4.69/2.02 YES 4.90/2.06 proof of /export/starexec/sandbox/benchmark/theBenchmark.itrs 4.90/2.06 # AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty 4.90/2.06 4.90/2.06 4.90/2.06 Termination of the given ITRS could be proven: 4.90/2.06 4.90/2.06 (0) ITRS 4.90/2.06 (1) ITRStoIDPProof [EQUIVALENT, 0 ms] 4.90/2.06 (2) IDP 4.90/2.06 (3) UsableRulesProof [EQUIVALENT, 0 ms] 4.90/2.06 (4) IDP 4.90/2.06 (5) IDependencyGraphProof [EQUIVALENT, 0 ms] 4.90/2.06 (6) IDP 4.90/2.06 (7) IDPNonInfProof [SOUND, 98 ms] 4.90/2.06 (8) AND 4.90/2.06 (9) IDP 4.90/2.06 (10) IDependencyGraphProof [EQUIVALENT, 0 ms] 4.90/2.06 (11) TRUE 4.90/2.06 (12) IDP 4.90/2.06 (13) IDependencyGraphProof [EQUIVALENT, 0 ms] 4.90/2.06 (14) TRUE 4.90/2.06 4.90/2.06 4.90/2.06 ---------------------------------------- 4.90/2.06 4.90/2.06 (0) 4.90/2.06 Obligation: 4.90/2.06 ITRS problem: 4.90/2.06 4.90/2.06 The following function symbols are pre-defined: 4.90/2.06 <<< 4.90/2.06 & ~ Bwand: (Integer, Integer) -> Integer 4.90/2.06 >= ~ Ge: (Integer, Integer) -> Boolean 4.90/2.06 | ~ Bwor: (Integer, Integer) -> Integer 4.90/2.06 / ~ Div: (Integer, Integer) -> Integer 4.90/2.06 != ~ Neq: (Integer, Integer) -> Boolean 4.90/2.06 && ~ Land: (Boolean, Boolean) -> Boolean 4.90/2.06 ! ~ Lnot: (Boolean) -> Boolean 4.90/2.06 = ~ Eq: (Integer, Integer) -> Boolean 4.90/2.06 <= ~ Le: (Integer, Integer) -> Boolean 4.90/2.06 ^ ~ Bwxor: (Integer, Integer) -> Integer 4.90/2.06 % ~ Mod: (Integer, Integer) -> Integer 4.90/2.06 + ~ Add: (Integer, Integer) -> Integer 4.90/2.06 > ~ Gt: (Integer, Integer) -> Boolean 4.90/2.06 -1 ~ UnaryMinus: (Integer) -> Integer 4.90/2.06 < ~ Lt: (Integer, Integer) -> Boolean 4.90/2.06 || ~ Lor: (Boolean, Boolean) -> Boolean 4.90/2.06 - ~ Sub: (Integer, Integer) -> Integer 4.90/2.06 ~ ~ Bwnot: (Integer) -> Integer 4.90/2.06 * ~ Mul: (Integer, Integer) -> Integer 4.90/2.06 >>> 4.90/2.06 4.90/2.06 The TRS R consists of the following rules: 4.90/2.06 sqrt(x) -> sqrtAcc(x, 0) 4.90/2.06 sqrtAcc(x, y) -> condAcc(y * y >= x || y < 0, x, y) 4.90/2.06 condAcc(TRUE, x, y) -> y 4.90/2.06 condAcc(FALSE, x, y) -> sqrtAcc(x, y + 1) 4.90/2.06 The set Q consists of the following terms: 4.90/2.06 sqrt(x0) 4.90/2.06 sqrtAcc(x0, x1) 4.90/2.06 condAcc(TRUE, x0, x1) 4.90/2.06 condAcc(FALSE, x0, x1) 4.90/2.06 4.90/2.06 ---------------------------------------- 4.90/2.06 4.90/2.06 (1) ITRStoIDPProof (EQUIVALENT) 4.90/2.06 Added dependency pairs 4.90/2.06 ---------------------------------------- 4.90/2.06 4.90/2.06 (2) 4.90/2.06 Obligation: 4.90/2.06 IDP problem: 4.90/2.06 The following function symbols are pre-defined: 4.90/2.06 <<< 4.90/2.06 & ~ Bwand: (Integer, Integer) -> Integer 