3.77/1.87 YES 3.77/1.88 proof of /export/starexec/sandbox/benchmark/theBenchmark.itrs 3.77/1.88 # AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty 3.77/1.88 3.77/1.88 3.77/1.88 Termination of the given ITRS could be proven: 3.77/1.88 3.77/1.88 (0) ITRS 3.77/1.88 (1) ITRStoIDPProof [EQUIVALENT, 0 ms] 3.77/1.88 (2) IDP 3.77/1.88 (3) UsableRulesProof [EQUIVALENT, 0 ms] 3.77/1.88 (4) IDP 3.77/1.88 (5) IDPNonInfProof [SOUND, 131 ms] 3.77/1.88 (6) IDP 3.77/1.88 (7) PisEmptyProof [EQUIVALENT, 0 ms] 3.77/1.88 (8) YES 3.77/1.88 3.77/1.88 3.77/1.88 ---------------------------------------- 3.77/1.88 3.77/1.88 (0) 3.77/1.88 Obligation: 3.77/1.88 ITRS problem: 3.77/1.88 3.77/1.88 The following function symbols are pre-defined: 3.77/1.88 <<< 3.77/1.88 & ~ Bwand: (Integer, Integer) -> Integer 3.77/1.88 >= ~ Ge: (Integer, Integer) -> Boolean 3.77/1.88 | ~ Bwor: (Integer, Integer) -> Integer 3.77/1.88 / ~ Div: (Integer, Integer) -> Integer 3.77/1.88 != ~ Neq: (Integer, Integer) -> Boolean 3.77/1.88 && ~ Land: (Boolean, Boolean) -> Boolean 3.77/1.88 ! ~ Lnot: (Boolean) -> Boolean 3.77/1.88 = ~ Eq: (Integer, Integer) -> Boolean 3.77/1.88 <= ~ Le: (Integer, Integer) -> Boolean 3.77/1.88 ^ ~ Bwxor: (Integer, Integer) -> Integer 3.77/1.88 % ~ Mod: (Integer, Integer) -> Integer 3.77/1.88 + ~ Add: (Integer, Integer) -> Integer 3.77/1.88 > ~ Gt: (Integer, Integer) -> Boolean 3.77/1.88 -1 ~ UnaryMinus: (Integer) -> Integer 3.77/1.88 < ~ Lt: (Integer, Integer) -> Boolean 3.77/1.88 || ~ Lor: (Boolean, Boolean) -> Boolean 3.77/1.88 - ~ Sub: (Integer, Integer) -> Integer 3.77/1.88 ~ ~ Bwnot: (Integer) -> Integer 3.77/1.88 * ~ Mul: (Integer, Integer) -> Integer 3.77/1.88 >>> 3.77/1.88 3.77/1.88 The TRS R consists of the following rules: 3.77/1.88 f(TRUE, x) -> f(1000 >= x, x + 1) 3.77/1.88 The set Q consists of the following terms: 3.77/1.88 f(TRUE, x0) 3.77/1.88 3.77/1.88 ---------------------------------------- 3.77/1.88 3.77/1.88 (1) ITRStoIDPProof (EQUIVALENT) 3.77/1.88 Added dependency pairs 3.77/1.88 ---------------------------------------- 3.77/1.88 3.77/1.88 (2) 3.77/1.88 Obligation: 3.77/1.88 IDP problem: 3.77/1.88 The following function symbols are pre-defined: 3.77/1.88 <<< 3.77/1.88 & ~ Bwand: (Integer, Integer) -> Integer 3.77/1.88 >= ~ Ge: (Integer, Integer) -> Boolean 3.77/1.88 | ~ Bwor: (Integer, Integer) -> Integer 3.77/1.88 / ~ Div: (Integer, Integer) -> Integer 3.77/1.88 != ~ Neq: (Integer, Integer) -> Boolean 3.77/1.88 && ~ Land: (Boolean, Boolean) -> Boolean 3.77/1.88 ! ~ Lnot: (Boolean) -> Boolean 3.77/1.88 = ~ Eq: (Integer, Integer) -> Boolean 3.77/1.88 <= ~ Le: (Integer, Integer) -> Boolean 3.77/1.88 ^ ~ Bwxor: (Integer, Integer) -> Integer 3.77/1.88 % ~ Mod: (Integer, Integer) -> Integer 3.77/1.88 + ~ Add: (Integer, Integer) -> Integer 3.77/1.88 > ~ Gt: (Integer, Integer) -> Boolean 3.77/1.88 -1 ~ UnaryMinus: (Integer) -> Integer 3.77/1.88 < ~ Lt: (Integer, Integer) -> Boolean 3.77/1.88 || ~ Lor: (Boolean, Boolean) -> Boolean 3.77/1.88 - ~ Sub: (Integer, Integer) -> Integer 3.77/1.88 ~ ~ Bwnot: (Integer) -> Integer 3.77/1.88 * ~ Mul: (Integer, Integer) -> Integer 3.77/1.88 >>> 3.77/1.88 3.77/1.88 3.77/1.88 The following domains are used: 3.77/1.88 Integer 3.77/1.88 3.77/1.88 The ITRS R consists of the following rules: 3.77/1.88 f(TRUE, x) -> f(1000 >= x, x + 1) 3.77/1.88 3.77/1.88 The integer pair graph contains the following rules and edges: 3.77/1.88 (0): F(TRUE, x[0]) -> F(1000 >= x[0], x[0] + 1) 3.77/1.88 3.77/1.88 (0) -> (0), if (1000 >= x[0] & x[0] + 1 ->^* x[0]') 3.77/1.88 3.77/1.88 The set Q consists of the following terms: 3.77/1.88 f(TRUE, x0) 3.77/1.88 3.77/1.88 ---------------------------------------- 3.77/1.88 3.77/1.88 (3) UsableRulesProof (EQUIVALENT) 3.77/1.88 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. 3.77/1.88 ---------------------------------------- 3.77/1.88 3.77/1.88 (4) 3.77/1.88 Obligation: 3.77/1.88 IDP problem: 3.77/1.88 The following function symbols are pre-defined: 3.77/1.88 <<< 3.77/1.88 & ~ Bwand: (Integer, Integer) -> Integer 3.77/1.88 >= ~ Ge: (Integer, Integer) -> Boolean 3.77/1.88 | ~ Bwor: (Integer, Integer) -> Integer 3.77/1.88 / ~ Div: (Integer, Integer) -> Integer 3.77/1.88 != ~ Neq: (Integer, Integer) -> Boolean 3.77/1.88 && ~ Land: (Boolean, Boolean) -> Boolean 3.77/1.88 ! ~ Lnot: (Boolean) -> Boolean 3.77/1.88 = ~ Eq: (Integer, Integer) -> Boolean 3.77/1.88 <= ~ Le: (Integer, Integer) -> Boolean 3.77/1.88 ^ ~ Bwxor: (Integer, Integer) -> Integer 3.77/1.88 % ~ Mod: (Integer, Integer) -> Integer 3.77/1.88 + ~ Add: (Integer, Integer) -> Integer 3.77/1.88 > ~ Gt: (Integer, Integer) -> Boolean 3.77/1.88 -1 ~ UnaryMinus: (Integer) -> Integer 3.77/1.88 < ~ Lt: (Integer, Integer) -> Boolean 3.77/1.88 || ~ Lor: (Boolean, Boolean) -> Boolean 3.77/1.88 - ~ Sub: (Integer, Integer) -> Integer 3.77/1.88 ~ ~ Bwnot: (Integer) -> Integer 3.77/1.88 * ~ Mul: (Integer, Integer) -> Integer 3.77/1.88 >>> 3.77/1.88 3.77/1.88 3.77/1.88 The following domains are used: 3.77/1.88 Integer 3.77/1.88 3.77/1.88 R is empty. 3.77/1.88 3.77/1.88 The integer pair graph contains the following rules and edges: 3.77/1.88 (0): F(TRUE, x[0]) -> F(1000 >= x[0], x[0] + 1) 3.77/1.88 3.77/1.88 (0) -> (0), if (1000 >= x[0] & x[0] + 1 ->^* x[0]') 3.77/1.88 3.77/1.88 The set Q consists of the following terms: 3.77/1.88 f(TRUE, x0) 3.77/1.88 3.77/1.88 ---------------------------------------- 3.77/1.88 3.77/1.88 (5) IDPNonInfProof (SOUND) 3.77/1.88 Used the following options for this NonInfProof: 3.77/1.88 3.77/1.88 IDPGPoloSolver: 3.77/1.88 Range: [(-1,2)] 3.77/1.88 IsNat: false 3.77/1.88 Interpretation Shape Heuristic: aprove.DPFramework.IDPProblem.Processors.nonInf.poly.IdpDefaultShapeHeuristic@63ec25cd 3.77/1.88 Constraint Generator: NonInfConstraintGenerator: 3.77/1.88 PathGenerator: MetricPathGenerator: 3.77/1.88 Max Left Steps: 1 3.77/1.88 Max Right Steps: 1 3.77/1.88 3.77/1.88 3.77/1.88 3.77/1.88 The constraints were generated the following way: 3.77/1.88 3.77/1.88 The DP