4.05/1.96 YES 4.05/1.97 proof of /export/starexec/sandbox/benchmark/theBenchmark.itrs 4.05/1.97 # AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty 4.05/1.97 4.05/1.97 4.05/1.97 Termination of the given ITRS could be proven: 4.05/1.97 4.05/1.97 (0) ITRS 4.05/1.97 (1) ITRStoIDPProof [EQUIVALENT, 0 ms] 4.05/1.97 (2) IDP 4.05/1.97 (3) UsableRulesProof [EQUIVALENT, 0 ms] 4.05/1.97 (4) IDP 4.05/1.97 (5) IDependencyGraphProof [EQUIVALENT, 0 ms] 4.05/1.97 (6) IDP 4.05/1.97 (7) IDPNonInfProof [SOUND, 154 ms] 4.05/1.97 (8) IDP 4.05/1.97 (9) IDependencyGraphProof [EQUIVALENT, 0 ms] 4.05/1.97 (10) TRUE 4.05/1.97 4.05/1.97 4.05/1.97 ---------------------------------------- 4.05/1.97 4.05/1.97 (0) 4.05/1.97 Obligation: 4.05/1.97 ITRS problem: 4.05/1.97 4.05/1.97 The following function symbols are pre-defined: 4.05/1.97 <<< 4.05/1.97 & ~ Bwand: (Integer, Integer) -> Integer 4.05/1.97 >= ~ Ge: (Integer, Integer) -> Boolean 4.05/1.97 | ~ Bwor: (Integer, Integer) -> Integer 4.05/1.97 / ~ Div: (Integer, Integer) -> Integer 4.05/1.97 != ~ Neq: (Integer, Integer) -> Boolean 4.05/1.97 && ~ Land: (Boolean, Boolean) -> Boolean 4.05/1.97 ! ~ Lnot: (Boolean) -> Boolean 4.05/1.97 = ~ Eq: (Integer, Integer) -> Boolean 4.05/1.97 <= ~ Le: (Integer, Integer) -> Boolean 4.05/1.97 ^ ~ Bwxor: (Integer, Integer) -> Integer 4.05/1.97 % ~ Mod: (Integer, Integer) -> Integer 4.05/1.97 > ~ Gt: (Integer, Integer) -> Boolean 4.05/1.97 + ~ Add: (Integer, Integer) -> Integer 4.05/1.97 -1 ~ UnaryMinus: (Integer) -> Integer 4.05/1.97 < ~ Lt: (Integer, Integer) -> Boolean 4.05/1.97 || ~ Lor: (Boolean, Boolean) -> Boolean 4.05/1.97 - ~ Sub: (Integer, Integer) -> Integer 4.05/1.97 ~ ~ Bwnot: (Integer) -> Integer 4.05/1.97 * ~ Mul: (Integer, Integer) -> Integer 4.05/1.97 >>> 4.05/1.97 4.05/1.97 The TRS R consists of the following rules: 4.05/1.97 random(x) -> Cond_random(x >= 0, x) 4.05/1.97 Cond_random(TRUE, x) -> rand(x, 0) 4.05/1.97 rand(x, y) -> Cond_rand(x = 0, x, y) 4.05/1.97 Cond_rand(TRUE, x, y) -> y 4.05/1.97 rand(x, y) -> Cond_rand1(x > 0, x, y) 4.05/1.97 Cond_rand1(TRUE, x, y) -> rand(x - 1, id_inc(y)) 4.05/1.97 id_inc(x) -> x 4.05/1.97 id_inc(x) -> x + 1 4.05/1.97 The set Q consists of the following terms: 4.05/1.97 random(x0) 4.05/1.97 Cond_random(TRUE, x0) 4.05/1.97 rand(x0, x1) 4.05/1.97 Cond_rand(TRUE, x0, x1) 4.05/1.97 Cond_rand1(TRUE, x0, x1) 4.05/1.97 id_inc(x0) 4.05/1.97 4.05/1.97 ---------------------------------------- 