8.42/6.57 YES 8.42/6.58 proof of /export/starexec/sandbox/benchmark/theBenchmark.itrs 8.42/6.58 # AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty 8.42/6.58 8.42/6.58 8.42/6.58 Termination of the given ITRS could be proven: 8.42/6.58 8.42/6.58 (0) ITRS 8.42/6.58 (1) ITRStoIDPProof [EQUIVALENT, 0 ms] 8.42/6.58 (2) IDP 8.42/6.58 (3) UsableRulesProof [EQUIVALENT, 0 ms] 8.42/6.58 (4) IDP 8.42/6.58 (5) IDependencyGraphProof [EQUIVALENT, 0 ms] 8.42/6.58 (6) IDP 8.42/6.58 (7) IDPNonInfProof [SOUND, 233 ms] 8.42/6.58 (8) IDP 8.42/6.58 (9) IDependencyGraphProof [EQUIVALENT, 0 ms] 8.42/6.58 (10) TRUE 8.42/6.58 8.42/6.58 8.42/6.58 ---------------------------------------- 8.42/6.58 8.42/6.58 (0) 8.42/6.58 Obligation: 8.42/6.58 ITRS problem: 8.42/6.58 8.42/6.58 The following function symbols are pre-defined: 8.42/6.58 <<< 8.42/6.58 & ~ Bwand: (Integer, Integer) -> Integer 8.42/6.58 >= ~ Ge: (Integer, Integer) -> Boolean 8.42/6.58 | ~ Bwor: (Integer, Integer) -> Integer 8.42/6.58 / ~ Div: (Integer, Integer) -> Integer 8.42/6.58 != ~ Neq: (Integer, Integer) -> Boolean 8.42/6.58 && ~ Land: (Boolean, Boolean) -> Boolean 8.42/6.58 ! ~ Lnot: (Boolean) -> Boolean 8.42/6.58 = ~ Eq: (Integer, Integer) -> Boolean 8.42/6.58 <= ~ Le: (Integer, Integer) -> Boolean 8.42/6.58 ^ ~ Bwxor: (Integer, Integer) -> Integer 8.42/6.58 % ~ Mod: (Integer, Integer) -> Integer 8.42/6.58 > ~ Gt: (Integer, Integer) -> Boolean 8.42/6.58 + ~ Add: (Integer, Integer) -> Integer 8.42/6.58 -1 ~ UnaryMinus: (Integer) -> Integer 8.42/6.58 < ~ Lt: (Integer, Integer) -> Boolean 8.42/6.58 || ~ Lor: (Boolean, Boolean) -> Boolean 8.42/6.58 - ~ Sub: (Integer, Integer) -> Integer 8.42/6.58 ~ ~ Bwnot: (Integer) -> Integer 8.42/6.58 * ~ Mul: (Integer, Integer) -> Integer 8.42/6.58 >>> 8.42/6.58 8.42/6.58 The TRS R consists of the following rules: 8.42/6.58 f(TRUE, x, y) -> fNat(x >= 0 && y >= 0, x, y) 8.42/6.58 fNat(TRUE, x, y) -> f(x > y, x, round(y + 1)) 8.42/6.58 round(x) -> if(x % 2 = 0, x, x + 1) 8.42/6.58 if(TRUE, u, v) -> u 8.42/6.58 if(FALSE, u, v) -> v 8.42/6.58 The set Q consists of the following terms: 8.42/6.58 f(TRUE, x0, x1) 8.42/6.58 fNat(TRUE, x0, x1) 8.42/6.58 round(x0) 8.42/6.58 if(TRUE, x0, x1) 8.42/6.58 if(FALSE, x0, x1) 8.42/6.58 8.42/6.58 ---------------------------------------- 8.42/6.58 8.42/6.58 (1) ITRStoIDPProof (EQUIVALENT) 8.42/6.58 Added dependency pairs 8.42/6.58 ---------------------------------------- 8.42/6.58 8.42/6.58 (2) 8.42/6.58 Obligation: 8.42/6.58 IDP problem: 8.42/6.58 The following function symbols are pre-defined: 8.42/6.58 <<< 8.42/6.58 & ~ Bwand: (Integer, Integer) -> Integer 8.42/6.58 >= ~ Ge: (Integer, Integer) -> Boolean 8.42/6.58 | ~ Bwor: (Integer, Integer) -> Integer 8.42/6.58 / ~ Div: (Integer, Integer) -> Integer 8.42/6.58 != ~ Neq: (Integer, Integer) -> Boolean 8.42/6.58 && ~ Land: (Boolean, Boolean) -> Boolean 8.42/6.58 ! ~ Lnot: (Boolean) -> Boolean 8.42/6.58 = ~ Eq: (Integer, Integer) -> Boolean 8.42/6.58 <= ~ Le: (Integer, Integer) -> Boolean 8.42/6.58 ^ ~ Bwxor: (Integer, Integer) -> Integer 8.42/6.58 % ~ Mod: (Integer, Integer) -> Integer 8.42/6.58 > ~ Gt: (Integer, Integer) -> Boolean 8.42/6.58 + ~ Add: (Integer, Integer) -> Integer 8.42/6.58 -1 ~ UnaryMinus: (Integer) -> Integer 8.42/6.58 < ~ Lt: (Integer, Integer) -> Boolean 8.42/6.58 || ~ Lor: (Boolean, Boolean) -> Boolean 8.42/6.58 - ~ Sub: (Integer, Integer) -> Integer 8.42/6.58 ~ ~ Bwnot: (Integer) -> Integer 8.42/6.58 * ~ Mul: (Integer, Integer) -> Integer 8.42/6.58 >>> 8.42/6.58 8.42/6.58 8.42/6.58 The following domains are used: 8.42/6.58 Boolean, Integer 8.42/6.58 8.42/6.58 The ITRS R consists of the following rules: 8.42/6.58 f(TRUE, x, y) -> fNat(x >= 0 && y >= 0, x, y) 8.42/6.58 fNat(TRUE, x, y) -> f(x > y, x, round(y + 1)) 8.42/6.58 round(x) -> if(x % 2 = 0, x, x + 1) 8.42/6.58 if(TRUE, u, v) -> u 8.42/6.58 if(FALSE, u, v) -> v 8.42/6.58 8.42/6.58 The integer pair graph contains the following rules and edges: 8.42/6.58 (0): F(TRUE, x[0], y[0]) -> FNAT(x[0] >= 0 && y[0] >= 0, x[0], y[0]) 8.42/6.58 (1): FNAT(TRUE, x[1], y[1]) -> F(x[1] > y[1], x[1], round(y[1] + 1)) 8.42/6.58 (2): FNAT(TRUE, x[2], y[2]) -> ROUND(y[2] + 1) 8.42/6.58 (3): ROUND(x[3]) -> IF(x[3] % 2 = 0, x[3], x[3] + 1) 8.42/6.58 8.42/6.58 (0) -> (1), if (x[0] >= 0 && y[0] >= 0 & x[0] ->^* x[1] & y[0] ->^* y[1]) 8.42/6.58 (0) -> (2), if (x[0] >= 0 && y[0] >= 0 & x[0] ->^* x[2] & y[0] ->^* y[2]) 8.42/6.58 (1) -> (0), if (x[1] > y[1] & x[1] ->^* x[0] & round(y[1] + 1) ->^* y[0]) 8.42/6.58 (2) -> (3), if (y[2] + 1 ->^* x[3]) 8.42/6.58 8.42/6.58 The set Q consists of the following terms: 8.42/6.58 f(TRUE, x0, x1) 8.42/6.58 fNat(TRUE, x0, x1) 8.42/6.58 round(x0) 8.42/6.58 if(TRUE, x0, x1) 8.42/6.58 if(FALSE, x0, x1) 8.42/6.58 8.42/6.58 ---------------------------------------- 8.42/6.58 8.42/6.58 (3) UsableRulesProof (EQUIVALENT) 8.42/6.58 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. 