Problem is simplified using the Induction Calculus [NONINF] with the following steps: 3.77/1.88 3.77/1.88 Note that final constraints are written in bold face. 3.77/1.88 3.77/1.88 3.77/1.88 3.77/1.88 For Pair F(TRUE, x) -> F(>=(1000, x), +(x, 1)) the following chains were created: 3.77/1.88 *We consider the chain F(TRUE, x[0]) -> F(>=(1000, x[0]), +(x[0], 1)), F(TRUE, x[0]) -> F(>=(1000, x[0]), +(x[0], 1)), F(TRUE, x[0]) -> F(>=(1000, x[0]), +(x[0], 1)) which results in the following constraint: 3.77/1.88 3.77/1.88 (1) (>=(1000, x[0])=TRUE & +(x[0], 1)=x[0]1 & >=(1000, x[0]1)=TRUE & +(x[0]1, 1)=x[0]2 ==> F(TRUE, x[0]1)_>=_NonInfC & F(TRUE, x[0]1)_>=_F(>=(1000, x[0]1), +(x[0]1, 1)) & (U^Increasing(F(>=(1000, x[0]1), +(x[0]1, 1))), >=)) 3.77/1.88 3.77/1.88 3.77/1.88 3.77/1.88 We simplified constraint (1) using rules (III), (IV) which results in the following new constraint: 3.77/1.88 3.77/1.88 (2) (>=(1000, x[0])=TRUE & >=(1000, +(x[0], 1))=TRUE ==> F(TRUE, +(x[0], 1))_>=_NonInfC & F(TRUE, +(x[0], 1))_>=_F(>=(1000, +(x[0], 1)), +(+(x[0], 1), 1)) & (U^Increasing(F(>=(1000, x[0]1), +(x[0]1, 1))), >=)) 3.77/1.88 3.77/1.88 3.77/1.88 3.77/1.88 We simplified constraint (2) using rule (POLY_CONSTRAINTS) which results in the following new constraint: 3.77/1.88 3.77/1.88 (3) ([1000] + [-1]x[0] >= 0 & [999] + [-1]x[0] >= 0 ==> (U^Increasing(F(>=(1000, x[0]1), +(x[0]1, 1))), >=) & [bni_7 + (-1)Bound*bni_7] + [(-1)bni_7]x[0] >= 0 & [1 + (-1)bso_8] >= 0) 3.77/1.88 3.77/1.88 3.77/1.88 3.77/1.88 We simplified constraint (3) using rule (IDP_POLY_SIMPLIFY) which results in the following new constraint: 3.77/1.88 3.77/1.88 (4) ([1000] + [-1]x[0] >= 0 & [999] + [-1]x[0] >= 0 ==> (U^Increasing(F(>=(1000, x[0]1), +(x[0]1, 1))), >=) & [bni_7 + (-1)Bound*bni_7] + [(-1)bni_7]x[0] >= 0 & [1 + (-1)bso_8] >= 0) 3.77/1.88 3.77/1.88 3.77/1.88 3.77/1.88 We simplified constraint (4) using rule (POLY_REMOVE_MIN_MAX) which results in the following new constraint: 3.77/1.88 3.77/1.88 (5) ([1000] + [-1]x[0] >= 0 & [999] + [-1]x[0] >= 0 ==> (U^Increasing(F(>=(1000, x[0]1), +(x[0]1, 1))), >=) & [bni_7 + (-1)Bound*bni_7] + [(-1)bni_7]x[0] >= 0 & [1 + (-1)bso_8] >= 0) 3.77/1.88 3.77/1.88 3.77/1.88 3.77/1.88 We simplified constraint (5) using rule (IDP_SMT_SPLIT) which results in the following new constraints: 3.77/1.88 3.77/1.88 (6) ([1000] + [-1]x[0] >= 0 & [999] + [-1]x[0] >= 0 & x[0] >= 0 ==> (U^Increasing(F(>=(1000, x[0]1), +(x[0]1, 1))), >=) & [bni_7 + (-1)Bound*bni_7] + [(-1)bni_7]x[0] >= 0 & [1 + (-1)bso_8] >= 0) 3.77/1.88 3.77/1.88 (7) ([1000] + x[0] >= 0 & [999] + x[0] >= 0 & x[0] >= 0 ==> (U^Increasing(F(>=(1000, x[0]1), +(x[0]1, 1))), >=) & [bni_7 + (-1)Bound*bni_7] + [bni_7]x[0] >= 0 & [1 + (-1)bso_8] >= 0) 3.77/1.88 3.77/1.88 3.77/1.88 3.77/1.88 3.77/1.88 3.77/1.88 3.77/1.88 3.77/1.88 3.77/1.88 To summarize, we get the following constraints P__>=_ for the following pairs. 