3.89/1.82 YES 3.89/1.83 proof of /export/starexec/sandbox2/benchmark/theBenchmark.itrs 3.89/1.83 # AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty 3.89/1.83 3.89/1.83 3.89/1.83 Termination of the given ITRS could be proven: 3.89/1.83 3.89/1.83 (0) ITRS 3.89/1.83 (1) ITRStoIDPProof [EQUIVALENT, 0 ms] 3.89/1.83 (2) IDP 3.89/1.83 (3) UsableRulesProof [EQUIVALENT, 0 ms] 3.89/1.83 (4) IDP 3.89/1.83 (5) IDPNonInfProof [SOUND, 154 ms] 3.89/1.83 (6) IDP 3.89/1.83 (7) IDependencyGraphProof [EQUIVALENT, 0 ms] 3.89/1.83 (8) TRUE 3.89/1.83 3.89/1.83 3.89/1.83 ---------------------------------------- 3.89/1.83 3.89/1.83 (0) 3.89/1.83 Obligation: 3.89/1.83 ITRS problem: 3.89/1.83 3.89/1.83 The following function symbols are pre-defined: 3.89/1.83 <<< 3.89/1.83 & ~ Bwand: (Integer, Integer) -> Integer 3.89/1.83 >= ~ Ge: (Integer, Integer) -> Boolean 3.89/1.83 | ~ Bwor: (Integer, Integer) -> Integer 3.89/1.83 / ~ Div: (Integer, Integer) -> Integer 3.89/1.83 != ~ Neq: (Integer, Integer) -> Boolean 3.89/1.83 && ~ Land: (Boolean, Boolean) -> Boolean 3.89/1.83 ! ~ Lnot: (Boolean) -> Boolean 3.89/1.83 = ~ Eq: (Integer, Integer) -> Boolean 3.89/1.83 <= ~ Le: (Integer, Integer) -> Boolean 3.89/1.83 ^ ~ Bwxor: (Integer, Integer) -> Integer 3.89/1.83 % ~ Mod: (Integer, Integer) -> Integer 3.89/1.83 + ~ Add: (Integer, Integer) -> Integer 3.89/1.83 > ~ Gt: (Integer, Integer) -> Boolean 3.89/1.83 -1 ~ UnaryMinus: (Integer) -> Integer 3.89/1.83 < ~ Lt: (Integer, Integer) -> Boolean 3.89/1.83 || ~ Lor: (Boolean, Boolean) -> Boolean 3.89/1.83 - ~ Sub: (Integer, Integer) -> Integer 3.89/1.83 ~ ~ Bwnot: (Integer) -> Integer 3.89/1.83 * ~ Mul: (Integer, Integer) -> Integer 3.89/1.83 >>> 3.89/1.83 3.89/1.83 The TRS R consists of the following rules: 3.89/1.83 f(x, y) -> Cond_f(x > y, x, y) 3.89/1.83 Cond_f(TRUE, x, y) -> f(x, y + 1) 3.89/1.83 The set Q consists of the following terms: 3.89/1.83 f(x0, x1) 3.89/1.83 Cond_f(TRUE, x0, x1) 3.89/1.83 3.89/1.83 ---------------------------------------- 3.89/1.83 3.89/1.83 (1) ITRStoIDPProof (EQUIVALENT) 3.89/1.83 Added dependency pairs 3.89/1.83 ---------------------------------------- 3.89/1.83 3.89/1.83 (2) 3.89/1.83 Obligation: 3.89/1.83 IDP problem: 3.89/1.83 The following function symbols are pre-defined: 3.89/1.83 <<< 3.89/1.83 & ~ Bwand: (Integer, Integer) -> Integer 3.89/1.83 >= ~ Ge: (Integer, Integer) -> Boolean 3.89/1.83 | ~ Bwor: (Integer, Integer) -> Integer 3.89/1.83 / ~ Div: (Integer, Integer) -> Integer 3.89/1.83 != ~ Neq: (Integer, Integer) -> Boolean 3.89/1.83 && ~ Land: (Boolean, Boolean) -> Boolean 3.89/1.83 ! ~ Lnot: (Boolean) -> Boolean 3.89/1.83 = ~ Eq: (Integer, Integer) -> Boolean 3.89/1.83 <= ~ Le: (Integer, Integer) -> Boolean 3.89/1.83 ^ ~ Bwxor: (Integer, Integer) -> Integer 3.89/1.83 % ~ Mod: (Integer, Integer) -> Integer 3.89/1.83 + ~ Add: (Integer, Integer) -> Integer 3.89/1.83 > ~ Gt: (Integer, Integer) -> Boolean 3.89/1.83 -1 ~ UnaryMinus: (Integer) -> Integer 3.89/1.83 < ~ Lt: (Integer, Integer) -> Boolean 3.89/1.83 || ~ Lor: (Boolean, Boolean) -> Boolean 3.89/1.83 - ~ Sub: (Integer, Integer) -> Integer 3.89/1.83 ~ ~ Bwnot: (Integer) -> Integer 3.89/1.83 * ~ Mul: (Integer, Integer) -> Integer 3.89/1.83 >>> 3.89/1.83 3.89/1.83 3.89/1.83 The following domains are used: 3.89/1.83 Integer 3.89/1.83 3.89/1.83 The ITRS R consists of the following rules: 3.89/1.83 f(x, y) -> Cond_f(x > y, x, y) 3.89/1.83 Cond_f(TRUE, x, y) -> f(x, y + 1) 3.89/1.83 3.89/1.83 The integer pair graph contains the following rules and edges: 3.89/1.83 (0): F(x[0], y[0]) -> COND_F(x[0] > y[0], x[0], y[0]) 3.89/1.83 (1): COND_F(TRUE, x[1], y[1]) -> F(x[1], y[1] + 1) 3.89/1.83 3.89/1.83 (0) -> (1), if (x[0] > y[0] & x[0] ->^* x[1] & y[0] ->^* y[1]) 3.89/1.83 (1) -> (0), if (x[1] ->^* x[0] & y[1] + 1 ->^* y[0]) 3.89/1.83 3.89/1.83 The set Q consists of the following terms: 3.89/1.83 f(x0, x1) 3.89/1.83 Cond_f(TRUE, x0, x1) 3.89/1.83 3.89/1.83 ---------------------------------------- 3.89/1.83 3.89/1.83 (3) UsableRulesProof (EQUIVALENT) 3.89/1.83 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. 3.89/1.83 ---------------------------------------- 3.89/1.83 3.89/1.83 (4) 3.89/1.83 Obligation: 3.89/1.83 IDP problem: 3.89/1.83 The following function symbols are pre-defined: 3.89/1.83 <<< 3.89/1.83 & ~ Bwand: (Integer, Integer) -> Integer 3.89/1.83 >= ~ Ge: (Integer, Integer) -> Boolean 3.89/1.83 | ~ Bwor: (Integer, Integer) -> Integer 3.89/1.83 / ~ Div: (Integer, Integer) -> Integer 3.89/1.83 != ~ Neq: (Integer, Integer) -> Boolean 3.89/1.83 && ~ Land: (Boolean, Boolean) -> Boolean 3.89/1.83 ! ~ Lnot: (Boolean) -> Boolean 3.89/1.83 = ~ Eq: (Integer, Integer) -> Boolean 3.89/1.83 <= ~ Le: (Integer, Integer) -> Boolean 3.89/1.83 ^ ~ Bwxor: (Integer, Integer) -> Integer 3.89/1.83 % ~ Mod: (Integer, Integer) -> Integer 3.89/1.83 + ~ Add: (Integer, Integer) -> Integer 3.89/1.83 > ~ Gt: (Integer, Integer) -> Boolean 3.89/1.83 -1 ~ UnaryMinus: (Integer) -> Integer 3.89/1.83 < ~ Lt: (Integer, Integer) -> Boolean 3.89/1.83 || ~ Lor: (Boolean, Boolean) -> Boolean 3.89/1.83 - ~ Sub: (Integer, Integer) -> Integer 3.89/1.83 ~ ~ Bwnot: (Integer) -> Integer 3.89/1.83 * ~ Mul: (Integer, Integer) -> Integer 3.89/1.83 >>> 3.89/1.83 3.89/1.83 3.89/1.83 The following domains are used: 3.89/1.83 Integer 3.89/1.83 3.89/1.83 R is empty. 