8.42/6.58 ---------------------------------------- 8.42/6.58 8.42/6.58 (4) 8.42/6.58 Obligation: 8.42/6.58 IDP problem: 8.42/6.58 The following function symbols are pre-defined: 8.42/6.58 <<< 8.42/6.58 & ~ Bwand: (Integer, Integer) -> Integer 8.42/6.59 >= ~ Ge: (Integer, Integer) -> Boolean 8.42/6.59 | ~ Bwor: (Integer, Integer) -> Integer 8.42/6.59 / ~ Div: (Integer, Integer) -> Integer 8.42/6.59 != ~ Neq: (Integer, Integer) -> Boolean 8.42/6.59 && ~ Land: (Boolean, Boolean) -> Boolean 8.42/6.59 ! ~ Lnot: (Boolean) -> Boolean 8.42/6.59 = ~ Eq: (Integer, Integer) -> Boolean 8.42/6.59 <= ~ Le: (Integer, Integer) -> Boolean 8.42/6.59 ^ ~ Bwxor: (Integer, Integer) -> Integer 8.42/6.59 % ~ Mod: (Integer, Integer) -> Integer 8.42/6.59 > ~ Gt: (Integer, Integer) -> Boolean 8.42/6.59 + ~ Add: (Integer, Integer) -> Integer 8.42/6.59 -1 ~ UnaryMinus: (Integer) -> Integer 8.42/6.59 < ~ Lt: (Integer, Integer) -> Boolean 8.42/6.59 || ~ Lor: (Boolean, Boolean) -> Boolean 8.42/6.59 - ~ Sub: (Integer, Integer) -> Integer 8.42/6.59 ~ ~ Bwnot: (Integer) -> Integer 8.42/6.59 * ~ Mul: (Integer, Integer) -> Integer 8.42/6.59 >>> 8.42/6.59 8.42/6.59 8.42/6.59 The following domains are used: 8.42/6.59 Integer, Boolean 8.42/6.59 8.42/6.59 The ITRS R consists of the following rules: 8.42/6.59 round(x) -> if(x % 2 = 0, x, x + 1) 8.42/6.59 if(TRUE, u, v) -> u 8.42/6.59 if(FALSE, u, v) -> v 8.42/6.59 8.42/6.59 The integer pair graph contains the following rules and edges: 8.42/6.59 (0): F(TRUE, x[0], y[0]) -> FNAT(x[0] >= 0 && y[0] >= 0, x[0], y[0]) 8.42/6.59 (1): FNAT(TRUE, x[1], y[1]) -> F(x[1] > y[1], x[1], round(y[1] + 1)) 8.42/6.59 (2): FNAT(TRUE, x[2], y[2]) -> ROUND(y[2] + 1) 8.42/6.59 (3): ROUND(x[3]) -> IF(x[3] % 2 = 0, x[3], x[3] + 1) 8.42/6.59 8.42/6.59 (0) -> (1), if (x[0] >= 0 && y[0] >= 0 & x[0] ->^* x[1] & y[0] ->^* y[1]) 8.42/6.59 (0) -> (2), if (x[0] >= 0 && y[0] >= 0 & x[0] ->^* x[2] & y[0] ->^* y[2]) 8.42/6.59 (1) -> (0), if (x[1] > y[1] & x[1] ->^* x[0] & round(y[1] + 1) ->^* y[0]) 8.42/6.59 (2) -> (3), if (y[2] + 1 ->^* x[3]) 8.42/6.59 8.42/6.59 The set Q consists of the following terms: 8.42/6.59 f(TRUE, x0, x1) 8.42/6.59 fNat(TRUE, x0, x1) 8.42/6.59 round(x0) 8.42/6.59 if(TRUE, x0, x1) 8.42/6.59 if(FALSE, x0, x1) 8.42/6.59 8.42/6.59 ---------------------------------------- 8.42/6.59 8.42/6.59 (5) IDependencyGraphProof (EQUIVALENT) 8.42/6.59 The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 2 less nodes. 