3.77/1.88 3.77/1.88 *F(TRUE, x) -> F(>=(1000, x), +(x, 1)) 3.77/1.88 3.77/1.88 *([1000] + [-1]x[0] >= 0 & [999] + [-1]x[0] >= 0 & x[0] >= 0 ==> (U^Increasing(F(>=(1000, x[0]1), +(x[0]1, 1))), >=) & [bni_7 + (-1)Bound*bni_7] + [(-1)bni_7]x[0] >= 0 & [1 + (-1)bso_8] >= 0) 3.77/1.88 3.77/1.88 3.77/1.88 *([1000] + x[0] >= 0 & [999] + x[0] >= 0 & x[0] >= 0 ==> (U^Increasing(F(>=(1000, x[0]1), +(x[0]1, 1))), >=) & [bni_7 + (-1)Bound*bni_7] + [bni_7]x[0] >= 0 & [1 + (-1)bso_8] >= 0) 3.77/1.88 3.77/1.88 3.77/1.88 3.77/1.88 3.77/1.88 3.77/1.88 3.77/1.88 3.77/1.88 3.77/1.88 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. 3.77/1.88 3.77/1.88 Using the following integer polynomial ordering the resulting constraints can be solved 3.77/1.88 3.77/1.88 Polynomial interpretation over integers[POLO]: 3.77/1.88 3.77/1.88 POL(TRUE) = 0 3.77/1.88 POL(FALSE) = 0 3.77/1.88 POL(F(x_1, x_2)) = [2] + [-1]x_2 3.77/1.88 POL(>=(x_1, x_2)) = [-1] 3.77/1.88 POL(1000) = [1000] 3.77/1.88 POL(+(x_1, x_2)) = x_1 + x_2 3.77/1.88 POL(1) = [1] 3.77/1.88 3.77/1.88 3.77/1.88 The following pairs are in P_>: 3.77/1.88 3.77/1.88 3.77/1.88 F(TRUE, x[0]) -> F(>=(1000, x[0]), +(x[0], 1)) 3.77/1.88 3.77/1.88 3.77/1.88 The following pairs are in P_bound: 3.77/1.88 3.77/1.88 3.77/1.88 F(TRUE, x[0]) -> F(>=(1000, x[0]), +(x[0], 1)) 3.77/1.88 3.77/1.88 3.77/1.88 The following pairs are in P_>=: 3.77/1.88 3.77/1.88 none 3.77/1.88 3.77/1.88 3.77/1.88 There are no usable rules. 3.77/1.88 ---------------------------------------- 3.77/1.88 3.77/1.88 (6) 3.77/1.88 Obligation: 3.77/1.88 IDP problem: 3.77/1.88 The following function symbols are pre-defined: 3.77/1.88 <<< 3.77/1.88 & ~ Bwand: (Integer, Integer) -> Integer 3.77/1.88 >= ~ Ge: (Integer, Integer) -> Boolean 3.77/1.88 | ~ Bwor: (Integer, Integer) -> Integer 3.77/1.88 / ~ Div: (Integer, Integer) -> Integer 3.77/1.88 != ~ Neq: (Integer, Integer) -> Boolean 3.77/1.88 && ~ Land: (Boolean, Boolean) -> Boolean 3.77/1.88 ! ~ Lnot: (Boolean) -> Boolean 3.77/1.88 = ~ Eq: (Integer, Integer) -> Boolean 3.77/1.88 <= ~ Le: (Integer, Integer) -> Boolean 3.77/1.88 ^ ~ Bwxor: (Integer, Integer) -> Integer 3.77/1.88 % ~ Mod: (Integer, Integer) -> Integer 3.77/1.88 + ~ Add: (Integer, Integer) -> Integer 3.77/1.88 > ~ Gt: (Integer, Integer) -> Boolean 3.77/1.88 -1 ~ UnaryMinus: (Integer) -> Integer 3.77/1.88 < ~ Lt: (Integer, Integer) -> Boolean 3.77/1.88 || ~ Lor: (Boolean, Boolean) -> Boolean 3.77/1.88 - ~ Sub: (Integer, Integer) -> Integer 3.77/1.88 ~ ~ Bwnot: (Integer) -> Integer 3.77/1.88 * ~ Mul: (Integer, Integer) -> Integer 3.77/1.88 >>> 3.77/1.88 3.77/1.88 3.77/1.88 The following domains are used: 3.77/1.88 none 3.77/1.88 3.77/1.88 R is empty. 3.77/1.88 3.77/1.88 The integer pair graph is empty. 3.77/1.88 3.77/1.88 The set Q consists of the following terms: 3.77/1.88 f(TRUE, x0) 3.77/1.88 3.77/1.88 ---------------------------------------- 3.77/1.88 3.77/1.88 (7) PisEmptyProof (EQUIVALENT) 3.77/1.88 The TRS P is empty. Hence, there is no (P,Q,R) chain. 3.77/1.88 ---------------------------------------- 3.77/1.88 3.77/1.88 (8) 3.77/1.88 YES 3.85/1.90 EOF