4.03/1.89 YES 4.03/1.90 proof of /export/starexec/sandbox2/benchmark/theBenchmark.itrs 4.03/1.90 # AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty 4.03/1.90 4.03/1.90 4.03/1.90 Termination of the given ITRS could be proven: 4.03/1.90 4.03/1.90 (0) ITRS 4.03/1.90 (1) ITRStoIDPProof [EQUIVALENT, 0 ms] 4.03/1.90 (2) IDP 4.03/1.90 (3) UsableRulesProof [EQUIVALENT, 0 ms] 4.03/1.90 (4) IDP 4.03/1.90 (5) IDependencyGraphProof [EQUIVALENT, 0 ms] 4.03/1.90 (6) IDP 4.03/1.90 (7) IDPNonInfProof [SOUND, 151 ms] 4.03/1.90 (8) IDP 4.03/1.90 (9) IDependencyGraphProof [EQUIVALENT, 0 ms] 4.03/1.90 (10) TRUE 4.03/1.90 4.03/1.90 4.03/1.90 ---------------------------------------- 4.03/1.90 4.03/1.90 (0) 4.03/1.90 Obligation: 4.03/1.90 ITRS problem: 4.03/1.90 4.03/1.90 The following function symbols are pre-defined: 4.03/1.90 <<< 4.03/1.90 & ~ Bwand: (Integer, Integer) -> Integer 4.03/1.90 >= ~ Ge: (Integer, Integer) -> Boolean 4.03/1.90 | ~ Bwor: (Integer, Integer) -> Integer 4.03/1.90 / ~ Div: (Integer, Integer) -> Integer 4.03/1.90 != ~ Neq: (Integer, Integer) -> Boolean 4.03/1.90 && ~ Land: (Boolean, Boolean) -> Boolean 4.03/1.90 ! ~ Lnot: (Boolean) -> Boolean 4.03/1.90 = ~ Eq: (Integer, Integer) -> Boolean 4.03/1.90 <= ~ Le: (Integer, Integer) -> Boolean 4.03/1.90 ^ ~ Bwxor: (Integer, Integer) -> Integer 4.03/1.90 % ~ Mod: (Integer, Integer) -> Integer 4.03/1.90 + ~ Add: (Integer, Integer) -> Integer 4.03/1.90 > ~ Gt: (Integer, Integer) -> Boolean 4.03/1.90 -1 ~ UnaryMinus: (Integer) -> Integer 4.03/1.90 < ~ Lt: (Integer, Integer) -> Boolean 4.03/1.90 || ~ Lor: (Boolean, Boolean) -> Boolean 4.03/1.90 - ~ Sub: (Integer, Integer) -> Integer 4.03/1.90 ~ ~ Bwnot: (Integer) -> Integer 4.03/1.90 * ~ Mul: (Integer, Integer) -> Integer 4.03/1.90 >>> 4.03/1.90 4.03/1.90 The TRS R consists of the following rules: 4.03/1.90 div(x, y) -> Cond_div(0 >= y, x, y) 4.03/1.90 Cond_div(TRUE, x, y) -> 0 4.03/1.90 div(x, y) -> Cond_div1(y >= x, x, y) 4.03/1.90 Cond_div1(TRUE, x, y) -> 0 4.03/1.90 div(x, y) -> Cond_div2(x > y && y > 0, x, y) 4.03/1.90 Cond_div2(TRUE, x, y) -> div(x - y, y) + 1 4.03/1.90 The set Q consists of the following terms: 4.03/1.90 div(x0, x1) 4.03/1.90 Cond_div(TRUE, x0, x1) 4.03/1.90 Cond_div1(TRUE, x0, x1) 4.03/1.90 Cond_div2(TRUE, x0, x1) 4.03/1.90 4.03/1.90 ---------------------------------------- 4.03/1.90 4.03/1.90 (1) ITRStoIDPProof (EQUIVALENT) 4.03/1.90 Added dependency pairs 4.03/1.90 ---------------------------------------- 4.03/1.90 4.03/1.90 (2) 4.03/1.90 Obligation: 4.03/1.90 IDP problem: 4.03/1.90 The following function symbols are pre-defined: 4.03/1.90 <<< 4.03/1.90 & ~ Bwand: (Integer, Integer) -> Integer 4.03/1.90 >= ~ Ge: (Integer, Integer) -> Boolean 4.03/1.90 | ~ Bwor: (Integer, Integer) -> Integer 4.03/1.90 / ~ Div: (Integer, Integer) -> Integer 4.03/1.90 != ~ Neq: (Integer, Integer) -> Boolean 4.03/1.90 && ~ Land: (Boolean, Boolean) -> Boolean 4.03/1.90 ! ~ Lnot: (Boolean) -> Boolean 4.03/1.90 = ~ Eq: (Integer, Integer) -> Boolean 4.03/1.90 <= ~ Le: (Integer, Integer) -> Boolean 4.03/1.90 ^ ~ Bwxor: (Integer, Integer) -> Integer 4.03/1.90 % ~ Mod: (Integer, Integer) -> Integer 4.03/1.90 + ~ Add: (Integer, Integer) -> Integer 4.03/1.90 > ~ Gt: (Integer, Integer) -> Boolean 4.03/1.90 -1 ~ UnaryMinus: (Integer) -> Integer 4.03/1.90 < ~ Lt: (Integer, Integer) -> Boolean 4.03/1.90 || ~ Lor: (Boolean, Boolean) -> Boolean 4.03/1.90 - ~ Sub: (Integer, Integer) -> Integer 4.03/1.90 ~ ~ Bwnot: (Integer) -> Integer 4.03/1.90 * ~ Mul: (Integer, Integer) -> Integer 4.03/1.90 >>> 4.03/1.90 4.03/1.90 4.03/1.90 The following domains are used: 4.03/1.90 Integer, Boolean 4.03/1.90 4.03/1.90 The ITRS R consists of the following rules: 4.03/1.90 div(x, y) -> Cond_div(0 >= y, x, y) 4.03/1.90 Cond_div(TRUE, x, y) -> 0 4.03/1.90 div(x, y) -> Cond_div1(y >= x, x, y) 4.03/1.90 Cond_div1(TRUE, x, y) -> 0 4.03/1.90 div(x, y) -> Cond_div2(x > y && y > 0, x, y) 4.03/1.90 Cond_div2(TRUE, x, y) -> div(x - y, y) + 1 4.03/1.90 4.03/1.90 The integer pair graph contains the following rules and edges: 4.03/1.90 (0): DIV(x[0], y[0]) -> COND_DIV(0 >= y[0], x[0], y[0]) 4.03/1.90 (1): DIV(x[1], y[1]) -> COND_DIV1(y[1] >= x[1], x[1], y[1]) 4.03/1.90 (2): DIV(x[2], y[2]) -> COND_DIV2(x[2] > y[2] && y[2] > 0, x[2], y[2]) 4.03/1.90 (3): COND_DIV2(TRUE, x[3], y[3]) -> DIV(x[3] - y[3], y[3]) 4.03/1.90 4.03/1.90 (2) -> (3), if (x[2] > y[2] && y[2] > 0 & x[2] ->^* x[3] & y[2] ->^* y[3]) 4.03/1.90 (3) -> (0), if (x[3] - y[3] ->^* x[0] & y[3] ->^* y[0]) 4.03/1.90 (3) -> (1), if (x[3] - y[3] ->^* x[1] & y[3] ->^* y[1]) 4.03/1.90 (3) -> (2), if (x[3] - y[3] ->^* x[2] & y[3] ->^* y[2]) 4.03/1.90 4.03/1.90 The set Q consists of the following terms: 4.03/1.90 div(x0, x1) 4.03/1.90 Cond_div(TRUE, x0, x1) 4.03/1.90 Cond_div1(TRUE, x0, x1) 4.03/1.90 Cond_div2(TRUE, x0, x1) 4.03/1.90 4.03/1.90 ---------------------------------------- 4.03/1.90 4.03/1.90 (3) UsableRulesProof (EQUIVALENT) 4.03/1.90 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. 