8.42/6.59 ---------------------------------------- 8.42/6.59 8.42/6.59 (6) 8.42/6.59 Obligation: 8.42/6.59 IDP problem: 8.42/6.59 The following function symbols are pre-defined: 8.42/6.59 <<< 8.42/6.59 & ~ Bwand: (Integer, Integer) -> Integer 8.42/6.59 >= ~ Ge: (Integer, Integer) -> Boolean 8.42/6.59 | ~ Bwor: (Integer, Integer) -> Integer 8.42/6.59 / ~ Div: (Integer, Integer) -> Integer 8.42/6.59 != ~ Neq: (Integer, Integer) -> Boolean 8.42/6.59 && ~ Land: (Boolean, Boolean) -> Boolean 8.42/6.59 ! ~ Lnot: (Boolean) -> Boolean 8.42/6.59 = ~ Eq: (Integer, Integer) -> Boolean 8.42/6.59 <= ~ Le: (Integer, Integer) -> Boolean 8.42/6.59 ^ ~ Bwxor: (Integer, Integer) -> Integer 8.42/6.59 % ~ Mod: (Integer, Integer) -> Integer 8.42/6.59 > ~ Gt: (Integer, Integer) -> Boolean 8.42/6.59 + ~ Add: (Integer, Integer) -> Integer 8.42/6.59 -1 ~ UnaryMinus: (Integer) -> Integer 8.42/6.59 < ~ Lt: (Integer, Integer) -> Boolean 8.42/6.59 || ~ Lor: (Boolean, Boolean) -> Boolean 8.42/6.59 - ~ Sub: (Integer, Integer) -> Integer 8.42/6.59 ~ ~ Bwnot: (Integer) -> Integer 8.42/6.59 * ~ Mul: (Integer, Integer) -> Integer 8.42/6.59 >>> 8.42/6.59 8.42/6.59 8.42/6.59 The following domains are used: 8.42/6.59 Integer, Boolean 8.42/6.59 8.42/6.59 The ITRS R consists of the following rules: 8.42/6.59 round(x) -> if(x % 2 = 0, x, x + 1) 8.42/6.59 if(TRUE, u, v) -> u 8.42/6.59 if(FALSE, u, v) -> v 8.42/6.59 8.42/6.59 The integer pair graph contains the following rules and edges: 8.42/6.59 (1): FNAT(TRUE, x[1], y[1]) -> F(x[1] > y[1], x[1], round(y[1] + 1)) 8.42/6.59 (0): F(TRUE, x[0], y[0]) -> FNAT(x[0] >= 0 && y[0] >= 0, x[0], y[0]) 8.42/6.59 8.42/6.59 (1) -> (0), if (x[1] > y[1] & x[1] ->^* x[0] & round(y[1] + 1) ->^* y[0]) 8.42/6.59 (0) -> (1), if (x[0] >= 0 && y[0] >= 0 & x[0] ->^* x[1] & y[0] ->^* y[1]) 8.42/6.59 8.42/6.59 The set Q consists of the following terms: 8.42/6.59 f(TRUE, x0, x1) 8.42/6.59 fNat(TRUE, x0, x1) 8.42/6.59 round(x0) 8.42/6.59 if(TRUE, x0, x1) 8.42/6.59 if(FALSE, x0, x1) 8.42/6.59 8.42/6.59 ---------------------------------------- 8.42/6.59 8.42/6.59 (7) IDPNonInfProof (SOUND) 8.42/6.59 Used the following options for this NonInfProof: 8.42/6.59 8.42/6.59 IDPGPoloSolver: 8.42/6.59 Range: [(-1,2)] 8.42/6.59 IsNat: false 8.42/6.59 Interpretation Shape Heuristic: aprove.DPFramework.IDPProblem.Processors.nonInf.poly.IdpCand1ShapeHeuristic@5853ba38 8.42/6.59 Constraint Generator: NonInfConstraintGenerator: 8.42/6.59 PathGenerator: MetricPathGenerator: 8.42/6.59 Max Left Steps: 2 8.42/6.59 Max Right Steps: 1 8.42/6.59 8.42/6.59 8.42/6.59 8.42/6.59 The constraints were generated the following way: 8.42/6.59 8.42/6.59 The DP Problem is simplified using the Induction Calculus [NONINF] with the following steps: 8.42/6.59 8.42/6.59 Note that final constraints are written in bold face. 8.42/6.59 8.42/6.59 8.42/6.59 8.42/6.59 For Pair FNAT(TRUE, x[1], y[1]) -> F(>(x[1], y[1]), x[1], round(+(y[1], 1))) the