4.90/2.06 >= ~ Ge: (Integer, Integer) -> Boolean 4.90/2.06 | ~ Bwor: (Integer, Integer) -> Integer 4.90/2.06 / ~ Div: (Integer, Integer) -> Integer 4.90/2.06 != ~ Neq: (Integer, Integer) -> Boolean 4.90/2.06 && ~ Land: (Boolean, Boolean) -> Boolean 4.90/2.06 ! ~ Lnot: (Boolean) -> Boolean 4.90/2.06 = ~ Eq: (Integer, Integer) -> Boolean 4.90/2.06 <= ~ Le: (Integer, Integer) -> Boolean 4.90/2.06 ^ ~ Bwxor: (Integer, Integer) -> Integer 4.90/2.06 % ~ Mod: (Integer, Integer) -> Integer 4.90/2.06 + ~ Add: (Integer, Integer) -> Integer 4.90/2.06 > ~ Gt: (Integer, Integer) -> Boolean 4.90/2.06 -1 ~ UnaryMinus: (Integer) -> Integer 4.90/2.06 < ~ Lt: (Integer, Integer) -> Boolean 4.90/2.06 || ~ Lor: (Boolean, Boolean) -> Boolean 4.90/2.06 - ~ Sub: (Integer, Integer) -> Integer 4.90/2.06 ~ ~ Bwnot: (Integer) -> Integer 4.90/2.06 * ~ Mul: (Integer, Integer) -> Integer 4.90/2.06 >>> 4.90/2.06 4.90/2.06 4.90/2.06 The following domains are used: 4.90/2.06 Boolean, Integer 4.90/2.06 4.90/2.06 The ITRS R consists of the following rules: 4.90/2.06 sqrt(x) -> sqrtAcc(x, 0) 4.90/2.06 sqrtAcc(x, y) -> condAcc(y * y >= x || y < 0, x, y) 4.90/2.06 condAcc(TRUE, x, y) -> y 4.90/2.06 condAcc(FALSE, x, y) -> sqrtAcc(x, y + 1) 4.90/2.06 4.90/2.06 The integer pair graph contains the following rules and edges: 4.90/2.06 (0): SQRT(x[0]) -> SQRTACC(x[0], 0) 4.90/2.06 (1): SQRTACC(x[1], y[1]) -> CONDACC(y[1] * y[1] >= x[1] || y[1] < 0, x[1], y[1]) 4.90/2.06 (2): CONDACC(FALSE, x[2], y[2]) -> SQRTACC(x[2], y[2] + 1) 4.90/2.06 4.90/2.06 (0) -> (1), if (x[0] ->^* x[1] & 0 ->^* y[1]) 4.90/2.06 (1) -> (2), if (y[1] * y[1] >= x[1] || y[1] < 0 ->^* FALSE & x[1] ->^* x[2] & y[1] ->^* y[2]) 4.90/2.06 (2) -> (1), if (x[2] ->^* x[1] & y[2] + 1 ->^* y[1]) 4.90/2.06 4.90/2.06 The set Q consists of the following terms: 4.90/2.06 sqrt(x0) 4.90/2.06 sqrtAcc(x0, x1) 4.90/2.06 condAcc(TRUE, x0, x1) 4.90/2.06 condAcc(FALSE, x0, x1) 4.90/2.06 4.90/2.06 ---------------------------------------- 4.90/2.06 4.90/2.06 (3) UsableRulesProof (EQUIVALENT) 4.90/2.06 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. 4.90/2.06 ---------------------------------------- 4.90/2.06 4.90/2.06 (4) 4.90/2.06 Obligation: 4.90/2.06 IDP problem: 4.90/2.06 The following function symbols are pre-defined: 4.90/2.06 <<< 4.90/2.06 & ~ Bwand: (Integer, Integer) -> Integer 4.90/2.06 >= ~ Ge: (Integer, Integer) -> Boolean 4.90/2.06 | ~ Bwor: (Integer, Integer) -> Integer 4.90/2.06 / ~ Div: (Integer, Integer) -> Integer 4.90/2.06 != ~ Neq: (Integer, Integer) -> Boolean 4.90/2.06 && ~ Land: (Boolean, Boolean) -> Boolean 4.90/2.06 ! ~ Lnot: (Boolean) -> Boolean 4.90/2.06 = ~ Eq: (Integer, Integer) -> Boolean 