4.28/2.04 YES 4.28/2.05 proof of /export/starexec/sandbox/benchmark/theBenchmark.itrs 4.28/2.05 # AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty 4.28/2.05 4.28/2.05 4.28/2.05 Termination of the given ITRS could be proven: 4.28/2.05 4.28/2.05 (0) ITRS 4.28/2.05 (1) ITRStoIDPProof [EQUIVALENT, 0 ms] 4.28/2.05 (2) IDP 4.28/2.05 (3) UsableRulesProof [EQUIVALENT, 0 ms] 4.28/2.05 (4) IDP 4.28/2.05 (5) IDependencyGraphProof [EQUIVALENT, 0 ms] 4.28/2.05 (6) IDP 4.28/2.05 (7) IDPNonInfProof [SOUND, 233 ms] 4.28/2.05 (8) IDP 4.28/2.05 (9) IDependencyGraphProof [EQUIVALENT, 0 ms] 4.28/2.05 (10) TRUE 4.28/2.05 4.28/2.05 4.28/2.05 ---------------------------------------- 4.28/2.05 4.28/2.05 (0) 4.28/2.05 Obligation: 4.28/2.05 ITRS problem: 4.28/2.05 4.28/2.05 The following function symbols are pre-defined: 4.28/2.05 <<< 4.28/2.05 & ~ Bwand: (Integer, Integer) -> Integer 4.28/2.05 >= ~ Ge: (Integer, Integer) -> Boolean 4.28/2.05 | ~ Bwor: (Integer, Integer) -> Integer 4.28/2.05 / ~ Div: (Integer, Integer) -> Integer 4.28/2.05 != ~ Neq: (Integer, Integer) -> Boolean 4.28/2.05 && ~ Land: (Boolean, Boolean) -> Boolean 4.28/2.05 ! ~ Lnot: (Boolean) -> Boolean 4.28/2.05 = ~ Eq: (Integer, Integer) -> Boolean 4.28/2.05 <= ~ Le: (Integer, Integer) -> Boolean 4.28/2.05 ^ ~ Bwxor: (Integer, Integer) -> Integer 4.28/2.05 % ~ Mod: (Integer, Integer) -> Integer 4.28/2.05 + ~ Add: (Integer, Integer) -> Integer 4.28/2.05 > ~ Gt: (Integer, Integer) -> Boolean 4.28/2.05 -1 ~ UnaryMinus: (Integer) -> Integer 4.28/2.05 < ~ Lt: (Integer, Integer) -> Boolean 4.28/2.05 || ~ Lor: (Boolean, Boolean) -> Boolean 4.28/2.05 - ~ Sub: (Integer, Integer) -> Integer 4.28/2.05 ~ ~ Bwnot: (Integer) -> Integer 4.28/2.05 * ~ Mul: (Integer, Integer) -> Integer 4.28/2.05 >>> 4.28/2.05 4.28/2.05 The TRS R consists of the following rules: 4.28/2.05 f(x, y) -> Cond_f(x >= 1 && y = x - 1, x, y) 4.28/2.05 Cond_f(TRUE, x, y) -> f(x, round(x)) 4.28/2.05 round(x) -> x 4.28/2.05 round(x) -> x + 1 4.28/2.05 The set Q consists of the following terms: 4.28/2.05 f(x0, x1) 4.28/2.05 Cond_f(TRUE, x0, x1) 4.28/2.05 round(x0) 4.28/2.05 4.28/2.05 ---------------------------------------- 4.28/2.05 4.28/2.05 (1) ITRStoIDPProof (EQUIVALENT) 4.28/2.05 Added dependency pairs 4.28/2.05 ---------------------------------------- 4.28/2.05 4.28/2.05 (2) 4.28/2.05 Obligation: 4.28/2.05 IDP problem: 4.28/2.05 The following function symbols are pre-defined: 4.28/2.05 <<< 4.28/2.05 & ~ Bwand: (Integer, Integer) -> Integer 4.28/2.05 >= ~ Ge: (Integer, Integer) -> Boolean 4.28/2.05 | ~ Bwor: (Integer, Integer) -> Integer 4.28/2.05 / ~ Div: (Integer, Integer) -> Integer 4.28/2.05 != ~ Neq: (Integer, Integer) -> Boolean 4.28/2.05 && ~ Land: (Boolean, Boolean) -> Boolean 4.28/2.05 ! ~ Lnot: (Boolean) -> Boolean 4.28/2.05 = ~ Eq: (Integer, Integer) -> Boolean 4.28/2.05 <= ~ Le: (Integer, Integer) -> Boolean 