4.15/1.98 YES 4.15/1.99 proof of /export/starexec/sandbox2/benchmark/theBenchmark.itrs 4.15/1.99 # AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty 4.15/1.99 4.15/1.99 4.15/1.99 Termination of the given ITRS could be proven: 4.15/1.99 4.15/1.99 (0) ITRS 4.15/1.99 (1) ITRStoIDPProof [EQUIVALENT, 0 ms] 4.15/1.99 (2) IDP 4.15/1.99 (3) UsableRulesProof [EQUIVALENT, 0 ms] 4.15/1.99 (4) IDP 4.15/1.99 (5) IDPNonInfProof [SOUND, 140 ms] 4.15/1.99 (6) IDP 4.15/1.99 (7) IDependencyGraphProof [EQUIVALENT, 0 ms] 4.15/1.99 (8) TRUE 4.15/1.99 4.15/1.99 4.15/1.99 ---------------------------------------- 4.15/1.99 4.15/1.99 (0) 4.15/1.99 Obligation: 4.15/1.99 ITRS problem: 4.15/1.99 4.15/1.99 The following function symbols are pre-defined: 4.15/1.99 <<< 4.15/1.99 & ~ Bwand: (Integer, Integer) -> Integer 4.15/1.99 >= ~ Ge: (Integer, Integer) -> Boolean 4.15/1.99 | ~ Bwor: (Integer, Integer) -> Integer 4.15/1.99 / ~ Div: (Integer, Integer) -> Integer 4.15/1.99 != ~ Neq: (Integer, Integer) -> Boolean 4.15/1.99 && ~ Land: (Boolean, Boolean) -> Boolean 4.15/1.99 ! ~ Lnot: (Boolean) -> Boolean 4.15/1.99 = ~ Eq: (Integer, Integer) -> Boolean 4.15/1.99 <= ~ Le: (Integer, Integer) -> Boolean 4.15/1.99 ^ ~ Bwxor: (Integer, Integer) -> Integer 4.15/1.99 % ~ Mod: (Integer, Integer) -> Integer 4.15/1.99 + ~ Add: (Integer, Integer) -> Integer 4.15/1.99 > ~ Gt: (Integer, Integer) -> Boolean 4.15/1.99 -1 ~ UnaryMinus: (Integer) -> Integer 4.15/1.99 < ~ Lt: (Integer, Integer) -> Boolean 4.15/1.99 || ~ Lor: (Boolean, Boolean) -> Boolean 4.15/1.99 - ~ Sub: (Integer, Integer) -> Integer 4.15/1.99 ~ ~ Bwnot: (Integer) -> Integer 4.15/1.99 * ~ Mul: (Integer, Integer) -> Integer 4.15/1.99 >>> 4.15/1.99 4.15/1.99 The TRS R consists of the following rules: 4.15/1.99 f(x, y) -> Cond_f(x > y, x, y) 4.15/1.99 Cond_f(TRUE, x, y) -> f(x + 1, y + 2) 4.15/1.99 The set Q consists of the following terms: 4.15/1.99 f(x0, x1) 4.15/1.99 Cond_f(TRUE, x0, x1) 4.15/1.99 4.15/1.99 ---------------------------------------- 4.15/1.99 4.15/1.99 (1) ITRStoIDPProof (EQUIVALENT) 4.15/1.99 Added dependency pairs 4.15/1.99 ---------------------------------------- 4.15/1.99 4.15/1.99 (2) 4.15/1.99 Obligation: 4.15/1.99 IDP problem: 4.15/1.99 The following function symbols are pre-defined: 4.15/1.99 <<< 4.15/1.99 & ~ Bwand: (Integer, Integer) -> Integer 4.15/1.99 >= ~ Ge: (Integer, Integer) -> Boolean 4.15/1.99 | ~ Bwor: (Integer, Integer) -> Integer 4.15/1.99 / ~ Div: (Integer, Integer) -> Integer 4.15/1.99 != ~ Neq: (Integer, Integer) -> Boolean 4.15/1.99 && ~ Land: (Boolean, Boolean) -> Boolean 4.15/1.99 ! ~ Lnot: (Boolean) -> Boolean 4.15/1.99 = ~ Eq: (Integer, Integer) -> Boolean 4.15/1.99 <= ~ Le: (Integer, Integer) -> Boolean 4.15/1.99 ^ ~ Bwxor: (Integer, Integer) -> Integer 4.15/1.99 % ~ Mod: (Integer, Integer) -> Integer 4.15/1.99 + ~ Add: (Integer, Integer) -> Integer 4.15/1.99 > ~ Gt: (Integer, Integer) -> Boolean 4.15/1.99 -1 ~ UnaryMinus: (Integer) -> Integer 4.15/1.99 < ~ Lt: (Integer, Integer) -> Boolean 4.15/1.99 || ~ Lor: (Boolean, Boolean) -> Boolean 4.15/1.99 - ~ Sub: (Integer, Integer) -> Integer 4.15/1.99 ~ ~ Bwnot: (Integer) -> Integer 4.15/1.99 * ~ Mul: (Integer, Integer) -> Integer 4.15/1.99 >>> 4.15/1.99 4.15/1.99 4.15/1.99 The following domains are used: 4.15/1.99 Integer 4.15/1.99 4.15/1.99 The ITRS R consists of the following rules: 4.15/1.99 f(x, y) -> Cond_f(x > y, x, y) 4.15/1.99 Cond_f(TRUE, x, y) -> f(x + 1, y + 2) 4.15/1.99 4.15/1.99 The integer pair graph contains the following rules and edges: 4.15/1.99 (0): F(x[0], y[0]) -> COND_F(x[0] > y[0], x[0], y[0]) 4.15/1.99 (1): COND_F(TRUE, x[1], y[1]) -> F(x[1] + 1, y[1] + 2) 4.15/1.99 4.15/1.99 (0) -> (1), if (x[0] > y[0] & x[0] ->^* x[1] & y[0] ->^* y[1]) 4.15/1.99 (1) -> (0), if (x[1] + 1 ->^* x[0] & y[1] + 2 ->^* y[0]) 4.15/1.99 4.15/1.99 The set Q consists of the following terms: 4.15/1.99 f(x0, x1) 4.15/1.99 Cond_f(TRUE, x0, x1) 4.15/1.99 4.15/1.99 ---------------------------------------- 4.15/1.99 4.15/1.99 (3) UsableRulesProof (EQUIVALENT) 4.15/1.99 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. 4.15/1.99 ---------------------------------------- 4.15/1.99 4.15/1.99 (4) 4.15/1.99 Obligation: 4.15/1.99 IDP problem: 4.15/1.99 The following function symbols are pre-defined: 4.15/1.99 <<< 4.15/1.99 & ~ Bwand: (Integer, Integer) -> Integer 4.15/1.99 >= ~ Ge: (Integer, Integer) -> Boolean 4.15/1.99 | ~ Bwor: (Integer, Integer) -> Integer 4.15/1.99 / ~ Div: (Integer, Integer) -> Integer 4.15/1.99 != ~ Neq: (Integer, Integer) -> Boolean 4.15/1.99 && ~ Land: (Boolean, Boolean) -> Boolean 4.15/1.99 ! ~ Lnot: (Boolean) -> Boolean 4.15/1.99 = ~ Eq: (Integer, Integer) -> Boolean 4.15/1.99 <= ~ Le: (Integer, Integer) -> Boolean 4.15/1.99 ^ ~ Bwxor: (Integer, Integer) -> Integer 4.15/1.99 % ~ Mod: (Integer, Integer) -> Integer 4.15/1.99 + ~ Add: (Integer, Integer) -> Integer 4.15/1.99 > ~ Gt: (Integer, Integer) -> Boolean 4.15/1.99 -1 ~ UnaryMinus: (Integer) -> Integer 4.15/1.99 < ~ Lt: (Integer, Integer) -> Boolean 4.15/1.99 || ~ Lor: (Boolean, Boolean) -> Boolean 4.15/1.99 - ~ Sub: (Integer, Integer) -> Integer 4.15/1.99 ~ ~ Bwnot: (Integer) -> Integer 4.15/1.99 * ~ Mul: (Integer, Integer) -> Integer 4.15/1.99 >>> 4.15/1.99 4.15/1.99 4.15/1.99 The following domains are used: 4.15/1.99 Integer 4.15/1.99 4.15/1.99 R is empty. 4.15/1.99 4.15/1.99 The integer pair graph contains the following rules and edges: 4.15/1.99 (0): F(x[0], y[0]) -> COND_F(x[0] > y[0], x[0], y[0]) 4.15/1.99 (1): COND_F(TRUE, x[1], y[1]) -> F(x[1] + 1, y[1] + 2) 4.15/1.99 4.15/1.99 (0) -> (1), if (x[0] > y[0] & x[0] ->^* x[1] & y[0] ->^* y[1]) 4.15/1.99 (1) -> (0), if (x[1] + 1 ->^* x[0] & y[1] + 2 ->^* y[0]) 4.15/1.99 4.15/1.99 The set Q consists of the following terms: 4.15/1.99 f(x0, x1) 4.15/1.99 Cond_f(TRUE, x0, x1) 4.15/1.99 4.15/1.99 ---------------------------------------- 4.15/1.99 4.15/1.99 (5) IDPNonInfProof (SOUND) 4.15/1.99 Used the following