4.90/2.06 <= ~ Le: (Integer, Integer) -> Boolean 4.90/2.06 ^ ~ Bwxor: (Integer, Integer) -> Integer 4.90/2.06 % ~ Mod: (Integer, Integer) -> Integer 4.90/2.06 + ~ Add: (Integer, Integer) -> Integer 4.90/2.06 > ~ Gt: (Integer, Integer) -> Boolean 4.90/2.06 -1 ~ UnaryMinus: (Integer) -> Integer 4.90/2.06 < ~ Lt: (Integer, Integer) -> Boolean 4.90/2.06 || ~ Lor: (Boolean, Boolean) -> Boolean 4.90/2.06 - ~ Sub: (Integer, Integer) -> Integer 4.90/2.06 ~ ~ Bwnot: (Integer) -> Integer 4.90/2.06 * ~ Mul: (Integer, Integer) -> Integer 4.90/2.06 >>> 4.90/2.06 4.90/2.06 4.90/2.06 The following domains are used: 4.90/2.06 Boolean, Integer 4.90/2.06 4.90/2.06 R is empty. 4.90/2.06 4.90/2.06 The integer pair graph contains the following rules and edges: 4.90/2.06 (0): SQRT(x[0]) -> SQRTACC(x[0], 0) 4.90/2.06 (1): SQRTACC(x[1], y[1]) -> CONDACC(y[1] * y[1] >= x[1] || y[1] < 0, x[1], y[1]) 4.90/2.06 (2): CONDACC(FALSE, x[2], y[2]) -> SQRTACC(x[2], y[2] + 1) 4.90/2.06 4.90/2.06 (0) -> (1), if (x[0] ->^* x[1] & 0 ->^* y[1]) 4.90/2.06 (1) -> (2), if (y[1] * y[1] >= x[1] || y[1] < 0 ->^* FALSE & x[1] ->^* x[2] & y[1] ->^* y[2]) 4.90/2.06 (2) -> (1), if (x[2] ->^* x[1] & y[2] + 1 ->^* y[1]) 4.90/2.06 4.90/2.06 The set Q consists of the following terms: 4.90/2.06 sqrt(x0) 4.90/2.06 sqrtAcc(x0, x1) 4.90/2.06 condAcc(TRUE, x0, x1) 4.90/2.06 condAcc(FALSE, x0, x1) 4.90/2.06 4.90/2.06 ---------------------------------------- 4.90/2.06 4.90/2.06 (5) IDependencyGraphProof (EQUIVALENT) 4.90/2.06 The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 1 less node. 4.90/2.06 ---------------------------------------- 4.90/2.06 4.90/2.06 (6) 4.90/2.06 Obligation: 4.90/2.06 IDP problem: 4.90/2.06 The following function symbols are pre-defined: 4.90/2.06 <<< 4.90/2.06 & ~ Bwand: (Integer, Integer) -> Integer 4.90/2.06 >= ~ Ge: (Integer, Integer) -> Boolean 4.90/2.06 | ~ Bwor: (Integer, Integer) -> Integer 4.90/2.06 / ~ Div: (Integer, Integer) -> Integer 4.90/2.06 != ~ Neq: (Integer, Integer) -> Boolean 4.90/2.06 && ~ Land: (Boolean, Boolean) -> Boolean 4.90/2.06 ! ~ Lnot: (Boolean) -> Boolean 4.90/2.06 = ~ Eq: (Integer, Integer) -> Boolean 4.90/2.06 <= ~ Le: (Integer, Integer) -> Boolean 4.90/2.06 ^ ~ Bwxor: (Integer, Integer) -> Integer 4.90/2.06 % ~ Mod: (Integer, Integer) -> Integer 4.90/2.06 + ~ Add: (Integer, Integer) -> Integer 4.90/2.06 > ~ Gt: (Integer, Integer) -> Boolean 4.90/2.06 -1 ~ UnaryMinus: (Integer) -> Integer 4.90/2.06 < ~ Lt: (Integer, Integer) -> Boolean 4.90/2.06 || ~ Lor: (Boolean, Boolean) -> Boolean 4.90/2.06 - ~ Sub: (Integer, Integer) -> Integer 4.90/2.06 ~ ~ Bwnot: (Integer) -> Integer 4.90/2.06 * ~ Mul: (Integer, Integer) -> Integer 4.90/2.06 >>> 4.90/2.06 4.90/2.06 4.90/2.06 The following domains are used: 4.90/2.06 Integer, Boolean 4.90/2.06 4.90/2.06 R is empty. 