following chains were created: 8.42/6.59 *We consider the chain FNAT(TRUE, x[1], y[1]) -> F(>(x[1], y[1]), x[1], round(+(y[1], 1))), F(TRUE, x[0], y[0]) -> FNAT(&&(>=(x[0], 0), >=(y[0], 0)), x[0], y[0]), FNAT(TRUE, x[1], y[1]) -> F(>(x[1], y[1]), x[1], round(+(y[1], 1))), F(TRUE, x[0], y[0]) -> FNAT(&&(>=(x[0], 0), >=(y[0], 0)), x[0], y[0]) which results in the following constraint: 8.42/6.59 8.42/6.59 (1) (>(x[1], y[1])=TRUE & x[1]=x[0] & round(+(y[1], 1))=y[0] & &&(>=(x[0], 0), >=(y[0], 0))=TRUE & x[0]=x[1]1 & y[0]=y[1]1 & >(x[1]1, y[1]1)=TRUE & x[1]1=x[0]1 & round(+(y[1]1, 1))=y[0]1 ==> FNAT(TRUE, x[1]1, y[1]1)_>=_NonInfC & FNAT(TRUE, x[1]1, y[1]1)_>=_F(>(x[1]1, y[1]1), x[1]1, round(+(y[1]1, 1))) & (U^Increasing(F(>(x[1]1, y[1]1), x[1]1, round(+(y[1]1, 1)))), >=)) 8.42/6.59 8.42/6.59 8.42/6.59 8.42/6.59 We simplified constraint (1) using rules (III), (IV), (VII), (IDP_BOOLEAN), (REWRITING) which results in the following new constraint: 8.42/6.59 8.42/6.59 (2) (>(x[1], y[1])=TRUE & >(x[1], y[0])=TRUE & =(%(+(y[1], 1), 2), 0)=x0 & +(y[1], 1)=x1 & +(+(y[1], 1), 1)=x2 & if(x0, x1, x2)=y[0] & >=(x[1], 0)=TRUE & >=(y[0], 0)=TRUE ==> FNAT(TRUE, x[1], y[0])_>=_NonInfC & FNAT(TRUE, x[1], y[0])_>=_F(>(x[1], y[0]), x[1], round(+(y[0], 1))) & (U^Increasing(F(>(x[1]1, y[1]1), x[1]1, round(+(y[1]1, 1)))), >=)) 8.42/6.59 8.42/6.59 8.42/6.59 8.42/6.59 We simplified constraint (2) using rule (POLY_CONSTRAINTS) which results in the following new constraint: 8.42/6.59 8.42/6.59 (3) (x[1] + [-1] + [-1]y[1] >= 0 & x[1] + [-1] + [-1]y[0] >= 0 & [-1] + [-1]x0 >= 0 & y[1] + [1] + [-1]x1 >= 0 & y[1] + [2] + [-1]x2 >= 0 & x[1] >= 0 & y[0] >= 0 ==> (U^Increasing(F(>(x[1]1, y[1]1), x[1]1, round(+(y[1]1, 1)))), >=) & [(-1)bni_20 + (-1)Bound*bni_20] + [bni_20]x[1] + [(-1)bni_20]y[0] >= 0 & [1 + (-1)bso_21] >= 0) 8.42/6.59 8.42/6.59 8.42/6.59 8.42/6.59 We simplified constraint (3) using rule (IDP_POLY_SIMPLIFY) which results in the following new constraint: 8.42/6.59 8.42/6.59 (4) (x[1] + [-1] + [-1]y[1] >= 0 & x[1] + [-1] + [-1]y[0] >= 0 & [-1] + [-1]x0 >= 0 & y[1] + [1] + [-1]x1 >= 0 & y[1] + [2] + [-1]x2 >= 0 & x[1] >= 0 & y[0] >= 0 ==> (U^Increasing(F(>(x[1]1, y[1]1), x[1]1, round(+(y[1]1, 1)))), >=) & [(-1)bni_20 + (-1)Bound*bni_20] + [bni_20]x[1] + [(-1)bni_20]y[0] >= 0 & [1 + (-1)bso_21] >= 0) 8.42/6.59 8.42/6.59 8.42/6.59 8.42/6.59 We simplified constraint (4) using rule (POLY_REMOVE_MIN_MAX) which results in the following new constraint: 8.42/6.59 8.42/6.59 (5) (x[1] + [-1] + [-1]y[1] >= 0 & x[1] + [-1] + [-1]y[0] >= 0 & [-1] + [-1]x0 >= 0 & y[1] + [1] + [-1]x1 >= 0 & y[1] + [2] + [-1]x2 >= 0 & x[1] >= 0 & y[0] >= 0 ==> (U^Increasing(F(>(x[1]1, y[1]1), x[1]1, round(+(y[1]1, 1)))), >=) & [(-1)bni_20 + (-1)Bound*bni_20] + [bni_20]x[1] + [(-1)bni_20]y[0] >= 0 & [1 + (-1)bso_21] >= 0) 8.42/6.59 8.42/6.59 8.42/6.59 8.42/6.59 We simplified constraint (5) using rule (IDP_SMT_SPLIT) which results in the following new constraint: 8.42/6.59 8.42/6.59 (6) (x[1] + [-1] + [-1]y[1] >= 0 & x[1] + [-1] + [-1]y[0] >= 0 & [-1] + x0 >= 0 & y[1] + [1] + [-1]x1 >= 0 & y[1] + [2] + [-1]x2 >= 0 & x[1] >= 0 & y[0] >= 0 ==> (U^Increasing(F(>(x[1]1, y[1]1), x[1]1, round(+(y[1]1, 1)))), >=) & [(-1)bni_20 + (-1)Bound*bni_20] + [bni_20]x[1] + [(-1)bni_20]y[0] >= 0 & [1 + (-1)bso_21] >= 0) 8.42/6.59 8.42/6.59 8.42/6.59 8.42/6.59 We simplified constraint (6) using rule (IDP_SMT_SPLIT) which results in the following new constraint: 8.42/6.59 8.42/6.59 (7) (x[1] + [-1] + [-1]y[0] >= 0 & x[1] >= 0 & y[0] >= 0 ==> (U^Increasing(F(>(x[1]1, y[1]1), x[1]1, round(+(y[1]1, 1)))), >=) & [(-1)bni_20 + (-1)Bound*bni_20] + [bni_20]x[1] + [(-1)bni_20]y[0] >= 0 & [1 + (-1)bso_21] >= 0) 8.42/6.59 8.42/6.59 8.42/6.59 8.42/6.59 8.42/6.59 8.42/6.59 8.42/6.59 8.42/6.59 8.42/6.59 For Pair F(TRUE, x[0], y[0]) -> FNAT(&&(>=(x[0], 0), >=(y[0], 0)), x[0], y[0]) the following chains were created: 8.42/6.59 *We consider the chain F(TRUE, x[0], y[0]) -> FNAT(&&(>=(x[0], 0), >=(y[0], 0)), x[0], y[0]), FNAT(TRUE, x[1], y[1]) -> F(>(x[1], y[1]), x[1], round(+(y[1], 1))) which results in the following constraint: 8.42/6.59 8.42/6.59 (1) (&&(>=(x[0], 0), >=(y[0], 0))=TRUE & x[0]=x[1] & y[0]=y[1] ==> F(TRUE, x[0], y[0])_>=_NonInfC & F(TRUE, x[0], y[0])_>=_FNAT(&&(>=(x[0], 0), >=(y[0], 0)), x[0], y[0]) & (U^Increasing(FNAT(&&(>=(x[0], 0), >=(y[0], 0)), x[0], y[0])), >=)) 8.42/6.59 8.42/6.59 8.42/6.59 8.42/6.59 We simplified constraint (1) using rules (IV), (IDP_BOOLEAN) which results in the following new constraint: 8.42/6.59 8.42/6.59 (2) (>=(x[0], 0)=TRUE & >=(y[0], 0)=TRUE ==> F(TRUE, x[0], y[0])_>=_NonInfC & F(TRUE, x[0], y[0])_>=_FNAT(&&(>=(x[0], 0), >=(y[0], 0)), x[0], y[0]) & (U^Increasing(FNAT(&&(>=(x[0], 0), >=(y[0], 0)), x[0], y[0])), >=)) 8.42/6.59 8.42/6.59 8.42/6.59 8.42/6.59 We simplified constraint (2) using rule (POLY_CONSTRAINTS) which results in the following new constraint: 8.42/6.59 8.42/6.59 (3) (x[0] >= 0 & y[0] >= 0 ==> (U^Increasing(FNAT(&&(>=(x[0], 0), >=(y[0], 0)), x[0], y[0])), >=) & [(-1)bni_22 + (-1)Bound*bni_22] + [(-1)bni_22]y[0] + [bni_22]x[0] >= 0 & [(-1)bso_23] >= 0) 8.42/6.59 8.42/6.59 8.42/6.59 8.42/6.59 