4.05/1.97 4.05/1.97 (1) ITRStoIDPProof (EQUIVALENT) 4.05/1.97 Added dependency pairs 4.05/1.97 ---------------------------------------- 4.05/1.97 4.05/1.97 (2) 4.05/1.97 Obligation: 4.05/1.97 IDP problem: 4.05/1.97 The following function symbols are pre-defined: 4.05/1.97 <<< 4.05/1.97 & ~ Bwand: (Integer, Integer) -> Integer 4.05/1.97 >= ~ Ge: (Integer, Integer) -> Boolean 4.05/1.97 | ~ Bwor: (Integer, Integer) -> Integer 4.05/1.97 / ~ Div: (Integer, Integer) -> Integer 4.05/1.97 != ~ Neq: (Integer, Integer) -> Boolean 4.05/1.97 && ~ Land: (Boolean, Boolean) -> Boolean 4.05/1.97 ! ~ Lnot: (Boolean) -> Boolean 4.05/1.97 = ~ Eq: (Integer, Integer) -> Boolean 4.05/1.97 <= ~ Le: (Integer, Integer) -> Boolean 4.05/1.97 ^ ~ Bwxor: (Integer, Integer) -> Integer 4.05/1.97 % ~ Mod: (Integer, Integer) -> Integer 4.05/1.97 > ~ Gt: (Integer, Integer) -> Boolean 4.05/1.97 + ~ Add: (Integer, Integer) -> Integer 4.05/1.97 -1 ~ UnaryMinus: (Integer) -> Integer 4.05/1.97 < ~ Lt: (Integer, Integer) -> Boolean 4.05/1.97 || ~ Lor: (Boolean, Boolean) -> Boolean 4.05/1.97 - ~ Sub: (Integer, Integer) -> Integer 4.05/1.97 ~ ~ Bwnot: (Integer) -> Integer 4.05/1.97 * ~ Mul: (Integer, Integer) -> Integer 4.05/1.97 >>> 4.05/1.97 4.05/1.97 4.05/1.97 The following domains are used: 4.05/1.97 Integer 4.05/1.97 4.05/1.97 The ITRS R consists of the following rules: 4.05/1.97 random(x) -> Cond_random(x >= 0, x) 4.05/1.97 Cond_random(TRUE, x) -> rand(x, 0) 4.05/1.97 rand(x, y) -> Cond_rand(x = 0, x, y) 4.05/1.97 Cond_rand(TRUE, x, y) -> y 4.05/1.97 rand(x, y) -> Cond_rand1(x > 0, x, y) 4.05/1.97 Cond_rand1(TRUE, x, y) -> rand(x - 1, id_inc(y)) 4.05/1.97 id_inc(x) -> x 4.05/1.97 id_inc(x) -> x + 1 4.05/1.97 4.05/1.97 The integer pair graph contains the following rules and edges: 4.05/1.97 (0): RANDOM(x[0]) -> COND_RANDOM(x[0] >= 0, x[0]) 4.05/1.97 (1): COND_RANDOM(TRUE, x[1]) -> RAND(x[1], 0) 4.05/1.97 (2): RAND(x[2], y[2]) -> COND_RAND(x[2] = 0, x[2], y[2]) 4.05/1.97 (3): RAND(x[3], y[3]) -> COND_RAND1(x[3] > 0, x[3], y[3]) 4.05/1.97 (4): COND_RAND1(TRUE, x[4], y[4]) -> RAND(x[4] - 1, id_inc(y[4])) 4.05/1.97 (5): COND_RAND1(TRUE, x[5], y[5]) -> ID_INC(y[5]) 4.05/1.97 4.05/1.97 (0) -> (1), if (x[0] >= 0 & x[0] ->^* x[1]) 4.05/1.97 (1) -> (2), if (x[1] ->^* x[2] & 0 ->^* y[2]) 4.05/1.97 (1) -> (3), if (x[1] ->^* x[3] & 0 ->^* y[3]) 4.05/1.97 (3) -> (4), if (x[3] > 0 & x[3] ->^* x[4] & y[3] ->^* y[4]) 4.05/1.97 (3) -> (5), if (x[3] > 0 & x[3] ->^* x[5] & y[3] ->^* y[5]) 4.05/1.97 (4) -> (2), if (x[4] - 1 ->^* x[2] & id_inc(y[4]) ->^* y[2]) 4.05/1.97 (4) -> (3), if (x[4] - 1 ->^* x[3] & id_inc(y[4]) ->^* y[3]) 4.05/1.97 4.05/1.97 The set Q consists of the following terms: 4.05/1.97 random(x0) 4.05/1.97 Cond_random(TRUE, x0) 4.05/1.97 rand(x0, x1) 4.05/1.97 Cond_rand(TRUE, x0, x1) 4.05/1.97 Cond_rand1(TRUE, x0, x1) 4.05/1.97 id_inc(x0) 4.05/1.97 4.05/1.97 ---------------------------------------- 4.05/1.97 4.05/1.97 (3) UsableRulesProof (EQUIVALENT) 4.05/1.97 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. 4.05/1.97 ---------------------------------------- 4.05/1.97 4.05/1.97 (4) 4.05/1.97 Obligation: 4.05/1.97 IDP problem: 4.05/1.97 The following function symbols are pre-defined: 4.05/1.97 <<< 4.05/1.97 & ~ Bwand: (Integer, Integer) -> Integer 4.05/1.97 >= ~ Ge: (Integer, Integer) -> Boolean 4.05/1.97 | ~ Bwor: (Integer, Integer) -> Integer 4.05/1.97 / ~ Div: (Integer, Integer) -> Integer 4.05/1.97 != ~ Neq: (Integer, Integer) -> Boolean 4.05/1.97 && ~ Land: (Boolean, Boolean) -> Boolean 4.05/1.97 ! ~ Lnot: (Boolean) -> Boolean 4.05/1.97 = ~ Eq: (Integer, Integer) -> Boolean 4.05/1.97 <= ~ Le: (Integer, Integer) -> Boolean 4.05/1.97 ^ ~ Bwxor: (Integer, Integer) -> Integer 4.05/1.97 % ~ Mod: (Integer, Integer) -> Integer 4.05/1.97 > ~ Gt: (Integer, Integer) -> Boolean 4.05/1.97 + ~ Add: (Integer, Integer) -> Integer 4.05/1.97 -1 ~ UnaryMinus: (Integer) -> Integer 4.05/1.97 < ~ Lt: (Integer, Integer) -> Boolean 4.05/1.97 || ~ Lor: (Boolean, Boolean) -> Boolean 4.05/1.97 - ~ Sub: (Integer, Integer) -> Integer 4.05/1.97 ~ ~ Bwnot: (Integer) -> Integer 4.05/1.97 * ~ Mul: (Integer, Integer) -> Integer 4.05/1.97 >>> 4.05/1.97 4.05/1.97 4.05/1.97 The following domains are used: 4.05/1.97 Integer 4.05/1.97 4.05/1.97 The ITRS R consists of the following rules: 4.05/1.97 id_inc(x) -> x 4.05/1.97 id_inc(x) -> x + 1 4.05/1.97 4.05/1.97 The integer pair graph contains the following rules and edges: 4.05/1.97 (0): RANDOM(x[0]) -> COND_RANDOM(x[0] >= 0, x[0]) 4.05/1.97 (1): COND_RANDOM(TRUE, x[1]) -> RAND(x[1], 0) 4.05/1.97 (2): RAND(x[2], y[2]) -> COND_RAND(x[2] = 0, x[2], y[2]) 4.05/1.97 (3): RAND(x[3], y[3]) -> COND_RAND1(x[3] > 0, x[3], y[3]) 4.05/1.97 (4): COND_RAND1(TRUE, x[4], y[4]) -> RAND(x[4] - 1, id_inc(y[4])) 4.05/1.97 (5): COND_RAND1(TRUE, x[5], y[5]) -> ID_INC(y[5]) 4.05/1.97 4.05/1.97 (0) -> (1), if (x[0] >= 