4.90/2.06 4.90/2.06 The integer pair graph contains the following rules and edges: 4.90/2.06 (2): CONDACC(FALSE, x[2], y[2]) -> SQRTACC(x[2], y[2] + 1) 4.90/2.06 (1): SQRTACC(x[1], y[1]) -> CONDACC(y[1] * y[1] >= x[1] || y[1] < 0, x[1], y[1]) 4.90/2.06 4.90/2.06 (2) -> (1), if (x[2] ->^* x[1] & y[2] + 1 ->^* y[1]) 4.90/2.06 (1) -> (2), if (y[1] * y[1] >= x[1] || y[1] < 0 ->^* FALSE & x[1] ->^* x[2] & y[1] ->^* y[2]) 4.90/2.06 4.90/2.06 The set Q consists of the following terms: 4.90/2.06 sqrt(x0) 4.90/2.06 sqrtAcc(x0, x1) 4.90/2.06 condAcc(TRUE, x0, x1) 4.90/2.06 condAcc(FALSE, x0, x1) 4.90/2.06 4.90/2.06 ---------------------------------------- 4.90/2.06 4.90/2.06 (7) IDPNonInfProof (SOUND) 4.90/2.06 Used the following options for this NonInfProof: 4.90/2.06 4.90/2.06 IDPGPoloSolver: 4.90/2.06 Range: [(-1,2)] 4.90/2.06 IsNat: false 4.90/2.06 Interpretation Shape Heuristic: aprove.DPFramework.IDPProblem.Processors.nonInf.poly.IdpCand1ShapeHeuristic@6fb2ff74 4.90/2.06 Constraint Generator: NonInfConstraintGenerator: 4.90/2.06 PathGenerator: MetricPathGenerator: 4.90/2.06 Max Left Steps: 2 4.90/2.06 Max Right Steps: 1 4.90/2.06 4.90/2.06 4.90/2.06 4.90/2.06 The constraints were generated the following way: 4.90/2.06 4.90/2.06 The DP Problem is simplified using the Induction Calculus [NONINF] with the following steps: 4.90/2.06 4.90/2.06 Note that final constraints are written in bold face. 4.90/2.06 4.90/2.06 4.90/2.06 4.90/2.06 For Pair CONDACC(FALSE, x[2], y[2]) -> SQRTACC(x[2], +(y[2], 1)) the following chains were created: 4.90/2.06 *We consider the chain CONDACC(FALSE, x[2], y[2]) -> SQRTACC(x[2], +(y[2], 1)), SQRTACC(x[1], y[1]) -> CONDACC(||(>=(*(y[1], y[1]), x[1]), <(y[1], 0)), x[1], y[1]), CONDACC(FALSE, x[2], y[2]) -> SQRTACC(x[2], +(y[2], 1)), SQRTACC(x[1], y[1]) -> CONDACC(||(>=(*(y[1], y[1]), x[1]), <(y[1], 0)), x[1], y[1]) which results in the following constraint: 4.90/2.06 4.90/2.06 (1) (x[2]=x[1] & +(y[2], 1)=y[1] & ||(>=(*(y[1], y[1]), x[1]), <(y[1], 0))=FALSE & x[1]=x[2]1 & y[1]=y[2]1 & x[2]1=x[1]1 & +(y[2]1, 1)=y[1]1 ==> CONDACC(FALSE, x[2]1, y[2]1)_>=_NonInfC & CONDACC(FALSE, x[2]1, y[2]1)_>=_SQRTACC(x[2]1, +(y[2]1, 1)) & (U^Increasing(SQRTACC(x[2]1, +(y[2]1, 1))), >=)) 4.90/2.06 4.90/2.06 4.90/2.06 4.90/2.06 We simplified constraint (1) using rules (III), (IV), (IDP_BOOLEAN) which results in the following new constraint: 4.90/2.06 4.90/2.06 (2) (>=(*(+(y[2], 1), +(y[2], 1)), x[1])=FALSE & <(+(y[2], 1), 0)=FALSE ==> CONDACC(FALSE, x[1], +(y[2], 1))_>=_NonInfC & CONDACC(FALSE, x[1], +(y[2], 1))_>=_SQRTACC(x[1], +(+(y[2], 1), 1)) & (U^Increasing(SQRTACC(x[2]1, +(y[2]1, 1))), >=)) 4.90/2.06 4.90/2.06 4.90/2.06 4.90/2.06 We simplified constraint (2) using rule (POLY_CONSTRAINTS) which results in the following new constraint: 4.90/2.06 4.90/2.06 (3) (x[1] + [-2] + [-1]y[2]^2 + [-2]y[2] >= 0 & y[2] + [1] >= 0 ==> (U^Increasing(SQRTACC(x[2]1, +(y[2]1, 1))), >=) & [(-2)bni_12 + (-1)Bound*bni_12] + [bni_12]x[1] + [(-1)bni_12]y[2] >= 0 & [(-1)bso_13] >= 0) 4.90/2.06 4.90/2.06 4.90/2.06 4.90/2.06 We simplified constraint (3) using rule (IDP_POLY_SIMPLIFY) which results in the following new constraint: 4.90/2.06 4.90/2.06 (4) (x[1] + [-2] + [-1]y[2]^2 + [-2]y[2] >= 0 & y[2] + [1] >= 0 ==> (U^Increasing(SQRTACC(x[2]1, +(y[2]1, 1))), >=) & [(-2)bni_12 + (-1)Bound*bni_12] + [bni_12]x[1] + [(-1)bni_12]y[2] >= 0 & [(-1)bso_13] >= 0) 4.90/2.06 4.90/2.06 4.90/2.06 4.90/2.06 We simplified constraint (4) using rule (POLY_REMOVE_MIN_MAX) which results in the following new constraint: 4.90/2.06 4.90/2.06 (5) (y[2] + [1] >= 0 & x[1] + [-2] + [-1]y[2]^2 + [-2]y[2] >= 0 ==> (U^Increasing(SQRTACC(x[2]1, +(y[2]1, 1))), >=) & [(-2)bni_12 + (-1)Bound*bni_12] + [bni_12]x[1] + [(-1)bni_12]y[2] >= 0 & [(-1)bso_13] >= 0) 4.90/2.06 4.90/2.06 4.90/2.06 4.90/2.06 We simplified constraint (5) using rule (IDP_SMT_SPLIT) which results in the following new constraint: 4.90/2.06 4.90/2.06 (6) (y[2] + [1] >= 0 & x[1] >= 0 ==> (U^Increasing(SQRTACC(x[2]1, +(y[2]1, 1))), >=) & [(-1)Bound*bni_12] + [bni_12]y[2]^2 + [bni_12]y[2] + [bni_12]x[1] >= 0 & [(-1)bso_13] >= 0) 4.90/2.06 4.90/2.06 4.90/2.06 4.90/2.06 We simplified constraint (6) using rule (IDP_SMT_SPLIT) which results in the following new constraints: 4.90/2.06 4.90/2.06 (7) (y[2] + [1] >= 0 & x[1] >= 0 & y[2] >= 0 ==> (U^Increasing(SQRTACC(x[2]1, +(y[2]1, 1))), >=) & [(-1)Bound*bni_12] + [bni_12]y[2]^2 + [bni_12]y[2] + [bni_12]x[1] >= 0 & [(-1)bso_13] >= 0) 4.90/2.06 4.90/2.06 (8) ([-1]y[2] + [1] >= 0 & x[1] >= 0 & y[2] >= 0 ==> (U^Increasing(SQRTACC(x[2]1, +(y[2]1, 1))), >=) & [(-1)Bound*bni_12] + [bni_12]y[2]^2 + [(-1)bni_12]y[2] + [bni_12]x[1] >= 0 & [(-1)bso_13] >= 0) 4.90/2.06 4.90/2.06 4.90/2.06 4.90/2.06 4.90/2.06 4.90/2.06 4.90/2.06 4.90/2.06 4.90/2.06 For Pair SQRTACC(x[1], y[1]) -> CONDACC(||(>=(*(y[1], y[1]), x[1]), <(y[1], 0)), x[1], y[1]) the following chains were created: 4.90/2.06 *We consider the chain SQRTACC(x[1], y[1]) -> CONDACC(||(>=(*(y[1], y[1]), x[1]), <(y[1], 0)), x[1], y[1]), CONDACC(FALSE, x[2], y[2]) -> SQRTACC(x[2], +(y[2], 1)) which results in the following constraint: 4.90/2.06 4.90/2.06 (1) (||(>=(*(y[1], y[1]), x[1]), <(y[1], 0))=FALSE & x[1]=x[2] & y[1]=y[2] ==> SQRTACC(x[1], y[1])_>=_NonInfC & SQRTACC(x[1], y[1])_>=_CONDACC(||(>=(*(y[1], y[1]), x[1]), <(y[1], 0)), x[1], y[1]) & (U^Increasing(CONDACC(||(>=(*(y[1], y[1]), x[1]), <(y[1], 0)), x[1], y[1])), >=)) 4.90/2.06 4.90/2.06 4.90/2.06 4.90/2.06 We simplified constraint (1) using rules (IV), (IDP_BOOLEAN) which results in the following new constraint: 4.90/2.06 4.90/2.06 (2) (>=(*(y[1], y[1]), x[1])=FALSE & <(y[1], 0)=FALSE ==> SQRTACC(x[1], y[1])_>=_NonInfC & SQRTACC(x[1], y[1])_>=_CONDACC(||(>=(*(y[1], y[1]), x[1]), <(y[1], 0)), x[1], y[1]) & (U^Increasing(CONDACC(||(>=(*(y[1], y[1]), x[1]), <(y[1], 0)), x[1], y[1])), >=)) 4.90/2.06 4.90/2.06 4.90/2.06 4.90/2.06 We simplified constraint (2) using rule (POLY_CONSTRAINTS) which results in the following new constraint: 4.90/2.06 4.90/2.06 (3) (x[1] + [-1] + [-1]y[1]^2 >= 0 & y[1] >= 0 ==> (U^Increasing(CONDACC(||(>=(*(y[1], y[1]), x[1]), <(y[1], 0)), x[1], y[1])), >=) & [(-1)Bound*bni_14] + [bni_14]x[1] + [(-1)bni_14]y[1] >= 0 & [1 + (-1)bso_15] >= 0) 4.90/2.06 4.90/2.06 4.90/2.06 4.90/2.06 We simplified constraint (3) using rule (IDP_POLY_SIMPLIFY) which results in the following new constraint: 4.90/2.06 4.90/2.06 (4) (x[1] + [-1] + [-1]y[1]^2 >= 0 & y[1] >= 0 ==> (U^Increasing(CONDACC(||(>=(*(y[1], y[1]), x[1]), <(y[1], 0)), x[1], y[1])), >=) & [(-1)Bound*bni_14] + [bni_14]x[1] + [(-1)bni_14]y[1] >= 0 & [1 + (-1)bso_15] >= 0) 4.90/2.06 4.90/2.06 4.90/2.06 4.90/2.06 We simplified constraint (4) using rule (POLY_REMOVE_MIN_MAX) which results in the following new constraint: 4.90/2.06 4.90/2.06 (5) (y[1] >= 0 & x[1] + [-1] + [-1]y[1]^2 >= 0 ==> (U^Increasing(CONDACC(||(>=(*(y[1], y[1]), x[1]), <(y[1], 0)), x[1], y[1])), >=) & [(-1)Bound*bni_14] + [bni_14]x[1] + [(-1)bni_14]y[1] >= 0 & [1 + (-1)bso_15] >= 0) 4.90/2.06 4.90/2.06 4.90/2.06 4.90/2.06 4.90/2.06 4.90/2.06 4.90/2.06 4.90/2.06 4.90/2.06 To summarize, we get the following constraints P__>=_ for the following pairs. 4.90/2.06 4.90/2.06 *CONDACC(FALSE, x[2], y[2]) -> SQRTACC(x[2], +(y[2], 1)) 4.90/2.06 4.90/2.06 *(y[2] + [1] >= 0 & x[1] >= 0 & y[2] >= 0 ==> (U^Increasing(SQRTACC(x[2]1, +(y[2]1, 1))), >=) & [(-1)Bound*bni_12] + [bni_12]y[2]^2 + [bni_12]y[2] + [bni_12]x[1] >= 0 & [(-1)bso_13] >= 0) 4.90/2.06 4.90/2.06 4.90/2.06 *([-1]y[2] + [1] >= 0 & x[1] >= 0 & y[2] >= 0 ==> (U^Increasing(SQRTACC(x[2]1, +(y[2]1, 1))), >=) & [(-1)Bound*bni_12] + [bni_12]y[2]^2 + [(-1)bni_12]y[2] + [bni_12]x[1] >= 0 & [(-1)bso_13] >= 0) 4.90/2.06 4.90/2.06 4.90/2.06 4.90/2.06 4.90/2.06 *SQRTACC(x[1], y[1]) -> CONDACC(||(>=(*(y[1], y[1]), x[1]), <(y[1], 0)), x[1], y[1]) 4.90/2.06 4.90/2.06 *(y[1] >= 0 & x[1] + [-1] + [-1]y[1]^2 >= 0 ==> (U^Increasing(CONDACC(||(>=(*(y[1], y[1]), x[1]), <(y[1], 0)), x[1], y[1])), >=) & [(-1)Bound*bni_14] + [bni_14]x[1] + [(-1)bni_14]y[1] >= 0 & [1 + (-1)bso_15] >= 0) 4.90/2.06 4.90/2.06 4.90/2.06 4.90/2.06 4.90/2.06 4.90/2.06 4.90/2.06 4.90/2.06 4.90/2.06 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. 