3.89/1.83 3.89/1.83 The integer pair graph contains the following rules and edges: 3.89/1.83 (0): F(x[0], y[0]) -> COND_F(x[0] > y[0], x[0], y[0]) 3.89/1.83 (1): COND_F(TRUE, x[1], y[1]) -> F(x[1], y[1] + 1) 3.89/1.83 3.89/1.83 (0) -> (1), if (x[0] > y[0] & x[0] ->^* x[1] & y[0] ->^* y[1]) 3.89/1.83 (1) -> (0), if (x[1] ->^* x[0] & y[1] + 1 ->^* y[0]) 3.89/1.83 3.89/1.83 The set Q consists of the following terms: 3.89/1.83 f(x0, x1) 3.89/1.83 Cond_f(TRUE, x0, x1) 3.89/1.83 3.89/1.83 ---------------------------------------- 3.89/1.83 3.89/1.83 (5) IDPNonInfProof (SOUND) 3.89/1.83 Used the following options for this NonInfProof: 3.89/1.83 3.89/1.83 IDPGPoloSolver: 3.89/1.83 Range: [(-1,2)] 3.89/1.83 IsNat: false 3.89/1.83 Interpretation Shape Heuristic: aprove.DPFramework.IDPProblem.Processors.nonInf.poly.IdpDefaultShapeHeuristic@1ccfbc04 3.89/1.83 Constraint Generator: NonInfConstraintGenerator: 3.89/1.83 PathGenerator: MetricPathGenerator: 3.89/1.83 Max Left Steps: 1 3.89/1.83 Max Right Steps: 1 3.89/1.83 3.89/1.83 3.89/1.83 3.89/1.83 The constraints were generated the following way: 3.89/1.83 3.89/1.83 The DP Problem is simplified using the Induction Calculus [NONINF] with the following steps: 3.89/1.83 3.89/1.83 Note that final constraints are written in bold face. 3.89/1.83 3.89/1.83 3.89/1.83 3.89/1.83 For Pair F(x, y) -> COND_F(>(x, y), x, y) the following chains were created: 3.89/1.83 *We consider the chain F(x[0], y[0]) -> COND_F(>(x[0], y[0]), x[0], y[0]), COND_F(TRUE, x[1], y[1]) -> F(x[1], +(y[1], 1)) which results in the following constraint: 3.89/1.83 3.89/1.83 (1) (>(x[0], y[0])=TRUE & x[0]=x[1] & y[0]=y[1] ==> F(x[0], y[0])_>=_NonInfC & F(x[0], y[0])_>=_COND_F(>(x[0], y[0]), x[0], y[0]) & (U^Increasing(COND_F(>(x[0], y[0]), x[0], y[0])), >=)) 3.89/1.83 3.89/1.83 3.89/1.83 3.89/1.83 We simplified constraint (1) using rule (IV) which results in the following new constraint: 3.89/1.83 3.89/1.83 (2) (>(x[0], y[0])=TRUE ==> F(x[0], y[0])_>=_NonInfC & F(x[0], y[0])_>=_COND_F(>(x[0], y[0]), x[0], y[0]) & (U^Increasing(COND_F(>(x[0], y[0]), x[0], y[0])), >=)) 3.89/1.83 3.89/1.83 3.89/1.83 3.89/1.83 We simplified constraint (2) using rule (POLY_CONSTRAINTS) which results in the following new constraint: 3.89/1.83 3.89/1.83 (3) (x[0] + [-1] + [-1]y[0] >= 0 ==> (U^Increasing(COND_F(>(x[0], y[0]), x[0], y[0])), >=) & [(-1)bni_11 + (-1)Bound*bni_11] + [(-1)bni_11]y[0] + [bni_11]x[0] >= 0 & [(-1)bso_12] >= 0) 3.89/1.83 3.89/1.83 3.89/1.83 3.89/1.83 We simplified constraint (3) using rule (IDP_POLY_SIMPLIFY) which results in the following new constraint: 3.89/1.83 3.89/1.83 (4) (x[0] + [-1] + [-1]y[0] >= 0 ==> (U^Increasing(COND_F(>(x[0], y[0]), x[0], y[0])), >=) & [(-1)bni_11 + (-1)Bound*bni_11] + [(-1)bni_11]y[0] + [bni_11]x[0] >= 0 & [(-1)bso_12] >= 0) 3.89/1.83 3.89/1.83 3.89/1.83 3.89/1.83 We simplified constraint (4) using rule (POLY_REMOVE_MIN_MAX) which results in the following new constraint: 3.89/1.83 3.89/1.83 (5) (x[0] + [-1] + [-1]y[0] >= 0 ==> (U^Increasing(COND_F(>(x[0], y[0]), x[0], y[0])), >=) & [(-1)bni_11 + (-1)Bound*bni_11] + [(-1)bni_11]y[0] + [bni_11]x[0] >= 0 & [(-1)bso_12] >= 0) 3.89/1.83 3.89/1.83 3.89/1.83 3.89/1.83 We simplified constraint (5) using rule (IDP_SMT_SPLIT) which results in the following new constraint: 3.89/1.83 3.89/1.83 (6) (x[0] >= 0 ==> (U^Increasing(COND_F(>(x[0], y[0]), x[0], y[0])), >=) & [(-1)Bound*bni_11] + [bni_11]x[0] >= 0 & [(-1)bso_12] >= 0) 3.89/1.83 3.89/1.83 3.89/1.83 3.89/1.83 We simplified constraint (6) using rule (IDP_SMT_SPLIT) which results in the following new constraints: 3.89/1.83 3.89/1.83 (7) (x[0] >= 0 & y[0] >= 0 ==> (U^Increasing(COND_F(>(x[0], y[0]), x[0], y[0])), >=) & [(-1)Bound*bni_11] + [bni_11]x[0] >= 0 & [(-1)bso_12] >= 0) 3.89/1.83 3.89/1.83 (8) (x[0] >= 0 & y[0] >= 0 ==> (U^Increasing(COND_F(>(x[0], y[0]), x[0], y[0])), >=) & [(-1)Bound*bni_11] + [bni_11]x[0] >= 0 & [(-1)bso_12] >= 0) 3.89/1.83 3.89/1.83 3.89/1.83 3.89/1.83 3.89/1.83 3.89/1.83 3.89/1.83 3.89/1.83 3.89/1.83 For Pair COND_F(TRUE, x, y) -> F(x, +(y, 1)) the following chains were created: 3.89/1.83 *We consider the chain F(x[0], y[0]) -> COND_F(>(x[0], y[0]), x[0], y[0]), COND_F(TRUE, x[1], y[1]) -> F(x[1], +(y[1], 1)), F(x[0], y[0]) -> COND_F(>(x[0], y[0]), x[0], y[0]) which results in the following constraint: 3.89/1.83 3.89/1.83 (1) (>(x[0], y[0])=TRUE & x[0]=x[1] & y[0]=y[1] & x[1]=x[0]1 & +(y[1], 1)=y[0]1 ==> COND_F(TRUE, x[1], y[1])_>=_NonInfC & COND_F(TRUE, x[1], y[1])_>=_F(x[1], +(y[1], 1)) & (U^Increasing(F(x[1], +(y[1], 1))), >=)) 3.89/1.83 3.89/1.83 3.89/1.83 3.89/1.83 We simplified constraint (1) using rules (III), (IV) which results in the following new constraint: 3.89/1.83 3.89/1.83 (2) (>(x[0], y[0])=TRUE ==> COND_F(TRUE, x[0], y[0])_>=_NonInfC & COND_F(TRUE, x[0], y[0])_>=_F(x[0], +(y[0], 1)) & (U^Increasing(F(x[1], +(y[1], 1))), >=)) 3.89/1.83 3.89/1.83 3.89/1.83 3.89/1.83 We simplified constraint (2) using rule (POLY_CONSTRAINTS) which results in the following new constraint: 3.89/1.83 3.89/1.83 (3) (x[0] + [-1] + [-1]y[0] >= 0 ==> (U^Increasing(F(x[1], +(y[1], 1))), >=) & [(-1)bni_13 + (-1)Bound*bni_13] + [(-1)bni_13]y[0] + [bni_13]x[0] >= 0 & [1 + (-1)bso_14] >= 0) 3.89/1.83 3.89/1.83 3.89/1.83 3.89/1.83 We