4.28/2.05 ^ ~ Bwxor: (Integer, Integer) -> Integer 4.28/2.05 % ~ Mod: (Integer, Integer) -> Integer 4.28/2.05 + ~ Add: (Integer, Integer) -> Integer 4.28/2.05 > ~ Gt: (Integer, Integer) -> Boolean 4.28/2.05 -1 ~ UnaryMinus: (Integer) -> Integer 4.28/2.05 < ~ Lt: (Integer, Integer) -> Boolean 4.28/2.05 || ~ Lor: (Boolean, Boolean) -> Boolean 4.28/2.05 - ~ Sub: (Integer, Integer) -> Integer 4.28/2.05 ~ ~ Bwnot: (Integer) -> Integer 4.28/2.05 * ~ Mul: (Integer, Integer) -> Integer 4.28/2.05 >>> 4.28/2.05 4.28/2.05 4.28/2.05 The following domains are used: 4.28/2.05 Boolean, Integer 4.28/2.05 4.28/2.05 The ITRS R consists of the following rules: 4.28/2.05 f(x, y) -> Cond_f(x >= 1 && y = x - 1, x, y) 4.28/2.05 Cond_f(TRUE, x, y) -> f(x, round(x)) 4.28/2.05 round(x) -> x 4.28/2.05 round(x) -> x + 1 4.28/2.05 4.28/2.05 The integer pair graph contains the following rules and edges: 4.28/2.05 (0): F(x[0], y[0]) -> COND_F(x[0] >= 1 && y[0] = x[0] - 1, x[0], y[0]) 4.28/2.05 (1): COND_F(TRUE, x[1], y[1]) -> F(x[1], round(x[1])) 4.28/2.05 (2): COND_F(TRUE, x[2], y[2]) -> ROUND(x[2]) 4.28/2.05 4.28/2.05 (0) -> (1), if (x[0] >= 1 && y[0] = x[0] - 1 & x[0] ->^* x[1] & y[0] ->^* y[1]) 4.28/2.05 (0) -> (2), if (x[0] >= 1 && y[0] = x[0] - 1 & x[0] ->^* x[2] & y[0] ->^* y[2]) 4.28/2.05 (1) -> (0), if (x[1] ->^* x[0] & round(x[1]) ->^* y[0]) 4.28/2.05 4.28/2.05 The set Q consists of the following terms: 4.28/2.05 f(x0, x1) 4.28/2.05 Cond_f(TRUE, x0, x1) 4.28/2.05 round(x0) 4.28/2.05 4.28/2.05 ---------------------------------------- 4.28/2.05 4.28/2.05 (3) UsableRulesProof (EQUIVALENT) 4.28/2.05 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. 4.28/2.05 ---------------------------------------- 4.28/2.05 4.28/2.05 (4) 4.28/2.05 Obligation: 4.28/2.05 IDP problem: 4.28/2.05 The following function symbols are pre-defined: 4.28/2.05 <<< 4.28/2.05 & ~ Bwand: (Integer, Integer) -> Integer 4.28/2.05 >= ~ Ge: (Integer, Integer) -> Boolean 4.28/2.05 | ~ Bwor: (Integer, Integer) -> Integer 4.28/2.05 / ~ Div: (Integer, Integer) -> Integer 4.28/2.05 != ~ Neq: (Integer, Integer) -> Boolean 4.28/2.05 && ~ Land: (Boolean, Boolean) -> Boolean 4.28/2.05 ! ~ Lnot: (Boolean) -> Boolean 4.28/2.05 = ~ Eq: (Integer, Integer) -> Boolean 4.28/2.05 <= ~ Le: (Integer, Integer) -> Boolean 4.28/2.05 ^ ~ Bwxor: (Integer, Integer) -> Integer 4.28/2.05 % ~ Mod: (Integer, Integer) -> Integer 4.28/2.05 + ~ Add: (Integer, Integer) -> Integer 4.28/2.05 > ~ Gt: (Integer, Integer) -> Boolean 4.28/2.05 -1 ~ UnaryMinus: (Integer) -> Integer 4.28/2.05 < ~ Lt: (Integer, Integer) -> Boolean 4.28/2.05 || ~ Lor: (Boolean, Boolean) -> Boolean 4.28/2.05 - ~ Sub: (Integer, Integer) -> Integer 4.28/2.05 ~ ~ Bwnot: (Integer) -> Integer 4.28/2.05 * ~ Mul: (Integer, Integer) -> Integer 4.28/2.05 >>> 4.28/2.05 4.28/2.05 4.28/2.05 The following domains are used: 4.28/2.05 Integer, Boolean 