0 & x[0] ->^* x[1]) 4.05/1.97 (1) -> (2), if (x[1] ->^* x[2] & 0 ->^* y[2]) 4.05/1.97 (1) -> (3), if (x[1] ->^* x[3] & 0 ->^* y[3]) 4.05/1.97 (3) -> (4), if (x[3] > 0 & x[3] ->^* x[4] & y[3] ->^* y[4]) 4.05/1.97 (3) -> (5), if (x[3] > 0 & x[3] ->^* x[5] & y[3] ->^* y[5]) 4.05/1.97 (4) -> (2), if (x[4] - 1 ->^* x[2] & id_inc(y[4]) ->^* y[2]) 4.05/1.97 (4) -> (3), if (x[4] - 1 ->^* x[3] & id_inc(y[4]) ->^* y[3]) 4.05/1.97 4.05/1.97 The set Q consists of the following terms: 4.05/1.97 random(x0) 4.05/1.97 Cond_random(TRUE, x0) 4.05/1.97 rand(x0, x1) 4.05/1.97 Cond_rand(TRUE, x0, x1) 4.05/1.97 Cond_rand1(TRUE, x0, x1) 4.05/1.97 id_inc(x0) 4.05/1.97 4.05/1.97 ---------------------------------------- 4.05/1.97 4.05/1.97 (5) IDependencyGraphProof (EQUIVALENT) 4.05/1.97 The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 4 less nodes. 4.05/1.97 ---------------------------------------- 4.05/1.97 4.05/1.97 (6) 4.05/1.97 Obligation: 4.05/1.97 IDP problem: 4.05/1.97 The following function symbols are pre-defined: 4.05/1.97 <<< 4.05/1.97 & ~ Bwand: (Integer, Integer) -> Integer 4.05/1.97 >= ~ Ge: (Integer, Integer) -> Boolean 4.05/1.97 | ~ Bwor: (Integer, Integer) -> Integer 4.05/1.97 / ~ Div: (Integer, Integer) -> Integer 4.05/1.97 != ~ Neq: (Integer, Integer) -> Boolean 4.05/1.97 && ~ Land: (Boolean, Boolean) -> Boolean 4.05/1.97 ! ~ Lnot: (Boolean) -> Boolean 4.05/1.97 = ~ Eq: (Integer, Integer) -> Boolean 4.05/1.97 <= ~ Le: (Integer, Integer) -> Boolean 4.05/1.97 ^ ~ Bwxor: (Integer, Integer) -> Integer 4.05/1.97 % ~ Mod: (Integer, Integer) -> Integer 4.05/1.97 > ~ Gt: (Integer, Integer) -> Boolean 4.05/1.97 + ~ Add: (Integer, Integer) -> Integer 4.05/1.97 -1 ~ UnaryMinus: (Integer) -> Integer 4.05/1.97 < ~ Lt: (Integer, Integer) -> Boolean 4.05/1.97 || ~ Lor: (Boolean, Boolean) -> Boolean 4.05/1.97 - ~ Sub: (Integer, Integer) -> Integer 4.05/1.97 ~ ~ Bwnot: (Integer) -> Integer 4.05/1.97 * ~ Mul: (Integer, Integer) -> Integer 4.05/1.97 >>> 4.05/1.97 4.05/1.97 4.05/1.97 The following domains are used: 4.05/1.97 Integer 4.05/1.97 4.05/1.97 The ITRS R consists of the following rules: 4.05/1.97 id_inc(x) -> x 4.05/1.97 id_inc(x) -> x + 1 4.05/1.97 4.05/1.97 The integer pair graph contains the following rules and edges: 4.05/1.97 (4): COND_RAND1(TRUE, x[4], y[4]) -> RAND(x[4] - 1, id_inc(y[4])) 4.05/1.97 (3): RAND(x[3], y[3]) -> COND_RAND1(x[3] > 0, x[3], y[3]) 4.05/1.97 4.05/1.97 (4) -> (3), if (x[4] - 1 ->^* x[3] & id_inc(y[4]) ->^* y[3]) 4.05/1.97 (3) -> (4), if (x[3] > 0 & x[3] ->^* x[4] & y[3] ->^* y[4]) 4.05/1.97 4.05/1.97 The set Q consists of the following terms: 4.05/1.97 random(x0) 4.05/1.97 Cond_random(TRUE, x0) 4.05/1.97 rand(x0, x1) 4.05/1.97 Cond_rand(TRUE, x0, x1) 4.05/1.97 Cond_rand1(TRUE, x0, x1) 4.05/1.97 id_inc(x0) 4.05/1.97 4.05/1.97 ---------------------------------------- 4.05/1.97 4.05/1.97 (7) IDPNonInfProof (SOUND) 4.05/1.97 Used the following options for this NonInfProof: 4.05/1.97 4.05/1.97 IDPGPoloSolver: 4.05/1.97 Range: [(-1,2)] 4.05/1.97 IsNat: false 4.05/1.97 Interpretation Shape Heuristic: aprove.DPFramework.IDPProblem.Processors.nonInf.poly.IdpDefaultShapeHeuristic@607274bc 4.05/1.97 Constraint Generator: NonInfConstraintGenerator: 4.05/1.97 PathGenerator: MetricPathGenerator: 4.05/1.97 Max Left Steps: 1 4.05/1.97 Max Right Steps: 1 4.05/1.97 4.05/1.97 4.05/1.97 4.05/1.97 The constraints were generated the following way: 4.05/1.97 4.05/1.97 The DP Problem is simplified using the Induction Calculus [NONINF] with the following steps: 4.05/1.97 4.05/1.97 Note that final constraints are written in bold face. 4.05/1.97 4.05/1.97 4.05/1.97 4.05/1.97 For Pair COND_RAND1(TRUE, x[4], y[4]) -> RAND(-(x[4], 1), id_inc(y[4])) the following chains were created: 4.05/1.97 *We consider the chain RAND(x[3], y[3]) -> COND_RAND1(>(x[3], 0), x[3], y[3]), COND_RAND1(TRUE, x[4], y[4]) -> RAND(-(x[4], 1), id_inc(y[4])), RAND(x[3], y[3]) -> COND_RAND1(>(x[3], 0), x[3], y[3]) which results in the following constraint: 4.05/1.97 4.05/1.97 (1) (>(x[3], 0)=TRUE & x[3]=x[4] & y[3]=y[4] & -(x[4], 1)=x[3]1 & id_inc(y[4])=y[3]1 ==> COND_RAND1(TRUE, x[4], y[4])_>=_NonInfC & COND_RAND1(TRUE, x[4], y[4])_>=_RAND(-(x[4], 1), id_inc(y[4])) & (U^Increasing(RAND(-(x[4], 1), id_inc(y[4]))), >=)) 4.05/1.97 4.05/1.97 4.05/1.97 4.05/1.97 We simplified constraint (1) using rules (III), (IV) which results in the following new constraint: 4.05/1.97 4.05/1.97 (2) (>(x[3], 0)=TRUE ==> COND_RAND1(TRUE, x[3], y[3])_>=_NonInfC & COND_RAND1(TRUE, x[3], y[3])_>=_RAND(-(x[3], 1), id_inc(y[3])) & (U^Increasing(RAND(-(x[4], 1), id_inc(y[4]))), >=)) 4.05/1.97 4.05/1.97 4.05/1.97 4.05/1.97 We simplified constraint (2) using rule (POLY_CONSTRAINTS) which results in the following new constraint: 4.05/1.97 4.05/1.97 (3) (x[3] + [-1] >= 0 ==> (U^Increasing(RAND(-(x[4], 1), id_inc(y[4]))), >=) & [(-1)bni_13 + (-1)Bound*bni_13] + [(2)bni_13]x[3] >= 0 & [(-1)bso_14] >= 0) 4.05/1.97 4.05/1.97 4.05/1.97 4.05/1.97 We simplified constraint (3) using rule (IDP_POLY_SIMPLIFY) which results in the following new constraint: 4.05/1.97 4.05/1.97 (4) (x[3] + [-1] >= 0 ==> (U^Increasing(RAND(-(x[4], 1), id_inc(y[4]))), >=) & [(-1)bni_13 + (-1)Bound*bni_13] + [(2)bni_13]x[3] >= 0 & [(-1)bso_14] >= 0) 4.05/1.97 4.05/1.97 4.05/1.97 4.05/1.97 We simplified constraint (4) using rule (POLY_REMOVE_MIN_MAX) which results in the following new constraint: 4.05/1.97 4.05/1.97 (5) (x[3] + [-1] >= 0 ==> (U^Increasing(RAND(-(x[4], 1), id_inc(y[4]))), >=) & [(-1)bni_13 + (-1)Bound*bni_13] + [(2)bni_13]x[3] >= 0 & [(-1)bso_14] >= 0) 4.05/1.97 4.05/1.97 4.05/1.97 4.05/1.97 We simplified constraint (5) using rule (IDP_UNRESTRICTED_VARS) which results in the following new constraint: 4.05/1.97 4.05/1.97 (6) (x[3] + [-1] >= 0 ==> (U^Increasing(RAND(-(x[4], 1), id_inc(y[4]))), >=) & 0 = 0 & [(-1)bni_13 + (-1)Bound*bni_13] + [(2)bni_13]x[3] >= 0 & 0 = 0 & [(-1)bso_14] >= 0) 4.05/1.97 4.05/1.97 4.05/1.97 4.05/1.97 4.05/1.97 4.05/1.97 4.05/1.97 4.05/1.97 4.05/1.97 For Pair RAND(x[3], y[3]) -> COND_RAND1(>(x[3], 0), x[3], y[3]) the following chains were created: 4.05/1.97 *We consider the chain RAND(x[3], y[3]) -> COND_RAND1(>(x[3], 0), x[3], y[3]), COND_RAND1(TRUE, x[4], y[4]) -> RAND(-(x[4], 1), id_inc(y[4])) which results in the following constraint: 4.05/1.97 4.05/1.97 (1) (>(x[3], 0)=TRUE & x[3]=x[4] & y[3]=y[4] ==> RAND(x[3], y[3])_>=_NonInfC & RAND(x[3], y[3])_>=_COND_RAND1(>(x[3], 0), x[3], y[3]) & (U^Increasing(COND_RAND1(>(x[3], 0), x[3], y[3])), >=)) 4.05/1.97 4.05/1.97 4.05/1.97 4.05/1.97 We simplified constraint (1) using rule (IV) which results in the following new constraint: 4.05/1.97 4.05/1.97 (2) (>(x[3], 0)=TRUE ==> RAND(x[3], y[3])_>=_NonInfC & RAND(x[3], y[3])_>=_COND_RAND1(>(x[3], 0), x[3], y[3]) & (U^Increasing(COND_RAND1(>(x[3], 0), x[3], y[3])), >=)) 4.05/1.97 4.05/1.97 4.05/1.97 4.05/1.97 We simplified constraint (2) using rule (POLY_CONSTRAINTS) which results in the following new constraint: 4.05/1.97 4.05/1.97 (3) (x[3] + [-1] >= 0 ==> (U^Increasing(COND_RAND1(>(x[3], 0), x[3], y[3])), >=) & [bni_15 + (-1)Bound*bni_15] + [(2)bni_15]x[3] >= 0 & [2 + (-1)bso_16] >= 0) 4.05/1.97 4.05/1.97 4.05/1.97 4.05/1.97 We simplified constraint (3) using rule (IDP_POLY_SIMPLIFY) which results in the following new constraint: 