simplified constraint (3) using rule (IDP_POLY_SIMPLIFY) which results in the following new constraint: 3.89/1.83 3.89/1.83 (4) (x[0] + [-1] + [-1]y[0] >= 0 ==> (U^Increasing(F(x[1], +(y[1], 1))), >=) & [(-1)bni_13 + (-1)Bound*bni_13] + [(-1)bni_13]y[0] + [bni_13]x[0] >= 0 & [1 + (-1)bso_14] >= 0) 3.89/1.83 3.89/1.83 3.89/1.83 3.89/1.83 We simplified constraint (4) using rule (POLY_REMOVE_MIN_MAX) which results in the following new constraint: 3.89/1.83 3.89/1.83 (5) (x[0] + [-1] + [-1]y[0] >= 0 ==> (U^Increasing(F(x[1], +(y[1], 1))), >=) & [(-1)bni_13 + (-1)Bound*bni_13] + [(-1)bni_13]y[0] + [bni_13]x[0] >= 0 & [1 + (-1)bso_14] >= 0) 3.89/1.83 3.89/1.83 3.89/1.83 3.89/1.83 We simplified constraint (5) using rule (IDP_SMT_SPLIT) which results in the following new constraint: 3.89/1.83 3.89/1.83 (6) (x[0] >= 0 ==> (U^Increasing(F(x[1], +(y[1], 1))), >=) & [(-1)Bound*bni_13] + [bni_13]x[0] >= 0 & [1 + (-1)bso_14] >= 0) 3.89/1.83 3.89/1.83 3.89/1.83 3.89/1.83 We simplified constraint (6) using rule (IDP_SMT_SPLIT) which results in the following new constraints: 3.89/1.83 3.89/1.83 (7) (x[0] >= 0 & y[0] >= 0 ==> (U^Increasing(F(x[1], +(y[1], 1))), >=) & [(-1)Bound*bni_13] + [bni_13]x[0] >= 0 & [1 + (-1)bso_14] >= 0) 3.89/1.83 3.89/1.83 (8) (x[0] >= 0 & y[0] >= 0 ==> (U^Increasing(F(x[1], +(y[1], 1))), >=) & [(-1)Bound*bni_13] + [bni_13]x[0] >= 0 & [1 + (-1)bso_14] >= 0) 3.89/1.83 3.89/1.83 3.89/1.83 3.89/1.83 3.89/1.83 3.89/1.83 3.89/1.83 3.89/1.83 3.89/1.83 To summarize, we get the following constraints P__>=_ for the following pairs. 3.89/1.83 3.89/1.83 *F(x, y) -> COND_F(>(x, y), x, y) 3.89/1.83 3.89/1.83 *(x[0] >= 0 & y[0] >= 0 ==> (U^Increasing(COND_F(>(x[0], y[0]), x[0], y[0])), >=) & [(-1)Bound*bni_11] + [bni_11]x[0] >= 0 & [(-1)bso_12] >= 0) 3.89/1.83 3.89/1.83 3.89/1.83 *(x[0] >= 0 & y[0] >= 0 ==> (U^Increasing(COND_F(>(x[0], y[0]), x[0], y[0])), >=) & [(-1)Bound*bni_11] + [bni_11]x[0] >= 0 & [(-1)bso_12] >= 0) 3.89/1.83 3.89/1.83 3.89/1.83 3.89/1.83 3.89/1.83 *COND_F(TRUE, x, y) -> F(x, +(y, 1)) 3.89/1.83 3.89/1.83 *(x[0] >= 0 & y[0] >= 0 ==> (U^Increasing(F(x[1], +(y[1], 1))), >=) & [(-1)Bound*bni_13] + [bni_13]x[0] >= 0 & [1 + (-1)bso_14] >= 0) 3.89/1.83 3.89/1.83 3.89/1.83 *(x[0] >= 0 & y[0] >= 0 ==> (U^Increasing(F(x[1], +(y[1], 1))), >=) & [(-1)Bound*bni_13] + [bni_13]x[0] >= 0 & [1 + (-1)bso_14] >= 0) 3.89/1.83 3.89/1.83 3.89/1.83 3.89/1.83 3.89/1.83 3.89/1.83 3.89/1.83 3.89/1.83 3.89/1.83 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. 