4.03/1.90 ---------------------------------------- 4.03/1.90 4.03/1.90 (4) 4.03/1.90 Obligation: 4.03/1.90 IDP problem: 4.03/1.90 The following function symbols are pre-defined: 4.03/1.90 <<< 4.03/1.90 & ~ Bwand: (Integer, Integer) -> Integer 4.03/1.90 >= ~ Ge: (Integer, Integer) -> Boolean 4.03/1.90 | ~ Bwor: (Integer, Integer) -> Integer 4.03/1.90 / ~ Div: (Integer, Integer) -> Integer 4.03/1.90 != ~ Neq: (Integer, Integer) -> Boolean 4.03/1.90 && ~ Land: (Boolean, Boolean) -> Boolean 4.03/1.90 ! ~ Lnot: (Boolean) -> Boolean 4.03/1.90 = ~ Eq: (Integer, Integer) -> Boolean 4.03/1.90 <= ~ Le: (Integer, Integer) -> Boolean 4.03/1.90 ^ ~ Bwxor: (Integer, Integer) -> Integer 4.03/1.90 % ~ Mod: (Integer, Integer) -> Integer 4.03/1.90 + ~ Add: (Integer, Integer) -> Integer 4.03/1.90 > ~ Gt: (Integer, Integer) -> Boolean 4.03/1.90 -1 ~ UnaryMinus: (Integer) -> Integer 4.03/1.90 < ~ Lt: (Integer, Integer) -> Boolean 4.03/1.90 || ~ Lor: (Boolean, Boolean) -> Boolean 4.03/1.90 - ~ Sub: (Integer, Integer) -> Integer 4.03/1.90 ~ ~ Bwnot: (Integer) -> Integer 4.03/1.90 * ~ Mul: (Integer, Integer) -> Integer 4.03/1.90 >>> 4.03/1.90 4.03/1.90 4.03/1.90 The following domains are used: 4.03/1.90 Integer, Boolean 4.03/1.90 4.03/1.90 R is empty. 4.03/1.90 4.03/1.90 The integer pair graph contains the following rules and edges: 4.03/1.90 (0): DIV(x[0], y[0]) -> COND_DIV(0 >= y[0], x[0], y[0]) 4.03/1.90 (1): DIV(x[1], y[1]) -> COND_DIV1(y[1] >= x[1], x[1], y[1]) 4.03/1.90 (2): DIV(x[2], y[2]) -> COND_DIV2(x[2] > y[2] && y[2] > 0, x[2], y[2]) 4.03/1.90 (3): COND_DIV2(TRUE, x[3], y[3]) -> DIV(x[3] - y[3], y[3]) 4.03/1.90 4.03/1.90 (2) -> (3), if (x[2] > y[2] && y[2] > 0 & x[2] ->^* x[3] & y[2] ->^* y[3]) 4.03/1.90 (3) -> (0), if (x[3] - y[3] ->^* x[0] & y[3] ->^* y[0]) 4.03/1.90 (3) -> (1), if (x[3] - y[3] ->^* x[1] & y[3] ->^* y[1]) 4.03/1.90 (3) -> (2), if (x[3] - y[3] ->^* x[2] & y[3] ->^* y[2]) 4.03/1.90 4.03/1.90 The set Q consists of the following terms: 4.03/1.90 div(x0, x1) 4.03/1.90 Cond_div(TRUE, x0, x1) 4.03/1.90 Cond_div1(TRUE, x0, x1) 4.03/1.90 Cond_div2(TRUE, x0, x1) 4.03/1.90 4.03/1.90 ---------------------------------------- 4.03/1.90 4.03/1.90 (5) IDependencyGraphProof (EQUIVALENT) 4.03/1.90 The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 2 less nodes. 4.03/1.90 ---------------------------------------- 4.03/1.90 4.03/1.90 (6) 4.03/1.90 Obligation: 4.03/1.90 IDP problem: 4.03/1.90 The following function symbols are pre-defined: 4.03/1.90 <<< 4.03/1.90 & ~ Bwand: (Integer, Integer) -> Integer 4.03/1.90 >= ~ Ge: (Integer, Integer) -> Boolean 4.03/1.90 | ~ Bwor: (Integer, Integer) -> Integer 4.03/1.90 / ~ Div: (Integer, Integer) -> Integer 4.03/1.90 != ~ Neq: (Integer, Integer) -> Boolean 4.03/1.90 && ~ Land: (Boolean, Boolean) -> Boolean 4.03/1.90 ! ~ Lnot: (Boolean) -> Boolean 4.03/1.90 = ~ Eq: (Integer, Integer) -> Boolean 4.03/1.90 <= ~ Le: (Integer, Integer) -> Boolean 4.03/1.90 ^ ~ Bwxor: (Integer, Integer) -> Integer 4.03/1.90 % ~ Mod: (Integer, Integer) -> Integer 4.03/1.90 + ~ Add: (Integer, Integer) -> Integer 4.03/1.90 > ~ Gt: (Integer, Integer) -> Boolean 4.03/1.90 -1 ~ UnaryMinus: (Integer) -> Integer 4.03/1.90 < ~ Lt: (Integer, Integer) -> Boolean 4.03/1.90 || ~ Lor: (Boolean, Boolean) -> Boolean 4.03/1.90 - ~ Sub: (Integer, Integer) -> Integer 4.03/1.90 ~ ~ Bwnot: (Integer) -> Integer 4.03/1.90 * ~ Mul: (Integer, Integer) -> Integer 4.03/1.90 >>> 4.03/1.90 4.03/1.90 4.03/1.90 The following domains are used: 4.03/1.90 Integer, Boolean 4.03/1.90 4.03/1.90 R is empty. 4.03/1.90 4.03/1.90 The integer pair graph contains the following rules and edges: 4.03/1.90 (3): COND_DIV2(TRUE, x[3], y[3]) -> DIV(x[3] - y[3], y[3]) 4.03/1.90 (2): DIV(x[2], y[2]) -> COND_DIV2(x[2] > y[2] && y[2] > 0, x[2], y[2]) 4.03/1.90 4.03/1.90 (3) -> (2), if (x[3] - y[3] ->^* x[2] & y[3] ->^* y[2]) 4.03/1.90 (2) -> (3), if (x[2] > y[2] && y[2] > 0 & x[2] ->^* x[3] & y[2] ->^* y[3]) 4.03/1.90 4.03/1.90 The set Q consists of the following terms: 4.03/1.90 div(x0, x1) 4.03/1.90 Cond_div(TRUE, x0, x1) 4.03/1.90 Cond_div1(TRUE, x0, x1) 4.03/1.90 Cond_div2(TRUE, x0, x1) 4.03/1.90 4.03/1.90 ---------------------------------------- 4.03/1.90 4.03/1.90 (7) IDPNonInfProof (SOUND) 4.03/1.90 Used the following options for this NonInfProof: 4.03/1.90 4.03/1.90 IDPGPoloSolver: 4.03/1.90 Range: [(-1,2)] 4.03/1.90 IsNat: false 4.03/1.90 Interpretation Shape Heuristic: aprove.DPFramework.IDPProblem.Processors.nonInf.poly.IdpDefaultShapeHeuristic@207bf53c 4.03/1.90 Constraint Generator: NonInfConstraintGenerator: 4.03/1.90 PathGenerator: MetricPathGenerator: 4.03/1.90 Max Left Steps: 1 4.03/1.90 Max Right Steps: 1 4.03/1.90 4.03/1.90 4.03/1.90 4.03/1.90 The constraints were generated the following way: 4.03/1.90 4.03/1.90 The DP Problem is simplified using the Induction Calculus [NONINF] with the following steps: 4.03/1.90 4.03/1.90 Note that final constraints are written in bold face. 4.03/1.90 4.03/1.90 4.03/1.90 4.03/1.90 For Pair COND_DIV2(TRUE, x[3], y[3]) -> DIV(-(x[3], y[3]), y[3]) the following chains were created: 4.03/1.90 *We consider the chain DIV(x[2], y[2]) -> COND_DIV2(&&(>(x[2], y[2]), >(y[2], 0)), x[2], y[2]), COND_DIV2(TRUE, x[3], y[3]) -> DIV(-(x[3], y[3]), y[3]), DIV(x[2], y[2]) -> COND_DIV2(&&(>(x[2], y[2]), >(y[2], 0)), x[2], y[2]) which results in the following constraint: 4.03/1.90 4.03/1.90 (1) (&&(>(x[2], y[2]), >(y[2], 0))=TRUE & x[2]=x[3] & y[2]=y[3] & -(x[3], y[3])=x[2]1 & y[3]=y[2]1 ==> COND_DIV2(TRUE, x[3], y[3])_>=_NonInfC & COND_DIV2(TRUE, x[3], y[3])_>=_DIV(-(x[3], y[3]), y[3]) & (U^Increasing(DIV(-(x[3], y[3]), y[3])), >=)) 4.03/1.90 4.03/1.90 4.03/1.90 4.03/1.90 We simplified constraint (1) using rules (III), (IV), (IDP_BOOLEAN) which results in the following new constraint: 4.03/1.90 4.03/1.90 (2) (>(x[2], y[2])=TRUE & >(y[2], 0)=TRUE ==> COND_DIV2(TRUE, x[2], y[2])_>=_NonInfC & COND_DIV2(TRUE, x[2], y[2])_>=_DIV(-(x[2], y[2]), y[2]) & (U^Increasing(DIV(-(x[3], y[3]), y[3])), >=)) 4.03/1.90 4.03/1.90 4.03/1.90 4.03/1.90 We simplified constraint (2) using rule (POLY_CONSTRAINTS) which results in the following new constraint: 4.03/1.90 4.03/1.90 (3) (x[2] + [-1] + [-1]y[2] >= 0 & y[2] + [-1] >= 0 ==> (U^Increasing(DIV(-(x[3], y[3]), y[3])), >=) & [(-1)bni_12 + (-1)Bound*bni_12] + [(-1)bni_12]y[2] + [bni_12]x[2] >= 0 & [(-1)bso_13] >= 0) 4.03/1.90 4.03/1.90 4.03/1.90 4.03/1.90 We simplified constraint (3) using rule (IDP_POLY_SIMPLIFY) which results in the following new constraint: 4.03/1.90 4.03/1.90 (4) (x[2] + [-1] + [-1]y[2] >= 0 & y[2] + [-1] >= 0 ==> (U^Increasing(DIV(-(x[3], y[3]), y[3])), >=) & [(-1)bni_12 + (-1)Bound*bni_12] + [(-1)bni_12]y[2] + [bni_12]x[2] >= 0 & [(-1)bso_13] >= 0) 4.03/1.90 4.03/1.90 4.03/1.90 4.03/1.90 We simplified constraint (4) using rule (POLY_REMOVE_MIN_MAX) which results in the following new constraint: 4.03/1.90 4.03/1.90 (5) (x[2] + [-1] + [-1]y[2] >= 0 & y[2] + [-1] >= 0 ==> (U^Increasing(DIV(-(x[3], y[3]), y[3])), >=) & [(-1)bni_12 + (-1)Bound*bni_12] + [(-1)bni_12]y[2] + [bni_12]x[2] >= 0 & [(-1)bso_13] >= 0) 4.03/1.90 4.03/1.90 4.03/1.90 4.03/1.90 4.03/1.90 4.03/1.90 4.03/1.90 4.03/1.90 4.03/1.90 For Pair DIV(x[2], y[2]) -> COND_DIV2(&&(>(x[2], y[2]), >(y[2], 0)), x[2], y[2]) the following chains were created: 4.03/1.90 *We consider the chain DIV(x[2], y[2]) -> COND_DIV2(&&(>(x[2], y[2]), >(y[2], 0)), x[2], y[2]), COND_DIV2(TRUE, x[3], y[3]) -> DIV(-(x[3], y[3]), y[3]) which results in the following constraint: 4.03/1.90 4.03/1.90 (1) (&&(>(x[2], y[2]), >(y[2], 0))=TRUE & x[2]=x[3] & y[2]=y[3] ==> DIV(x[2], y[2])_>=_NonInfC & DIV(x[2], y[2])_>=_COND_DIV2(&&(>(x[2], y[2]), >(y[2], 0)), x[2], y[2]) & (U^Increasing(COND_DIV2(&&(>(x[2], y[2]), >(y[2], 0)), x[2], y[2])), >=)) 4.03/1.90 4.03/1.90 4.03/1.90 4.03/1.90 We simplified constraint (1) using rules (IV), (IDP_BOOLEAN) which results in the following new constraint: 4.03/1.90 4.03/1.90 (2) (>(x[2], y[2])=TRUE & >(y[2], 0)=TRUE ==> DIV(x[2], y[2])_>=_NonInfC & DIV(x[2], y[2])_>=_COND_DIV2(&&(>(x[2], y[2]), >(y[2], 0)), x[2], y[2]) & (U^Increasing(COND_DIV2(&&(>(x[2], y[2]), >(y[2], 0)), x[2], y[2])), >=)) 4.03/1.90 4.03/1.90 4.03/1.90 