Eq: (Integer, Integer) -> Boolean 3.61/1.88 <= ~ Le: (Integer, Integer) -> Boolean 3.61/1.88 ^ ~ Bwxor: (Integer, Integer) -> Integer 3.61/1.88 % ~ Mod: (Integer, Integer) -> Integer 3.61/1.88 > ~ Gt: (Integer, Integer) -> Boolean 3.61/1.88 + ~ Add: (Integer, Integer) -> Integer 3.61/1.88 -1 ~ UnaryMinus: (Integer) -> Integer 3.61/1.88 < ~ Lt: (Integer, Integer) -> Boolean 3.61/1.88 || ~ Lor: (Boolean, Boolean) -> Boolean 3.61/1.88 - ~ Sub: (Integer, Integer) -> Integer 3.61/1.88 ~ ~ Bwnot: (Integer) -> Integer 3.61/1.88 * ~ Mul: (Integer, Integer) -> Integer 3.61/1.88 >>> 3.61/1.88 3.61/1.88 3.61/1.88 The following domains are used: 3.61/1.88 Boolean, Integer 3.61/1.88 3.61/1.88 R is empty. 3.61/1.88 3.61/1.88 The integer pair graph contains the following rules and edges: 3.61/1.88 (0): EVAL(x[0]) -> COND_EVAL(x[0] % 2 > 0 && x[0] > 0, x[0]) 3.61/1.88 (1): COND_EVAL(TRUE, x[1]) -> EVAL(x[1] - 1) 3.61/1.88 3.61/1.88 (0) -> (1), if (x[0] % 2 > 0 && x[0] > 0 & x[0] ->^* x[1]) 3.61/1.88 (1) -> (0), if (x[1] - 1 ->^* x[0]) 3.61/1.88 3.61/1.88 The set Q consists of the following terms: 3.61/1.88 eval(x0) 3.61/1.88 Cond_eval(TRUE, x0) 3.61/1.88 3.61/1.88 ---------------------------------------- 3.61/1.88 3.61/1.88 (5) IDPNonInfProof (SOUND) 3.61/1.88 Used the following options for this NonInfProof: 3.61/1.88 3.61/1.88 IDPGPoloSolver: 3.61/1.88 Range: [(-1,2)] 3.61/1.88 IsNat: false 3.61/1.88 Interpretation Shape Heuristic: aprove.DPFramework.IDPProblem.Processors.nonInf.poly.IdpDefaultShapeHeuristic@404bf09f 3.61/1.88 Constraint Generator: NonInfConstraintGenerator: 3.61/1.88 PathGenerator: MetricPathGenerator: 3.61/1.88 Max Left Steps: 1 3.61/1.88 Max Right Steps: 1 3.61/1.88 3.61/1.88 3.61/1.88 3.61/1.88 The constraints were generated the following way: 3.61/1.88 3.61/1.88 The DP Problem is simplified using the Induction Calculus [NONINF] with the following steps: 3.61/1.89 3.61/1.89 Note that final constraints are written in bold face. 3.61/1.89 3.61/1.89 3.61/1.89 3.61/1.89 For Pair EVAL(x) -> COND_EVAL(&&(>(%(x, 2), 0), >(x, 0)), x) the following chains were created: 3.61/1.89 *We consider the chain EVAL(x[0]) -> COND_EVAL(&&(>(%(x[0], 2), 0), >(x[0], 0)), x[0]), COND_EVAL(TRUE, x[1]) -> EVAL(-(x[1], 1)) which results in the following constraint: 3.61/1.89 3.61/1.89 (1) (&&(>(%(x[0], 2), 0), >(x[0], 0))=TRUE & x[0]=x[1] ==> EVAL(x[0])_>=_NonInfC & EVAL(x[0])_>=_COND_EVAL(&&(>(%(x[0], 2), 0), >(x[0], 0)), x[0]) & (U^Increasing(COND_EVAL(&&(>(%(x[0], 2), 0), >(x[0], 0)), x[0])), >=)) 3.61/1.89 3.61/1.89 3.61/1.89 3.61/1.89 We simplified constraint (1) using rules (IV), (IDP_BOOLEAN) which results in the following new constraint: 3.61/1.89 3.61/1.89 (2) (>(%(x[0], 