4.28/2.05 4.28/2.05 The ITRS R consists of the following rules: 4.28/2.05 round(x) -> x 4.28/2.05 round(x) -> x + 1 4.28/2.05 4.28/2.05 The integer pair graph contains the following rules and edges: 4.28/2.05 (0): F(x[0], y[0]) -> COND_F(x[0] >= 1 && y[0] = x[0] - 1, x[0], y[0]) 4.28/2.05 (1): COND_F(TRUE, x[1], y[1]) -> F(x[1], round(x[1])) 4.28/2.05 (2): COND_F(TRUE, x[2], y[2]) -> ROUND(x[2]) 4.28/2.05 4.28/2.05 (0) -> (1), if (x[0] >= 1 && y[0] = x[0] - 1 & x[0] ->^* x[1] & y[0] ->^* y[1]) 4.28/2.05 (0) -> (2), if (x[0] >= 1 && y[0] = x[0] - 1 & x[0] ->^* x[2] & y[0] ->^* y[2]) 4.28/2.05 (1) -> (0), if (x[1] ->^* x[0] & round(x[1]) ->^* y[0]) 4.28/2.05 4.28/2.05 The set Q consists of the following terms: 4.28/2.05 f(x0, x1) 4.28/2.05 Cond_f(TRUE, x0, x1) 4.28/2.05 round(x0) 4.28/2.05 4.28/2.05 ---------------------------------------- 4.28/2.05 4.28/2.05 (5) IDependencyGraphProof (EQUIVALENT) 4.28/2.05 The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 1 less node. 4.28/2.05 ---------------------------------------- 4.28/2.05 4.28/2.05 (6) 4.28/2.05 Obligation: 4.28/2.05 IDP problem: 4.28/2.05 The following function symbols are pre-defined: 4.28/2.05 <<< 4.28/2.05 & ~ Bwand: (Integer, Integer) -> Integer 4.28/2.05 >= ~ Ge: (Integer, Integer) -> Boolean 4.28/2.05 | ~ Bwor: (Integer, Integer) -> Integer 4.28/2.05 / ~ Div: (Integer, Integer) -> Integer 4.28/2.05 != ~ Neq: (Integer, Integer) -> Boolean 4.28/2.05 && ~ Land: (Boolean, Boolean) -> Boolean 4.28/2.05 ! ~ Lnot: (Boolean) -> Boolean 4.28/2.05 = ~ Eq: (Integer, Integer) -> Boolean 4.28/2.05 <= ~ Le: (Integer, Integer) -> Boolean 4.28/2.05 ^ ~ Bwxor: (Integer, Integer) -> Integer 4.28/2.05 % ~ Mod: (Integer, Integer) -> Integer 4.28/2.05 + ~ Add: (Integer, Integer) -> Integer 4.28/2.05 > ~ Gt: (Integer, Integer) -> Boolean 4.28/2.05 -1 ~ UnaryMinus: (Integer) -> Integer 4.28/2.05 < ~ Lt: (Integer, Integer) -> Boolean 4.28/2.05 || ~ Lor: (Boolean, Boolean) -> Boolean 4.28/2.05 - ~ Sub: (Integer, Integer) -> Integer 4.28/2.05 ~ ~ Bwnot: (Integer) -> Integer 4.28/2.05 * ~ Mul: (Integer, Integer) -> Integer 4.28/2.05 >>> 4.28/2.05 4.28/2.05 4.28/2.05 The following domains are used: 4.28/2.05 Integer, Boolean 4.28/2.05 4.28/2.05 The ITRS R consists of the following rules: 4.28/2.05 round(x) -> x 4.28/2.05 round(x) -> x + 1 4.28/2.05 4.28/2.05 The integer pair graph contains the following rules and edges: 4.28/2.05 (1): COND_F(TRUE, x[1], y[1]) -> F(x[1], round(x[1])) 4.28/2.05 (0): F(x[0], y[0]) -> COND_F(x[0] >= 1 && y[0] = x[0] - 1, x[0], y[0]) 4.28/2.05 4.28/2.05 (1) -> (0), if (x[1] ->^* x[0] & round(x[1]) ->^* y[0]) 4.28/2.05 (0) -> (1), if (x[0] >= 1 && y[0] = x[0] - 1 & x[0] ->^* x[1] & y[0] ->^* y[1]) 4.28/2.05 4.28/2.05 The set Q consists of the following terms: 4.28/2.05 f(x0, x1) 4.28/2.05 Cond_f(TRUE, x0, x1) 4.28/2.05 round(x0) 4.28/2.05 4.28/2.05 ---------------------------------------- 4.28/2.05 4.28/2.05 (7) IDPNonInfProof (SOUND) 4.28/2.05 Used the following options for this NonInfProof: 4.28/2.05 4.28/2.05 IDPGPoloSolver: 4.28/2.05 Range: [(-1,2)] 4.28/2.05 IsNat: false 4.28/2.05 Interpretation Shape Heuristic: aprove.DPFramework.IDPProblem.Processors.nonInf.poly.IdpDefaultShapeHeuristic@dd9abe 4.28/2.05 Constraint Generator: NonInfConstraintGenerator: 4.28/2.06 PathGenerator: MetricPathGenerator: 4.28/2.06 Max Left Steps: 1 4.28/2.06 Max Right Steps: 1 4.28/2.06 4.28/2.06 4.28/2.06 4.28/2.06 The constraints were generated the following way: 4.28/2.06 4.28/2.06 The DP Problem is simplified using the Induction Calculus [NONINF] with the following steps: 4.28/2.06 4.28/2.06 Note that final constraints are written in bold face. 4.28/2.06 4.28/2.06 4.28/2.06 4.28/2.06 For Pair COND_F(TRUE, x[1], y[1]) -> F(x[1], round(x[1])) the following chains were created: 4.28/2.06 *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: 4.28/2.06 4.28/2.06 (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]))), >=)) 4.28/2.06 4.28/2.06 4.28/2.06 4.28/2.06 We simplified constraint (1) using rules (III), (IV), (IDP_BOOLEAN) which results in the following new constraint: 4.28/2.06 4.28/2.06 (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]))), >=)) 4.28/2.06 4.28/2.06 4.28/2.06 4.28/2.06 We simplified constraint (2) using rule (POLY_CONSTRAINTS) which results in the following new constraint: 4.28/2.06 4.28/2.06 (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] + [(-1)bni_15]y[0] + [(2)bni_15]x[0] >= 0 & [(-1)bso_16] + [-1]y[0] + x[0] >= 0) 4.28/2.06 4.28/2.06 4.28/2.06 4.28/2.06 We simplified constraint (3) using rule (IDP_POLY_SIMPLIFY) which results in the following new constraint: 4.28/2.06 4.28/2.06 (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] + [(-1)bni_15]y[0] + [(2)bni_15]x[0] >= 0 & [(-1)bso_16] + [-1]y[0] + x[0] >= 0) 4.28/2.06 4.28/2.06 4.28/2.06 4.28/2.06 We simplified constraint (4) using rule (POLY_REMOVE_MIN_MAX) which results in the following new constraint: 4.28/2.06 4.28/2.06 (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] + [(-1)bni_15]y[0] + [(2)bni_15]x[0] >= 0 & [(-1)bso_16] + [-1]y[0] + x[0] >= 0) 4.28/2.06 4.28/2.06 4.28/2.06 4.28/2.06 We simplified constraint (5) using rule (IDP_SMT_SPLIT) which results in the following new constraints: 4.28/2.06 4.28/2.06 (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] + [(-1)bni_15]y[0] + [(2)bni_15]x[0] >= 0 & [(-1)bso_16] + [-1]y[0] + x[0] >= 0) 4.28/2.06 4.28/2.06 (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]y[0] + [(2)bni_15]x[0] >= 0 & [(-1)bso_16] + y[0] + x[0] >= 0) 4.28/2.06 4.28/2.06 4.28/2.06 4.28/2.06 4.28/2.06 4.28/2.06 4.28/2.06 4.28/2.06 4.28/2.06 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: 4.28/2.06 *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: 