4.90/2.06 4.90/2.06 Using the following integer polynomial ordering the resulting constraints can be solved 4.90/2.06 4.90/2.06 Polynomial interpretation over integers[POLO]: 4.90/2.06 4.90/2.06 POL(TRUE) = 0 4.90/2.06 POL(FALSE) = 0 4.90/2.06 POL(CONDACC(x_1, x_2, x_3)) = [-1] + x_2 + [-1]x_3 4.90/2.06 POL(SQRTACC(x_1, x_2)) = x_1 + [-1]x_2 4.90/2.06 POL(+(x_1, x_2)) = x_1 + x_2 4.90/2.06 POL(1) = [1] 4.90/2.06 POL(||(x_1, x_2)) = [-1] 4.90/2.06 POL(>=(x_1, x_2)) = [-1] 4.90/2.06 POL(*(x_1, x_2)) = x_1*x_2 4.90/2.06 POL(<(x_1, x_2)) = [-1] 4.90/2.06 POL(0) = 0 4.90/2.06 4.90/2.06 4.90/2.06 The following pairs are in P_>: 4.90/2.06 4.90/2.06 4.90/2.06 SQRTACC(x[1], y[1]) -> CONDACC(||(>=(*(y[1], y[1]), x[1]), <(y[1], 0)), x[1], y[1]) 4.90/2.06 4.90/2.06 4.90/2.06 The following pairs are in P_bound: 4.90/2.06 4.90/2.06 4.90/2.06 CONDACC(FALSE, x[2], y[2]) -> SQRTACC(x[2], +(y[2], 1)) 4.90/2.06 4.90/2.06 4.90/2.06 The following pairs are in P_>=: 4.90/2.06 4.90/2.06 4.90/2.06 CONDACC(FALSE, x[2], y[2]) -> SQRTACC(x[2], +(y[2], 1)) 4.90/2.06 4.90/2.06 4.90/2.06 There are no usable rules. 4.90/2.06 ---------------------------------------- 4.90/2.06 4.90/2.06 (8) 4.90/2.06 Complex Obligation (AND) 4.90/2.06 4.90/2.06 ---------------------------------------- 4.90/2.06 4.90/2.06 (9) 4.90/2.06 Obligation: 4.90/2.06 IDP problem: 4.90/2.06 The following function symbols are pre-defined: 4.90/2.06 <<< 4.90/2.06 & ~ Bwand: (Integer, Integer) -> Integer 4.90/2.06 >= ~ Ge: (Integer, Integer) -> Boolean 4.90/2.06 | ~ Bwor: (Integer, Integer) -> Integer 4.90/2.06 / ~ Div: (Integer, Integer) -> Integer 4.90/2.06 != ~ Neq: (Integer, Integer) -> Boolean 4.90/2.06 && ~ Land: (Boolean, Boolean) -> Boolean 4.90/2.06 ! ~ Lnot: (Boolean) -> Boolean 4.90/2.06 = ~ Eq: (Integer, Integer) -> Boolean 4.90/2.06 <= ~ Le: (Integer, Integer) -> Boolean 4.90/2.06 ^ ~ Bwxor: (Integer, Integer) -> Integer 4.90/2.06 % ~ Mod: (Integer, Integer) -> Integer 4.90/2.06 + ~ Add: (Integer, Integer) -> Integer 4.90/2.06 > ~ Gt: (Integer, Integer) -> Boolean 4.90/2.06 -1 ~ UnaryMinus: (Integer) -> Integer 4.90/2.06 < ~ Lt: (Integer, Integer) -> Boolean 4.90/2.06 || ~ Lor: (Boolean, Boolean) -> Boolean 4.90/2.06 - ~ Sub: (Integer, Integer) -> Integer 4.90/2.06 ~ ~ Bwnot: (Integer) -> Integer 4.90/2.06 * ~ Mul: (Integer, Integer) -> Integer 4.90/2.06 >>> 4.90/2.06 4.90/2.06 4.90/2.06 The following domains are used: 4.90/2.06 Integer 4.90/2.06 4.90/2.06 R is empty. 