2), 0)=TRUE & >(x[0], 0)=TRUE ==> EVAL(x[0])_>=_NonInfC & EVAL(x[0])_>=_COND_EVAL(&&(>(%(x[0], 2), 0), >(x[0], 0)), x[0]) & (U^Increasing(COND_EVAL(&&(>(%(x[0], 2), 0), >(x[0], 0)), x[0])), >=)) 3.61/1.89 3.61/1.89 3.61/1.89 3.61/1.89 We simplified constraint (2) using rule (POLY_CONSTRAINTS) which results in the following new constraint: 3.61/1.89 3.61/1.89 (3) (max{[2], [-2]} + [-1] >= 0 & x[0] + [-1] >= 0 ==> (U^Increasing(COND_EVAL(&&(>(%(x[0], 2), 0), >(x[0], 0)), x[0])), >=) & [(-1)Bound*bni_10] + [(2)bni_10]x[0] >= 0 & [(-1)bso_11] >= 0) 3.61/1.89 3.61/1.89 3.61/1.89 3.61/1.89 We simplified constraint (3) using rule (IDP_POLY_SIMPLIFY) which results in the following new constraint: 3.61/1.89 3.61/1.89 (4) (max{[2], [-2]} + [-1] >= 0 & x[0] + [-1] >= 0 ==> (U^Increasing(COND_EVAL(&&(>(%(x[0], 2), 0), >(x[0], 0)), x[0])), >=) & [(-1)Bound*bni_10] + [(2)bni_10]x[0] >= 0 & [(-1)bso_11] >= 0) 3.61/1.89 3.61/1.89 3.61/1.89 3.61/1.89 We simplified constraint (4) using rule (POLY_REMOVE_MIN_MAX) which results in the following new constraint: 3.61/1.89 3.61/1.89 (5) (x[0] + [-1] >= 0 & [4] >= 0 & [1] >= 0 ==> (U^Increasing(COND_EVAL(&&(>(%(x[0], 2), 0), >(x[0], 0)), x[0])), >=) & [(-1)Bound*bni_10] + [(2)bni_10]x[0] >= 0 & [(-1)bso_11] >= 0) 3.61/1.89 3.61/1.89 3.61/1.89 3.61/1.89 We simplified constraint (5) using rule (IDP_POLY_GCD) which results in the following new constraint: 3.61/1.89 3.61/1.89 (6) (x[0] + [-1] >= 0 & [1] >= 0 & [1] >= 0 ==> (U^Increasing(COND_EVAL(&&(>(%(x[0], 2), 0), >(x[0], 0)), x[0])), >=) & [(-1)Bound*bni_10] + [(2)bni_10]x[0] >= 0 & [(-1)bso_11] >= 0) 3.61/1.89 3.61/1.89 3.61/1.89 3.61/1.89 3.61/1.89 3.61/1.89 3.61/1.89 3.61/1.89 3.61/1.89 For Pair COND_EVAL(TRUE, x) -> EVAL(-(x, 1)) the following chains were created: 3.61/1.89 *We consider the chain EVAL(x[0]) -> COND_EVAL(&&(>(%(x[0], 2), 0), >(x[0], 0)), x[0]), COND_EVAL(TRUE, x[1]) -> EVAL(-(x[1], 1)), EVAL(x[0]) -> COND_EVAL(&&(>(%(x[0], 2), 0), >(x[0], 0)), x[0]) which results in the following constraint: 3.61/1.89 3.61/1.89 (1) (&&(>(%(x[0], 2), 0), >(x[0], 0))=TRUE & x[0]=x[1] & -(x[1], 1)=x[0]1 ==> COND_EVAL(TRUE, x[1])_>=_NonInfC & COND_EVAL(TRUE, x[1])_>=_EVAL(-(x[1], 1)) & (U^Increasing(EVAL(-(x[1], 1))), >=)) 3.61/1.89 3.61/1.89 3.61/1.89 3.61/1.89 We simplified constraint (1) using rules (III), (IV), (IDP_BOOLEAN) which results in the following new constraint: 3.61/1.89 3.61/1.89 (2) (>(%(x[0], 2), 0)=TRUE & >(x[0], 0)=TRUE ==> COND_EVAL(TRUE, x[0])_>=_NonInfC & COND_EVAL(TRUE, x[0])_>=_EVAL(-(x[0], 1)) & (U^Increasing(EVAL(-(x[1], 1))), >=)) 3.61/1.89 3.61/1.89 3.61/1.89 3.61/1.89 We simplified constraint (2) using rule (POLY_CONSTRAINTS) which results in the following