4.05/1.97 4.05/1.97 (4) (x[3] + [-1] >= 0 ==> (U^Increasing(COND_RAND1(>(x[3], 0), x[3], y[3])), >=) & [bni_15 + (-1)Bound*bni_15] + [(2)bni_15]x[3] >= 0 & [2 + (-1)bso_16] >= 0) 4.05/1.97 4.05/1.97 4.05/1.97 4.05/1.97 We simplified constraint (4) using rule (POLY_REMOVE_MIN_MAX) which results in the following new constraint: 4.05/1.97 4.05/1.97 (5) (x[3] + [-1] >= 0 ==> (U^Increasing(COND_RAND1(>(x[3], 0), x[3], y[3])), >=) & [bni_15 + (-1)Bound*bni_15] + [(2)bni_15]x[3] >= 0 & [2 + (-1)bso_16] >= 0) 4.05/1.97 4.05/1.97 4.05/1.97 4.05/1.97 We simplified constraint (5) using rule (IDP_UNRESTRICTED_VARS) which results in the following new constraint: 4.05/1.97 4.05/1.97 (6) (x[3] + [-1] >= 0 ==> (U^Increasing(COND_RAND1(>(x[3], 0), x[3], y[3])), >=) & 0 = 0 & [bni_15 + (-1)Bound*bni_15] + [(2)bni_15]x[3] >= 0 & [2 + (-1)bso_16] >= 0) 4.05/1.97 4.05/1.97 4.05/1.97 4.05/1.97 4.05/1.97 4.05/1.97 4.05/1.97 4.05/1.97 4.05/1.97 To summarize, we get the following constraints P__>=_ for the following pairs. 4.05/1.97 4.05/1.97 *COND_RAND1(TRUE, x[4], y[4]) -> RAND(-(x[4], 1), id_inc(y[4])) 4.05/1.97 4.05/1.97 *(x[3] + [-1] >= 0 ==> (U^Increasing(RAND(-(x[4], 1), id_inc(y[4]))), >=) & 0 = 0 & [(-1)bni_13 + (-1)Bound*bni_13] + [(2)bni_13]x[3] >= 0 & 0 = 0 & [(-1)bso_14] >= 0) 4.05/1.97 4.05/1.97 4.05/1.97 4.05/1.97 4.05/1.97 *RAND(x[3], y[3]) -> COND_RAND1(>(x[3], 0), x[3], y[3]) 4.05/1.97 4.05/1.97 *(x[3] + [-1] >= 0 ==> (U^Increasing(COND_RAND1(>(x[3], 0), x[3], y[3])), >=) & 0 = 0 & [bni_15 + (-1)Bound*bni_15] + [(2)bni_15]x[3] >= 0 & [2 + (-1)bso_16] >= 0) 4.05/1.97 4.05/1.97 4.05/1.97 4.05/1.97 4.05/1.97 4.05/1.97 4.05/1.97 4.05/1.97 4.05/1.97 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. 4.05/1.97 4.05/1.97 Using the following integer polynomial ordering the resulting constraints can be solved 4.05/1.97 4.05/1.97 Polynomial interpretation over integers[POLO]: 4.05/1.97 4.05/1.97 POL(TRUE) = 0 4.05/1.97 POL(FALSE) = 0 4.05/1.97 POL(id_inc(x_1)) = [-1] + [-1]x_1 4.05/1.97 POL(+(x_1, x_2)) = x_1 + x_2 4.05/1.97 POL(1) = [1] 4.05/1.97 POL(COND_RAND1(x_1, x_2, x_3)) = [-1] + [2]x_2 4.05/1.97 POL(RAND(x_1, x_2)) = [1] + [2]x_1 4.05/1.97 POL(-(x_1, x_2)) = x_1 + [-1]x_2 4.05/1.97 POL(>(x_1, x_2)) = [1] 4.05/1.97 POL(0) = 0 4.05/1.97 4.05/1.97 4.05/1.97 The following pairs are in P_>: 4.05/1.97 4.05/1.97 4.05/1.97 RAND(x[3], y[3]) -> COND_RAND1(>(x[3], 0), x[3], y[3]) 4.05/1.97 4.05/1.97 4.05/1.97 