We simplified constraint (3) using rule (IDP_POLY_SIMPLIFY) which results in the following new constraint: 8.42/6.59 8.42/6.59 (4) (x[0] >= 0 & y[0] >= 0 ==> (U^Increasing(FNAT(&&(>=(x[0], 0), >=(y[0], 0)), x[0], y[0])), >=) & [(-1)bni_22 + (-1)Bound*bni_22] + [(-1)bni_22]y[0] + [bni_22]x[0] >= 0 & [(-1)bso_23] >= 0) 8.42/6.59 8.42/6.59 8.42/6.59 8.42/6.59 We simplified constraint (4) using rule (POLY_REMOVE_MIN_MAX) which results in the following new constraint: 8.42/6.59 8.42/6.59 (5) (x[0] >= 0 & y[0] >= 0 ==> (U^Increasing(FNAT(&&(>=(x[0], 0), >=(y[0], 0)), x[0], y[0])), >=) & [(-1)bni_22 + (-1)Bound*bni_22] + [(-1)bni_22]y[0] + [bni_22]x[0] >= 0 & [(-1)bso_23] >= 0) 8.42/6.59 8.42/6.59 8.42/6.59 8.42/6.59 8.42/6.59 8.42/6.59 8.42/6.59 8.42/6.59 8.42/6.59 To summarize, we get the following constraints P__>=_ for the following pairs. 8.42/6.59 8.42/6.59 *FNAT(TRUE, x[1], y[1]) -> F(>(x[1], y[1]), x[1], round(+(y[1], 1))) 8.42/6.59 8.42/6.59 *(x[1] + [-1] + [-1]y[0] >= 0 & x[1] >= 0 & y[0] >= 0 ==> (U^Increasing(F(>(x[1]1, y[1]1), x[1]1, round(+(y[1]1, 1)))), >=) & [(-1)bni_20 + (-1)Bound*bni_20] + [bni_20]x[1] + [(-1)bni_20]y[0] >= 0 & [1 + (-1)bso_21] >= 0) 8.42/6.59 8.42/6.59 8.42/6.59 8.42/6.59 8.42/6.59 *F(TRUE, x[0], y[0]) -> FNAT(&&(>=(x[0], 0), >=(y[0], 0)), x[0], y[0]) 8.42/6.59 8.42/6.59 *(x[0] >= 0 & y[0] >= 0 ==> (U^Increasing(FNAT(&&(>=(x[0], 0), >=(y[0], 0)), x[0], y[0])), >=) & [(-1)bni_22 + (-1)Bound*bni_22] + [(-1)bni_22]y[0] + [bni_22]x[0] >= 0 & [(-1)bso_23] >= 0) 8.42/6.59 8.42/6.59 8.42/6.59 8.42/6.59 8.42/6.59 8.42/6.59 8.42/6.59 8.42/6.59 8.42/6.59 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. 8.42/6.59 8.42/6.59 Using the following integer polynomial ordering the resulting constraints can be solved 8.42/6.59 8.42/6.59 Polynomial interpretation over integers[POLO]: 8.42/6.59 8.42/6.59 POL(TRUE) = 0 8.42/6.59 POL(FALSE) = 0 8.42/6.59 POL(round(x_1)) = x_1 8.42/6.59 POL(if(x_1, x_2, x_3)) = [-1]max{[-1]x_3, [-1]x_2} 8.42/6.59 POL(=(x_1, x_2)) = [-1] 8.42/6.59 POL(2) = [2] 8.42/6.59 POL(0) = 0 8.42/6.59 POL(+(x_1, x_2)) = x_1 + x_2 8.42/6.59 POL(1) = [1] 8.42/6.59 POL(FNAT(x_1, x_2, x_3)) = [-1] + x_2 + [-1]x_3 8.42/6.59 POL(F(x_1, x_2, x_3)) = [-1] + [-1]x_3 + x_2 8.42/6.59 POL(>(x_1, x_2)) = [-1] 8.42/6.59 POL(&&(x_1, x_2)) = [-1] 8.42/6.59 POL(>=(x_1, x_2)) = [-1] 8.42/6.59 8.42/6.59 8.42/6.59 The following pairs are in P_>: 8.42/6.59 8.42/6.59 8.42/6.59 FNAT(TRUE, x[1], y[1]) -> F(>(x[1], y[1]), x[1], round(+(y[1], 1))) 8.42/6.59 8.42/6.59 8.42/6.59 The following pairs are in P_bound: 8.42/6.59 