4.03/1.90 We simplified constraint (2) using rule (POLY_CONSTRAINTS) which results in the following new constraint: 4.03/1.90 4.03/1.90 (3) (x[2] + [-1] + [-1]y[2] >= 0 & y[2] + [-1] >= 0 ==> (U^Increasing(COND_DIV2(&&(>(x[2], y[2]), >(y[2], 0)), x[2], y[2])), >=) & [(-1)bni_14 + (-1)Bound*bni_14] + [bni_14]x[2] >= 0 & [(-1)bso_15] + y[2] >= 0) 4.03/1.90 4.03/1.90 4.03/1.90 4.03/1.90 We simplified constraint (3) using rule (IDP_POLY_SIMPLIFY) which results in the following new constraint: 4.03/1.90 4.03/1.90 (4) (x[2] + [-1] + [-1]y[2] >= 0 & y[2] + [-1] >= 0 ==> (U^Increasing(COND_DIV2(&&(>(x[2], y[2]), >(y[2], 0)), x[2], y[2])), >=) & [(-1)bni_14 + (-1)Bound*bni_14] + [bni_14]x[2] >= 0 & [(-1)bso_15] + y[2] >= 0) 4.03/1.90 4.03/1.90 4.03/1.90 4.03/1.90 We simplified constraint (4) using rule (POLY_REMOVE_MIN_MAX) which results in the following new constraint: 4.03/1.90 4.03/1.90 (5) (x[2] + [-1] + [-1]y[2] >= 0 & y[2] + [-1] >= 0 ==> (U^Increasing(COND_DIV2(&&(>(x[2], y[2]), >(y[2], 0)), x[2], y[2])), >=) & [(-1)bni_14 + (-1)Bound*bni_14] + [bni_14]x[2] >= 0 & [(-1)bso_15] + y[2] >= 0) 4.03/1.90 4.03/1.90 4.03/1.90 4.03/1.90 4.03/1.90 4.03/1.90 4.03/1.90 4.03/1.90 4.03/1.90 To summarize, we get the following constraints P__>=_ for the following pairs. 4.03/1.90 4.03/1.90 *COND_DIV2(TRUE, x[3], y[3]) -> DIV(-(x[3], y[3]), y[3]) 4.03/1.90 4.03/1.90 *(x[2] + [-1] + [-1]y[2] >= 0 & y[2] + [-1] >= 0 ==> (U^Increasing(DIV(-(x[3], y[3]), y[3])), >=) & [(-1)bni_12 + (-1)Bound*bni_12] + [(-1)bni_12]y[2] + [bni_12]x[2] >= 0 & [(-1)bso_13] >= 0) 4.03/1.90 4.03/1.90 4.03/1.90 4.03/1.90 4.03/1.90 *DIV(x[2], y[2]) -> COND_DIV2(&&(>(x[2], y[2]), >(y[2], 0)), x[2], y[2]) 4.03/1.90 4.03/1.90 *(x[2] + [-1] + [-1]y[2] >= 0 & y[2] + [-1] >= 0 ==> (U^Increasing(COND_DIV2(&&(>(x[2], y[2]), >(y[2], 0)), x[2], y[2])), >=) & [(-1)bni_14 + (-1)Bound*bni_14] + [bni_14]x[2] >= 0 & [(-1)bso_15] + y[2] >= 0) 4.03/1.90 4.03/1.90 4.03/1.90 4.03/1.90 4.03/1.90 4.03/1.90 4.03/1.90 4.03/1.90 4.03/1.90 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. 4.03/1.90 4.03/1.90 Using the following integer polynomial ordering the resulting constraints can be solved 4.03/1.90 4.03/1.90 Polynomial interpretation over integers[POLO]: 4.03/1.90 4.03/1.90 POL(TRUE) = 0 4.03/1.90 POL(FALSE) = [1] 4.03/1.90 POL(COND_DIV2(x_1, x_2, x_3)) = [-1] + [-1]x_3 + x_2 + [-1]x_1 4.03/1.90 POL(DIV(x_1, x_2)) = [-1] + x_1 4.03/1.90 POL(-(x_1, x_2)) = x_1 + [-1]x_2 4.03/1.90 POL(&&(x_1, x_2)) = 0 4.03/1.90 POL(>(x_1, x_2)) = [-1] 4.03/1.90 POL(0) = 0 4.03/1.90 4.03/1.90 4.03/1.90 The following pairs are in P_>: 4.03/1.90 4.03/1.90 4.03/1.90 