The following pairs are in P_bound: 4.05/1.97 4.05/1.97 4.05/1.97 COND_RAND1(TRUE, x[4], y[4]) -> RAND(-(x[4], 1), id_inc(y[4])) 4.05/1.97 RAND(x[3], y[3]) -> COND_RAND1(>(x[3], 0), x[3], y[3]) 4.05/1.97 4.05/1.97 4.05/1.97 The following pairs are in P_>=: 4.05/1.97 4.05/1.97 4.05/1.97 COND_RAND1(TRUE, x[4], y[4]) -> RAND(-(x[4], 1), id_inc(y[4])) 4.05/1.97 4.05/1.97 4.05/1.97 There are no usable rules. 4.05/1.97 ---------------------------------------- 4.05/1.97 4.05/1.97 (8) 4.05/1.97 Obligation: 4.05/1.97 IDP problem: 4.05/1.97 The following function symbols are pre-defined: 4.05/1.97 <<< 4.05/1.97 & ~ Bwand: (Integer, Integer) -> Integer 4.05/1.97 >= ~ Ge: (Integer, Integer) -> Boolean 4.05/1.97 | ~ Bwor: (Integer, Integer) -> Integer 4.05/1.97 / ~ Div: (Integer, Integer) -> Integer 4.05/1.97 != ~ Neq: (Integer, Integer) -> Boolean 4.05/1.97 && ~ Land: (Boolean, Boolean) -> Boolean 4.05/1.97 ! ~ Lnot: (Boolean) -> Boolean 4.05/1.97 = ~ Eq: (Integer, Integer) -> Boolean 4.05/1.97 <= ~ Le: (Integer, Integer) -> Boolean 4.05/1.97 ^ ~ Bwxor: (Integer, Integer) -> Integer 4.05/1.97 % ~ Mod: (Integer, Integer) -> Integer 4.05/1.97 > ~ Gt: (Integer, Integer) -> Boolean 4.05/1.97 + ~ Add: (Integer, Integer) -> Integer 4.05/1.97 -1 ~ UnaryMinus: (Integer) -> Integer 4.05/1.97 < ~ Lt: (Integer, Integer) -> Boolean 4.05/1.97 || ~ Lor: (Boolean, Boolean) -> Boolean 4.05/1.97 - ~ Sub: (Integer, Integer) -> Integer 4.05/1.97 ~ ~ Bwnot: (Integer) -> Integer 4.05/1.97 * ~ Mul: (Integer, Integer) -> Integer 4.05/1.97 >>> 4.05/1.97 4.05/1.97 4.05/1.97 The following domains are used: 4.05/1.97 Integer 4.05/1.97 4.05/1.97 The ITRS R consists of the following rules: 4.05/1.97 id_inc(x) -> x 4.05/1.97 id_inc(x) -> x + 1 4.05/1.97 4.05/1.97 The integer pair graph contains the following rules and edges: 4.05/1.97 (4): COND_RAND1(TRUE, x[4], y[4]) -> RAND(x[4] - 1, id_inc(y[4])) 4.05/1.97 4.05/1.97 4.05/1.97 The set Q consists of the following terms: 4.05/1.97 random(x0) 4.05/1.97 Cond_random(TRUE, x0) 4.05/1.97 rand(x0, x1) 4.05/1.97 Cond_rand(TRUE, x0, x1) 4.05/1.97 Cond_rand1(TRUE, x0, x1) 4.05/1.97 id_inc(x0) 4.05/1.97 4.05/1.97 ---------------------------------------- 4.05/1.97 4.05/1.97 (9) IDependencyGraphProof (EQUIVALENT) 4.05/1.97 The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 1 less node. 4.05/1.97 ---------------------------------------- 4.05/1.97 4.05/1.97 (10) 4.05/1.97 TRUE 4.05/1.98 EOF