options for this NonInfProof: 4.15/1.99 4.15/1.99 IDPGPoloSolver: 4.15/1.99 Range: [(-1,2)] 4.15/1.99 IsNat: false 4.15/1.99 Interpretation Shape Heuristic: aprove.DPFramework.IDPProblem.Processors.nonInf.poly.IdpDefaultShapeHeuristic@3c402468 4.15/1.99 Constraint Generator: NonInfConstraintGenerator: 4.15/1.99 PathGenerator: MetricPathGenerator: 4.15/1.99 Max Left Steps: 1 4.15/1.99 Max Right Steps: 1 4.15/1.99 4.15/1.99 4.15/1.99 4.15/1.99 The constraints were generated the following way: 4.15/1.99 4.15/1.99 The DP Problem is simplified using the Induction Calculus [NONINF] with the following steps: 4.15/1.99 4.15/1.99 Note that final constraints are written in bold face. 4.15/1.99 4.15/1.99 4.15/1.99 4.15/1.99 For Pair F(x, y) -> COND_F(>(x, y), x, y) the following chains were created: 4.15/1.99 *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], 1), +(y[1], 2)) which results in the following constraint: 4.15/1.99 4.15/1.99 (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])), >=)) 4.15/1.99 4.15/1.99 4.15/1.99 4.15/1.99 We simplified constraint (1) using rule (IV) which results in the following new constraint: 4.15/1.99 4.15/1.99 (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])), >=)) 4.15/1.99 4.15/1.99 4.15/1.99 4.15/1.99 We simplified constraint (2) using rule (POLY_CONSTRAINTS) which results in the following new constraint: 4.15/1.99 4.15/1.99 (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) 4.15/1.99 4.15/1.99 4.15/1.99 4.15/1.99 We simplified constraint (3) using rule (IDP_POLY_SIMPLIFY) which results in the following new constraint: 4.15/1.99 4.15/1.99 (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) 4.15/1.99 4.15/1.99 4.15/1.99 4.15/1.99 We simplified constraint (4) using rule (POLY_REMOVE_MIN_MAX) which results in the following new constraint: 4.15/1.99 4.15/1.99 (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) 4.15/1.99 4.15/1.99 4.15/1.99 4.15/1.99 We simplified constraint (5) using rule (IDP_SMT_SPLIT) which results in the following new constraint: 4.15/1.99 4.15/1.99 (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) 4.15/1.99 4.15/1.99 4.15/1.99 4.15/1.99 We simplified constraint (6) using rule (IDP_SMT_SPLIT) which results in the following new constraints: 4.15/1.99 4.15/1.99 (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) 4.15/1.99 4.15/1.99 (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) 4.15/1.99 4.15/1.99 4.15/1.99 4.15/1.99 4.15/1.99 4.15/1.99 4.15/1.99 4.15/1.99 4.15/1.99 For Pair COND_F(TRUE, x, y) -> F(+(x, 1), +(y, 2)) the following chains were created: 4.15/1.99 *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], 1), +(y[1], 2)), F(x[0], y[0]) -> COND_F(>(x[0], y[0]), x[0], y[0]) which results in the following constraint: 4.15/1.99 4.15/1.99 (1) (>(x[0], y[0])=TRUE & x[0]=x[1] & y[0]=y[1] & +(x[1], 1)=x[0]1 & +(y[1], 2)=y[0]1 ==> COND_F(TRUE, x[1], y[1])_>=_NonInfC & COND_F(TRUE, x[1], y[1])_>=_F(+(x[1], 1), +(y[1], 2)) & (U^Increasing(F(+(x[1], 