new constraint: 3.61/1.89 3.61/1.89 (3) (max{[2], [-2]} + [-1] >= 0 & x[0] + [-1] >= 0 ==> (U^Increasing(EVAL(-(x[1], 1))), >=) & [(-1)bni_12 + (-1)Bound*bni_12] + [(2)bni_12]x[0] >= 0 & [1 + (-1)bso_13] >= 0) 3.61/1.89 3.61/1.89 3.61/1.89 3.61/1.89 We simplified constraint (3) using rule (IDP_POLY_SIMPLIFY) which results in the following new constraint: 3.61/1.89 3.61/1.89 (4) (max{[2], [-2]} + [-1] >= 0 & x[0] + [-1] >= 0 ==> (U^Increasing(EVAL(-(x[1], 1))), >=) & [(-1)bni_12 + (-1)Bound*bni_12] + [(2)bni_12]x[0] >= 0 & [1 + (-1)bso_13] >= 0) 3.61/1.89 3.61/1.89 3.61/1.89 3.61/1.89 We simplified constraint (4) using rule (POLY_REMOVE_MIN_MAX) which results in the following new constraint: 3.61/1.89 3.61/1.89 (5) (x[0] + [-1] >= 0 & [4] >= 0 & [1] >= 0 ==> (U^Increasing(EVAL(-(x[1], 1))), >=) & [(-1)bni_12 + (-1)Bound*bni_12] + [(2)bni_12]x[0] >= 0 & [1 + (-1)bso_13] >= 0) 3.61/1.89 3.61/1.89 3.61/1.89 3.61/1.89 We simplified constraint (5) using rule (IDP_POLY_GCD) which results in the following new constraint: 3.61/1.89 3.61/1.89 (6) (x[0] + [-1] >= 0 & [1] >= 0 & [1] >= 0 ==> (U^Increasing(EVAL(-(x[1], 1))), >=) & [(-1)bni_12 + (-1)Bound*bni_12] + [(2)bni_12]x[0] >= 0 & [1 + (-1)bso_13] >= 0) 3.61/1.89 3.61/1.89 3.61/1.89 3.61/1.89 3.61/1.89 3.61/1.89 3.61/1.89 3.61/1.89 3.61/1.89 To summarize, we get the following constraints P__>=_ for the following pairs. 3.61/1.89 3.61/1.89 *EVAL(x) -> COND_EVAL(&&(>(%(x, 2), 0), >(x, 0)), x) 3.61/1.89 3.61/1.89 *(x[0] + [-1] >= 0 & [1] >= 0 & [1] >= 0 ==> (U^Increasing(COND_EVAL(&&(>(%(x[0], 2), 0), >(x[0], 0)), x[0])), >=) & [(-1)Bound*bni_10] + [(2)bni_10]x[0] >= 0 & [(-1)bso_11] >= 0) 3.61/1.89 3.61/1.89 3.61/1.89 3.61/1.89 3.61/1.89 *COND_EVAL(TRUE, x) -> EVAL(-(x, 1)) 3.61/1.89 3.61/1.89 *(x[0] + [-1] >= 0 & [1] >= 0 & [1] >= 0 ==> (U^Increasing(EVAL(-(x[1], 1))), >=) & [(-1)bni_12 + (-1)Bound*bni_12] + [(2)bni_12]x[0] >= 0 & [1 + (-1)bso_13] >= 0) 3.61/1.89 3.61/1.89 3.61/1.89 3.61/1.89 3.61/1.89 3.61/1.89 3.61/1.89 3.61/1.89 3.61/1.89 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. 3.61/1.89 3.61/1.89 Using the following integer polynomial ordering the resulting constraints can be solved 3.61/1.89 3.61/1.89 Polynomial interpretation over integers[POLO]: 3.61/1.89 3.61/1.89 POL(TRUE) = 0 3.61/1.89 POL(FALSE) = [1] 3.61/1.89 POL(EVAL(x_1)) = [2]x_1 3.61/1.89 POL(COND_EVAL(x_1, x_2)) = [-1] + [2]x_2 + [-1]x_1 3.61/1.89 POL(&&(x_1, x_2)) = [-1] 3.61/1.89 POL(>(x_1, x_2)) = [-1] 3.61/1.89 POL(2) = [2] 3.61/1.89 POL(0) = 0 3.61/1.89 POL(-(x_1, x_2)) = x_1 + [-1]x_2 3.61/1.89 POL(1) = [1] 3.61/1.89 3.61/1.89 Polynomial Interpretations with Context Sensitive Arithemetic Replacement 3.61/1.89 POL(Term^CSAR-Mode @ Context) 3.61/1.89 3.61/1.89 POL(%(x_1, 2)^1 @ {}) = max{x_2, [-1]x_2} 3.61/1.89 3.61/1.89 3.61/1.89 The following pairs are in P_>: 3.61/1.89 3.61/1.89 3.61/1.89 COND_EVAL(TRUE, x[1]) -> EVAL(-(x[1], 1)) 3.61/1.89 3.61/1.89 3.61/1.89 The following pairs are in P_bound: 3.61/1.89 3.61/1.89 3.61/1.89 EVAL(x[0]) -> COND_EVAL(&&(>(%(x[0], 2), 0), >(x[0], 0)), x[0]) 3.61/1.89 COND_EVAL(TRUE, x[1]) -> EVAL(-(x[1], 1)) 3.61/1.89 3.61/1.89 3.61/1.89 The following pairs are in P_>=: 3.61/1.89 3.61/1.89 3.61/1.89 EVAL(x[0]) -> COND_EVAL(&&(>(%(x[0], 2), 0), >(x[0], 0)), x[0]) 3.61/1.89 3.61/1.89 3.61/1.89 At least the following rules have been oriented under context sensitive arithmetic replacement: 3.61/1.89 3.61/1.89 TRUE^1 -> &&(TRUE, TRUE)^1 3.61/1.89 FALSE^1 -> &&(TRUE, FALSE)^1 3.61/1.89 FALSE^1 -> &&(FALSE, TRUE)^1 3.61/1.89 FALSE^1 -> &&(FALSE, FALSE)^1 3.61/1.89 3.61/1.89 ---------------------------------------- 3.61/1.89 3.61/1.89 (6) 3.61/1.89 Obligation: 3.61/1.89 IDP problem: 3.61/1.89 The following function symbols are pre-defined: 3.61/1.89 <<< 3.61/1.89 & ~ Bwand: (Integer, Integer) -> Integer 3.61/1.89 >= ~ Ge: (Integer, Integer) -> Boolean 3.61/1.89 | ~ Bwor: (Integer, Integer) -> Integer 3.61/1.89 / ~ Div: (Integer, Integer) -> Integer 3.61/1.89 != ~ Neq: (Integer, Integer) -> Boolean 3.61/1.89 && ~ Land: (Boolean, Boolean) -> Boolean 3.61/1.89 ! ~ Lnot: (Boolean) -> Boolean 3.61/1.89 = ~ Eq: (Integer, Integer) -> Boolean 3.61/1.89 <= ~ Le: (Integer, Integer) -> Boolean 3.61/1.89 ^ ~ Bwxor: (Integer, Integer) -> Integer 3.61/1.89 % ~ Mod: (Integer, Integer) -> Integer 3.61/1.89 > ~ Gt: (Integer, Integer) -> Boolean 3.61/1.89 + ~ Add: (Integer, Integer) -> Integer 3.61/1.89 -1 ~ UnaryMinus: (Integer) -> Integer 3.61/1.89 < ~ Lt: (Integer, Integer) -> Boolean 3.61/1.89 || ~ Lor: (Boolean, Boolean) -> Boolean 3.61/1.89 - ~ Sub: (Integer, Integer) -> Integer 3.61/1.89 ~ ~ Bwnot: (Integer) -> Integer 3.61/1.89 * ~ Mul: (Integer, Integer) -> Integer 3.61/1.89 >>> 3.61/1.89 3.61/1.89 3.61/1.89 The following domains are used: 3.61/1.89 Boolean, Integer 3.61/1.89 3.61/1.89 R is empty. 3.61/1.89 3.61/1.89 The integer pair graph contains the following rules and edges: 3.61/1.89 (0): EVAL(x[0]) -> COND_EVAL(x[0] % 2 > 0 && x[0] > 0, x[0]) 3.61/1.89 3.61/1.89 3.61/1.89 The set Q consists of the following terms: 3.61/1.89 eval(x0) 3.61/1.89 Cond_eval(TRUE, x0) 3.61/1.89 3.61/1.89 ---------------------------------------- 3.61/1.89 3.61/1.89 (7) IDependencyGraphProof (EQUIVALENT) 3.61/1.89 The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 1 less node. 3.61/1.89 ---------------------------------------- 3.61/1.89 3.61/1.89 (8) 3.61/1.89 TRUE 3.61/1.90 EOF