8.42/6.59 8.42/6.59 FNAT(TRUE, x[1], y[1]) -> F(>(x[1], y[1]), x[1], round(+(y[1], 1))) 8.42/6.59 8.42/6.59 8.42/6.59 The following pairs are in P_>=: 8.42/6.59 8.42/6.59 8.42/6.59 F(TRUE, x[0], y[0]) -> FNAT(&&(>=(x[0], 0), >=(y[0], 0)), x[0], y[0]) 8.42/6.59 8.42/6.59 8.42/6.59 At least the following rules have been oriented under context sensitive arithmetic replacement: 8.42/6.59 8.42/6.59 if(=(%(x, 2), 0), x, +(x, 1))^1 -> round(x)^1 8.42/6.59 u^1 -> if(TRUE, u, v)^1 8.42/6.59 v^1 -> if(FALSE, u, v)^1 8.42/6.59 8.42/6.59 ---------------------------------------- 8.42/6.59 8.42/6.59 (8) 8.42/6.59 Obligation: 8.42/6.59 IDP problem: 8.42/6.59 The following function symbols are pre-defined: 8.42/6.59 <<< 8.42/6.59 & ~ Bwand: (Integer, Integer) -> Integer 8.42/6.59 >= ~ Ge: (Integer, Integer) -> Boolean 8.42/6.59 | ~ Bwor: (Integer, Integer) -> Integer 8.42/6.59 / ~ Div: (Integer, Integer) -> Integer 8.42/6.59 != ~ Neq: (Integer, Integer) -> Boolean 8.42/6.59 && ~ Land: (Boolean, Boolean) -> Boolean 8.42/6.59 ! ~ Lnot: (Boolean) -> Boolean 8.42/6.59 = ~ Eq: (Integer, Integer) -> Boolean 8.42/6.59 <= ~ Le: (Integer, Integer) -> Boolean 8.42/6.59 ^ ~ Bwxor: (Integer, Integer) -> Integer 8.42/6.59 % ~ Mod: (Integer, Integer) -> Integer 8.42/6.59 > ~ Gt: (Integer, Integer) -> Boolean 8.42/6.59 + ~ Add: (Integer, Integer) -> Integer 8.42/6.59 -1 ~ UnaryMinus: (Integer) -> Integer 8.42/6.59 < ~ Lt: (Integer, Integer) -> Boolean 8.42/6.59 || ~ Lor: (Boolean, Boolean) -> Boolean 8.42/6.59 - ~ Sub: (Integer, Integer) -> Integer 8.42/6.59 ~ ~ Bwnot: (Integer) -> Integer 8.42/6.59 * ~ Mul: (Integer, Integer) -> Integer 8.42/6.59 >>> 8.42/6.59 8.42/6.59 8.42/6.59 The following domains are used: 8.42/6.59 Integer, Boolean 8.42/6.59 8.42/6.59 The ITRS R consists of the following rules: 8.42/6.59 round(x) -> if(x % 2 = 0, x, x + 1) 8.42/6.59 if(TRUE, u, v) -> u 8.42/6.59 if(FALSE, u, v) -> v 8.42/6.59 8.42/6.59 The integer pair graph contains the following rules and edges: 8.42/6.59 (0): F(TRUE, x[0], y[0]) -> FNAT(x[0] >= 0 && y[0] >= 0, x[0], y[0]) 8.42/6.59 8.42/6.59 8.42/6.59 The set Q consists of the following terms: 8.42/6.59 f(TRUE, x0, x1) 8.42/6.59 fNat(TRUE, x0, x1) 8.42/6.59 round(x0) 8.42/6.59 if(TRUE, x0, x1) 8.42/6.59 if(FALSE, x0, x1) 8.42/6.59 8.42/6.59 ---------------------------------------- 8.42/6.59 8.42/6.59 (9) IDependencyGraphProof (EQUIVALENT) 8.42/6.59 The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 1 less node. 8.42/6.59 ---------------------------------------- 8.42/6.59 8.42/6.59 (10) 8.42/6.59 TRUE 8.68/6.67 EOF