1), +(y[1], 2))), >=)) 4.15/1.99 4.15/1.99 4.15/1.99 4.15/1.99 We simplified constraint (1) using rules (III), (IV) which results in the following new constraint: 4.15/1.99 4.15/1.99 (2) (>(x[0], y[0])=TRUE ==> COND_F(TRUE, x[0], y[0])_>=_NonInfC & COND_F(TRUE, x[0], y[0])_>=_F(+(x[0], 1), +(y[0], 2)) & (U^Increasing(F(+(x[1], 1), +(y[1], 2))), >=)) 4.15/1.99 4.15/1.99 4.15/1.99 4.15/1.99 We simplified constraint (2) using rule (POLY_CONSTRAINTS) which results in the following new constraint: 4.15/1.99 4.15/1.99 (3) (x[0] + [-1] + [-1]y[0] >= 0 ==> (U^Increasing(F(+(x[1], 1), +(y[1], 2))), >=) & [(-1)bni_13 + (-1)Bound*bni_13] + [(-1)bni_13]y[0] + [bni_13]x[0] >= 0 & [1 + (-1)bso_14] >= 0) 4.15/1.99 4.15/1.99 4.15/1.99 4.15/1.99 We simplified constraint (3) using rule (IDP_POLY_SIMPLIFY) which results in the following new constraint: 4.15/1.99 4.15/1.99 (4) (x[0] + [-1] + [-1]y[0] >= 0 ==> (U^Increasing(F(+(x[1], 1), +(y[1], 2))), >=) & [(-1)bni_13 + (-1)Bound*bni_13] + [(-1)bni_13]y[0] + [bni_13]x[0] >= 0 & [1 + (-1)bso_14] >= 0) 4.15/1.99 4.15/1.99 4.15/1.99 4.15/1.99 We simplified constraint (4) using rule (POLY_REMOVE_MIN_MAX) which results in the following new constraint: 4.15/1.99 4.15/1.99 (5) (x[0] + [-1] + [-1]y[0] >= 0 ==> (U^Increasing(F(+(x[1], 1), +(y[1], 2))), >=) & [(-1)bni_13 + (-1)Bound*bni_13] + [(-1)bni_13]y[0] + [bni_13]x[0] >= 0 & [1 + (-1)bso_14] >= 0) 4.15/1.99 4.15/1.99 4.15/1.99 4.15/1.99 We simplified constraint (5) using rule (IDP_SMT_SPLIT) which results in the following new constraint: 4.15/1.99 4.15/1.99 (6) (x[0] >= 0 ==> (U^Increasing(F(+(x[1], 1), +(y[1], 2))), >=) & [(-1)Bound*bni_13] + [bni_13]x[0] >= 0 & [1 + (-1)bso_14] >= 0) 4.15/1.99 4.15/1.99 4.15/1.99 4.15/1.99 We simplified constraint (6) using rule (IDP_SMT_SPLIT) which results in the following new constraints: 4.15/1.99 4.15/1.99 (7) (x[0] >= 0 & y[0] >= 0 ==> (U^Increasing(F(+(x[1], 1), +(y[1], 2))), >=) & [(-1)Bound*bni_13] + [bni_13]x[0] >= 0 & [1 + (-1)bso_14] >= 0) 4.15/1.99 4.15/1.99 (8) (x[0] >= 0 & y[0] >= 0 ==> (U^Increasing(F(+(x[1], 1), +(y[1], 2))), >=) & [(-1)Bound*bni_13] + [bni_13]x[0] >= 0 & [1 + (-1)bso_14] >= 0) 4.15/1.99 4.15/1.99 4.15/1.99 4.15/1.99 4.15/1.99 4.15/1.99 4.15/1.99 4.15/1.99 4.15/1.99 To summarize, we get the following constraints P__>=_ for the following pairs. 4.15/1.99 4.15/1.99 *F(x, y) -> COND_F(>(x, y), x, y) 4.15/1.99 4.15/1.99 *(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) 4.15/1.99 4.15/1.99 4.15/1.99 *(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) 4.15/1.99 4.15/1.99 4.15/1.99 4.15/1.99 4.15/1.99 *COND_F(TRUE, x, y) -> F(+(x, 1), +(y, 2)) 4.15/1.99 4.15/1.99 *(x[0] >= 0 & y[0] >= 0 ==> (U^Increasing(F(+(x[1], 1), +(y[1], 2))), >=) & [(-1)Bound*bni_13] + [bni_13]x[0] >= 0 & [1 + (-1)bso_14] >= 0) 4.15/1.99 4.15/1.99 4.15/1.99 *(x[0] >= 0 & y[0] >= 0 ==> (U^Increasing(F(+(x[1], 1), +(y[1], 2))), >=) & [(-1)Bound*bni_13] + [bni_13]x[0] >= 0 & [1 + (-1)bso_14] >= 0) 4.15/1.99 4.15/1.99 4.15/1.99 4.15/1.99 4.15/1.99 4.15/1.99 4.15/1.99 4.15/1.99 4.15/1.99 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. 