3.89/1.83 3.89/1.83 Using the following integer polynomial ordering the resulting constraints can be solved 3.89/1.83 3.89/1.83 Polynomial interpretation over integers[POLO]: 3.89/1.83 3.89/1.83 POL(TRUE) = 0 3.89/1.83 POL(FALSE) = 0 3.89/1.83 POL(F(x_1, x_2)) = [-1] + [-1]x_2 + x_1 3.89/1.83 POL(COND_F(x_1, x_2, x_3)) = [-1] + [-1]x_3 + x_2 3.89/1.83 POL(>(x_1, x_2)) = [-1] 3.89/1.83 POL(+(x_1, x_2)) = x_1 + x_2 3.89/1.83 POL(1) = [1] 3.89/1.83 3.89/1.83 3.89/1.83 The following pairs are in P_>: 3.89/1.83 3.89/1.83 3.89/1.83 COND_F(TRUE, x[1], y[1]) -> F(x[1], +(y[1], 1)) 3.89/1.83 3.89/1.83 3.89/1.83 The following pairs are in P_bound: 3.89/1.83 3.89/1.83 3.89/1.83 F(x[0], y[0]) -> COND_F(>(x[0], y[0]), x[0], y[0]) 3.89/1.84 COND_F(TRUE, x[1], y[1]) -> F(x[1], +(y[1], 1)) 3.89/1.84 3.89/1.84 3.89/1.84 The following pairs are in P_>=: 3.89/1.84 3.89/1.84 3.89/1.84 F(x[0], y[0]) -> COND_F(>(x[0], y[0]), x[0], y[0]) 3.89/1.84 3.89/1.84 3.89/1.84 There are no usable rules. 3.89/1.84 ---------------------------------------- 3.89/1.84 3.89/1.84 (6) 3.89/1.84 Obligation: 3.89/1.84 IDP problem: 3.89/1.84 The following function symbols are pre-defined: 3.89/1.84 <<< 3.89/1.84 & ~ Bwand: (Integer, Integer) -> Integer 3.89/1.84 >= ~ Ge: (Integer, Integer) -> Boolean 3.89/1.84 | ~ Bwor: (Integer, Integer) -> Integer 3.89/1.84 / ~ Div: (Integer, Integer) -> Integer 3.89/1.84 != ~ Neq: (Integer, Integer) -> Boolean 3.89/1.84 && ~ Land: (Boolean, Boolean) -> Boolean 3.89/1.84 ! ~ Lnot: (Boolean) -> Boolean 3.89/1.84 = ~ Eq: (Integer, Integer) -> Boolean 3.89/1.84 <= ~ Le: (Integer, Integer) -> Boolean 3.89/1.84 ^ ~ Bwxor: (Integer, Integer) -> Integer 3.89/1.84 % ~ Mod: (Integer, Integer) -> Integer 3.89/1.84 + ~ Add: (Integer, Integer) -> Integer 3.89/1.84 > ~ Gt: (Integer, Integer) -> Boolean 3.89/1.84 -1 ~ UnaryMinus: (Integer) -> Integer 3.89/1.84 < ~ Lt: (Integer, Integer) -> Boolean 3.89/1.84 || ~ Lor: (Boolean, Boolean) -> Boolean 3.89/1.84 - ~ Sub: (Integer, Integer) -> Integer 3.89/1.84 ~ ~ Bwnot: (Integer) -> Integer 3.89/1.84 * ~ Mul: (Integer, Integer) -> Integer 3.89/1.84 >>> 3.89/1.84 3.89/1.84 3.89/1.84 The following domains are used: 3.89/1.84 Integer 3.89/1.84 3.89/1.84 R is empty. 3.89/1.84 3.89/1.84 The integer pair graph contains the following rules and edges: 3.89/1.84 (0): F(x[0], y[0]) -> COND_F(x[0] > y[0], x[0], y[0]) 3.89/1.84 3.89/1.84 3.89/1.84 The set Q consists of the following terms: 3.89/1.84 f(x0, x1) 3.89/1.84 Cond_f(TRUE, x0, x1) 3.89/1.84 3.89/1.84 ---------------------------------------- 3.89/1.84 3.89/1.84 (7) IDependencyGraphProof (EQUIVALENT) 3.89/1.84 The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 1 less node. 3.89/1.84 ---------------------------------------- 3.89/1.84 3.89/1.84 (8) 3.89/1.84 TRUE 3.89/1.86 EOF