4.28/2.06 4.28/2.06 (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])), >=)) 4.28/2.06 4.28/2.06 4.28/2.06 4.28/2.06 We simplified constraint (1) using rules (IV), (IDP_BOOLEAN) which results in the following new constraint: 4.28/2.06 4.28/2.06 (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])), >=)) 4.28/2.06 4.28/2.06 4.28/2.06 4.28/2.06 We simplified constraint (2) using rule (POLY_CONSTRAINTS) which results in the following new constraint: 4.28/2.06 4.28/2.06 (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] >= 0) 4.28/2.06 4.28/2.06 4.28/2.06 4.28/2.06 We simplified constraint (3) using rule (IDP_POLY_SIMPLIFY) which results in the following new constraint: 4.28/2.06 4.28/2.06 (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] >= 0) 4.28/2.06 4.28/2.06 4.28/2.06 4.28/2.06 We simplified constraint (4) using rule (POLY_REMOVE_MIN_MAX) which results in the following new constraint: 4.28/2.06 4.28/2.06 (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] >= 0) 4.28/2.06 4.28/2.06 4.28/2.06 4.28/2.06 We simplified constraint (5) using rule (IDP_SMT_SPLIT) which results in the following new constraints: 4.28/2.06 4.28/2.06 (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] >= 0) 4.28/2.06 4.28/2.06 (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] >= 0) 4.28/2.06 4.28/2.06 4.28/2.06 4.28/2.06 4.28/2.06 4.28/2.06 4.28/2.06 4.28/2.06 4.28/2.06 To summarize, we get the following constraints P__>=_ for the following pairs. 4.28/2.06 4.28/2.06 *COND_F(TRUE, x[1], y[1]) -> F(x[1], round(x[1])) 4.28/2.06 4.28/2.06 *(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] + [(-1)bni_15]y[0] + [(2)bni_15]x[0] >= 0 & [(-1)bso_16] + [-1]y[0] + x[0] >= 0) 4.28/2.06 4.28/2.06 4.28/2.06 *(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]y[0] + [(2)bni_15]x[0] >= 0 & [(-1)bso_16] + y[0] + x[0] >= 0) 4.28/2.06 4.28/2.06 4.28/2.06 4.28/2.06 4.28/2.06 *F(x[0], y[0]) -> COND_F(&&(>=(x[0], 1), =(y[0], -(x[0], 1))), x[0], y[0]) 4.28/2.06 4.28/2.06 *(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] >= 0) 4.28/2.06 4.28/2.06 4.28/2.06 *(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] >= 0) 4.28/2.06 4.28/2.06 4.28/2.06 4.28/2.06 4.28/2.06 4.28/2.06 4.28/2.06 4.28/2.06 4.28/2.06 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. 4.28/2.06 4.28/2.06 Using the following integer polynomial ordering the resulting constraints can be solved 4.28/2.06 4.28/2.06 Polynomial interpretation over integers[POLO]: 4.28/2.06 4.28/2.06 POL(TRUE) = 0 4.28/2.06 POL(FALSE) = [3] 4.28/2.06 POL(round(x_1)) = x_1 4.28/2.06 POL(+(x_1, x_2)) = x_1 + x_2 4.28/2.06 POL(1) = [1] 4.28/2.06 POL(COND_F(x_1, x_2, x_3)) = [-1] + [-1]x_3 + [2]x_2 + [-1]x_1 4.28/2.06 POL(F(x_1, x_2)) = [-1] + [-1]x_2 + [2]x_1 4.28/2.06 POL(&&(x_1, x_2)) = 0 4.28/2.06 POL(>=(x_1, x_2)) = [-1] 4.28/2.06 POL(=(x_1, x_2)) = [-1] 4.28/2.06 