DIV(x[2], y[2]) -> COND_DIV2(&&(>(x[2], y[2]), >(y[2], 0)), x[2], y[2]) 4.03/1.90 4.03/1.90 4.03/1.90 The following pairs are in P_bound: 4.03/1.90 4.03/1.90 4.03/1.90 COND_DIV2(TRUE, x[3], y[3]) -> DIV(-(x[3], y[3]), y[3]) 4.03/1.90 DIV(x[2], y[2]) -> COND_DIV2(&&(>(x[2], y[2]), >(y[2], 0)), x[2], y[2]) 4.03/1.90 4.03/1.90 4.03/1.90 The following pairs are in P_>=: 4.03/1.90 4.03/1.90 4.03/1.90 COND_DIV2(TRUE, x[3], y[3]) -> DIV(-(x[3], y[3]), y[3]) 4.03/1.90 4.03/1.90 4.03/1.90 At least the following rules have been oriented under context sensitive arithmetic replacement: 4.03/1.90 4.03/1.90 TRUE^1 -> &&(TRUE, TRUE)^1 4.03/1.90 FALSE^1 -> &&(TRUE, FALSE)^1 4.03/1.90 FALSE^1 -> &&(FALSE, TRUE)^1 4.03/1.90 FALSE^1 -> &&(FALSE, FALSE)^1 4.03/1.90 4.03/1.90 ---------------------------------------- 4.03/1.90 4.03/1.90 (8) 4.03/1.90 Obligation: 4.03/1.90 IDP problem: 4.03/1.90 The following function symbols are pre-defined: 4.03/1.90 <<< 4.03/1.90 & ~ Bwand: (Integer, Integer) -> Integer 4.03/1.90 >= ~ Ge: (Integer, Integer) -> Boolean 4.03/1.90 | ~ Bwor: (Integer, Integer) -> Integer 4.03/1.90 / ~ Div: (Integer, Integer) -> Integer 4.03/1.90 != ~ Neq: (Integer, Integer) -> Boolean 4.03/1.90 && ~ Land: (Boolean, Boolean) -> Boolean 4.03/1.90 ! ~ Lnot: (Boolean) -> Boolean 4.03/1.90 = ~ Eq: (Integer, Integer) -> Boolean 4.03/1.90 <= ~ Le: (Integer, Integer) -> Boolean 4.03/1.90 ^ ~ Bwxor: (Integer, Integer) -> Integer 4.03/1.90 % ~ Mod: (Integer, Integer) -> Integer 4.03/1.90 + ~ Add: (Integer, Integer) -> Integer 4.03/1.90 > ~ Gt: (Integer, Integer) -> Boolean 4.03/1.90 -1 ~ UnaryMinus: (Integer) -> Integer 4.03/1.90 < ~ Lt: (Integer, Integer) -> Boolean 4.03/1.90 || ~ Lor: (Boolean, Boolean) -> Boolean 4.03/1.90 - ~ Sub: (Integer, Integer) -> Integer 4.03/1.90 ~ ~ Bwnot: (Integer) -> Integer 4.03/1.90 * ~ Mul: (Integer, Integer) -> Integer 4.03/1.90 >>> 4.03/1.90 4.03/1.90 4.03/1.90 The following domains are used: 4.03/1.90 Integer 4.03/1.90 4.03/1.90 R is empty. 4.03/1.90 4.03/1.90 The integer pair graph contains the following rules and edges: 4.03/1.90 (3): COND_DIV2(TRUE, x[3], y[3]) -> DIV(x[3] - y[3], y[3]) 4.03/1.90 4.03/1.90 4.03/1.90 The set Q consists of the following terms: 4.03/1.90 div(x0, x1) 4.03/1.90 Cond_div(TRUE, x0, x1) 4.03/1.90 Cond_div1(TRUE, x0, x1) 4.03/1.90 Cond_div2(TRUE, x0, x1) 4.03/1.90 4.03/1.90 ---------------------------------------- 4.03/1.90 4.03/1.90 (9) IDependencyGraphProof (EQUIVALENT) 4.03/1.90 The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 1 less node. 4.03/1.90 ---------------------------------------- 4.03/1.90 4.03/1.90 (10) 4.03/1.90 TRUE 4.13/1.92 EOF