4.15/1.99 4.15/1.99 Using the following integer polynomial ordering the resulting constraints can be solved 4.15/1.99 4.15/1.99 Polynomial interpretation over integers[POLO]: 4.15/1.99 4.15/1.99 POL(TRUE) = 0 4.15/1.99 POL(FALSE) = 0 4.15/1.99 POL(F(x_1, x_2)) = [-1] + [-1]x_2 + x_1 4.15/1.99 POL(COND_F(x_1, x_2, x_3)) = [-1] + [-1]x_3 + x_2 4.15/1.99 POL(>(x_1, x_2)) = [-1] 4.15/1.99 POL(+(x_1, x_2)) = x_1 + x_2 4.15/1.99 POL(1) = [1] 4.15/1.99 POL(2) = [2] 4.15/1.99 4.15/1.99 4.15/1.99 The following pairs are in P_>: 4.15/1.99 4.15/1.99 4.15/1.99 COND_F(TRUE, x[1], y[1]) -> F(+(x[1], 1), +(y[1], 2)) 4.15/1.99 4.15/1.99 4.15/1.99 The following pairs are in P_bound: 4.15/1.99 4.15/1.99 4.15/1.99 F(x[0], y[0]) -> COND_F(>(x[0], y[0]), x[0], y[0]) 4.15/1.99 COND_F(TRUE, x[1], y[1]) -> F(+(x[1], 1), +(y[1], 2)) 4.15/1.99 4.15/1.99 4.15/1.99 The following pairs are in P_>=: 4.15/1.99 4.15/1.99 4.15/1.99 F(x[0], y[0]) -> COND_F(>(x[0], y[0]), x[0], y[0]) 4.15/1.99 4.15/1.99 4.15/1.99 There are no usable rules. 4.15/1.99 ---------------------------------------- 4.15/1.99 4.15/1.99 (6) 4.15/1.99 Obligation: 4.15/1.99 IDP problem: 4.15/1.99 The following function symbols are pre-defined: 4.15/1.99 <<< 4.15/1.99 & ~ Bwand: (Integer, Integer) -> Integer 4.15/1.99 >= ~ Ge: (Integer, Integer) -> Boolean 4.15/1.99 | ~ Bwor: (Integer, Integer) -> Integer 4.15/1.99 / ~ Div: (Integer, Integer) -> Integer 4.15/1.99 != ~ Neq: (Integer, Integer) -> Boolean 4.15/1.99 && ~ Land: (Boolean, Boolean) -> Boolean 4.15/1.99 ! ~ Lnot: (Boolean) -> Boolean 4.15/1.99 = ~ Eq: (Integer, Integer) -> Boolean 4.15/1.99 <= ~ Le: (Integer, Integer) -> Boolean 4.15/1.99 ^ ~ Bwxor: (Integer, Integer) -> Integer 4.15/1.99 % ~ Mod: (Integer, Integer) -> Integer 4.15/1.99 + ~ Add: (Integer, Integer) -> Integer 4.15/1.99 > ~ Gt: (Integer, Integer) -> Boolean 4.15/1.99 -1 ~ UnaryMinus: (Integer) -> Integer 4.15/1.99 < ~ Lt: (Integer, Integer) -> Boolean 4.15/1.99 || ~ Lor: (Boolean, Boolean) -> Boolean 4.15/1.99 - ~ Sub: (Integer, Integer) -> Integer 4.15/1.99 ~ ~ Bwnot: (Integer) -> Integer 4.15/1.99 * ~ Mul: (Integer, Integer) -> Integer 4.15/1.99 >>> 4.15/1.99 4.15/1.99 4.15/1.99 The following domains are used: 4.15/1.99 Integer 4.15/1.99 4.15/1.99 R is empty. 4.15/1.99 4.15/1.99 The integer pair graph contains the following rules and edges: 4.15/1.99 (0): F(x[0], y[0]) -> COND_F(x[0] > y[0], x[0], y[0]) 4.15/1.99 4.15/1.99 4.15/1.99 The set Q consists of the following terms: 4.15/1.99 f(x0, x1) 4.15/1.99 Cond_f(TRUE, x0, x1) 4.15/1.99 4.15/1.99 ---------------------------------------- 4.15/1.99 4.15/1.99 (7) IDependencyGraphProof (EQUIVALENT) 4.15/1.99 The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 1 less node. 4.15/1.99 ---------------------------------------- 4.15/1.99 4.15/1.99 (8) 4.15/1.99 TRUE 4.15/2.01 EOF