POL(-(x_1, x_2)) = x_1 + [-1]x_2 4.28/2.06 4.28/2.06 4.28/2.06 The following pairs are in P_>: 4.28/2.06 4.28/2.06 4.28/2.06 COND_F(TRUE, x[1], y[1]) -> F(x[1], round(x[1])) 4.28/2.06 4.28/2.06 4.28/2.06 The following pairs are in P_bound: 4.28/2.06 4.28/2.06 4.28/2.06 COND_F(TRUE, x[1], y[1]) -> F(x[1], round(x[1])) 4.28/2.06 F(x[0], y[0]) -> COND_F(&&(>=(x[0], 1), =(y[0], -(x[0], 1))), x[0], y[0]) 4.28/2.06 4.28/2.06 4.28/2.06 The following pairs are in P_>=: 4.28/2.06 4.28/2.06 4.28/2.06 F(x[0], y[0]) -> COND_F(&&(>=(x[0], 1), =(y[0], -(x[0], 1))), x[0], y[0]) 4.28/2.06 4.28/2.06 4.28/2.06 At least the following rules have been oriented under context sensitive arithmetic replacement: 4.28/2.06 4.28/2.06 x^1 -> round(x)^1 4.28/2.06 +(x, 1)^1 -> round(x)^1 4.28/2.06 TRUE^1 -> &&(TRUE, TRUE)^1 4.28/2.06 FALSE^1 -> &&(TRUE, FALSE)^1 4.28/2.06 FALSE^1 -> &&(FALSE, TRUE)^1 4.28/2.06 FALSE^1 -> &&(FALSE, FALSE)^1 4.28/2.06 4.28/2.06 ---------------------------------------- 4.28/2.06 4.28/2.06 (8) 4.28/2.06 Obligation: 4.28/2.06 IDP problem: 4.28/2.06 The following function symbols are pre-defined: 4.28/2.06 <<< 4.28/2.06 & ~ Bwand: (Integer, Integer) -> Integer 4.28/2.06 >= ~ Ge: (Integer, Integer) -> Boolean 4.28/2.06 | ~ Bwor: (Integer, Integer) -> Integer 4.28/2.06 / ~ Div: (Integer, Integer) -> Integer 4.28/2.06 != ~ Neq: (Integer, Integer) -> Boolean 4.28/2.06 && ~ Land: (Boolean, Boolean) -> Boolean 4.28/2.06 ! ~ Lnot: (Boolean) -> Boolean 4.28/2.06 = ~ Eq: (Integer, Integer) -> Boolean 4.28/2.06 <= ~ Le: (Integer, Integer) -> Boolean 4.28/2.06 ^ ~ Bwxor: (Integer, Integer) -> Integer 4.28/2.06 % ~ Mod: (Integer, Integer) -> Integer 4.28/2.06 + ~ Add: (Integer, Integer) -> Integer 4.28/2.06 > ~ Gt: (Integer, Integer) -> Boolean 4.28/2.06 -1 ~ UnaryMinus: (Integer) -> Integer 4.28/2.06 < ~ Lt: (Integer, Integer) -> Boolean 4.28/2.06 || ~ Lor: (Boolean, Boolean) -> Boolean 4.28/2.06 - ~ Sub: (Integer, Integer) -> Integer 4.28/2.06 ~ ~ Bwnot: (Integer) -> Integer 4.28/2.06 * ~ Mul: (Integer, Integer) -> Integer 4.28/2.06 >>> 4.28/2.06 4.28/2.06 4.28/2.06 The following domains are used: 4.28/2.06 Integer, Boolean 4.28/2.06 4.28/2.06 The ITRS R consists of the following rules: 4.28/2.06 round(x) -> x 4.28/2.06 round(x) -> x + 1 4.28/2.06 4.28/2.06 The integer pair graph contains the following rules and edges: 4.28/2.06 (0): F(x[0], y[0]) -> COND_F(x[0] >= 1 && y[0] = x[0] - 1, x[0], y[0]) 4.28/2.06 4.28/2.06 4.28/2.06 The set Q consists of the following terms: 4.28/2.06 f(x0, x1) 4.28/2.06 Cond_f(TRUE, x0, x1) 4.28/2.06 round(x0) 4.28/2.06 4.28/2.06 ---------------------------------------- 4.28/2.06 4.28/2.06 (9) IDependencyGraphProof (EQUIVALENT) 4.28/2.06 The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 1 less node. 4.28/2.06 ---------------------------------------- 4.28/2.06 4.28/2.06 (10) 4.28/2.06 TRUE 4.58/2.09 EOF