4.90/2.06 4.90/2.06 The integer pair graph contains the following rules and edges: 4.90/2.06 (2): CONDACC(FALSE, x[2], y[2]) -> SQRTACC(x[2], y[2] + 1) 4.90/2.06 4.90/2.06 4.90/2.06 The set Q consists of the following terms: 4.90/2.06 sqrt(x0) 4.90/2.06 sqrtAcc(x0, x1) 4.90/2.06 condAcc(TRUE, x0, x1) 4.90/2.06 condAcc(FALSE, x0, x1) 4.90/2.06 4.90/2.06 ---------------------------------------- 4.90/2.06 4.90/2.06 (10) IDependencyGraphProof (EQUIVALENT) 4.90/2.06 The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 1 less node. 4.90/2.06 ---------------------------------------- 4.90/2.06 4.90/2.06 (11) 4.90/2.06 TRUE 4.90/2.06 4.90/2.06 ---------------------------------------- 4.90/2.06 4.90/2.06 (12) 4.90/2.06 Obligation: 4.90/2.06 IDP problem: 4.90/2.06 The following function symbols are pre-defined: 4.90/2.06 <<< 4.90/2.06 & ~ Bwand: (Integer, Integer) -> Integer 4.90/2.06 >= ~ Ge: (Integer, Integer) -> Boolean 4.90/2.06 | ~ Bwor: (Integer, Integer) -> Integer 4.90/2.06 / ~ Div: (Integer, Integer) -> Integer 4.90/2.06 != ~ Neq: (Integer, Integer) -> Boolean 4.90/2.06 && ~ Land: (Boolean, Boolean) -> Boolean 4.90/2.06 ! ~ Lnot: (Boolean) -> Boolean 4.90/2.06 = ~ Eq: (Integer, Integer) -> Boolean 4.90/2.06 <= ~ Le: (Integer, Integer) -> Boolean 4.90/2.06 ^ ~ Bwxor: (Integer, Integer) -> Integer 4.90/2.06 % ~ Mod: (Integer, Integer) -> Integer 4.90/2.06 + ~ Add: (Integer, Integer) -> Integer 4.90/2.06 > ~ Gt: (Integer, Integer) -> Boolean 4.90/2.06 -1 ~ UnaryMinus: (Integer) -> Integer 4.90/2.06 < ~ Lt: (Integer, Integer) -> Boolean 4.90/2.06 || ~ Lor: (Boolean, Boolean) -> Boolean 4.90/2.06 - ~ Sub: (Integer, Integer) -> Integer 4.90/2.06 ~ ~ Bwnot: (Integer) -> Integer 4.90/2.06 * ~ Mul: (Integer, Integer) -> Integer 4.90/2.06 >>> 4.90/2.06 4.90/2.06 4.90/2.06 The following domains are used: 4.90/2.06 Boolean, Integer 4.90/2.06 4.90/2.06 R is empty. 4.90/2.06 4.90/2.06 The integer pair graph contains the following rules and edges: 4.90/2.06 (1): SQRTACC(x[1], y[1]) -> CONDACC(y[1] * y[1] >= x[1] || y[1] < 0, x[1], y[1]) 4.90/2.06 4.90/2.06 4.90/2.06 The set Q consists of the following terms: 4.90/2.06 sqrt(x0) 4.90/2.06 sqrtAcc(x0, x1) 4.90/2.06 condAcc(TRUE, x0, x1) 4.90/2.06 condAcc(FALSE, x0, x1) 4.90/2.06 4.90/2.06 ---------------------------------------- 4.90/2.06 4.90/2.06 (13) IDependencyGraphProof (EQUIVALENT) 4.90/2.06 The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 1 less node. 4.90/2.06 ---------------------------------------- 4.90/2.06 4.90/2.06 (14) 4.90/2.06 TRUE 5.10/2.18 EOF