12.64/12.58 MAYBE 12.64/12.58 12.64/12.58 DP problem for innermost termination. 12.64/12.58 P = 12.64/12.58 f7#(x1, x2, x3, x4, x5) -> f6#(x1, x2, x3, x4, x5) 12.64/12.58 f6#(I0, I1, I2, I3, I4) -> f2#(I0, I1, I2, rnd4, I4) [1 <= rnd4 /\ rnd4 = rnd4] 12.64/12.58 f5#(I5, I6, I7, I8, I9) -> f1#(I5, I6, I7, I8, rnd5) [rnd5 = rnd5] 12.64/12.58 f4#(I10, I11, I12, I13, I14) -> f5#(I10, I11, I12, I13, I14) [I12 = I12] 12.64/12.58 f3#(I15, I16, I17, I18, I19) -> f4#(I15, I16, I17, I18, I19) [I16 = I16] 12.64/12.58 f2#(I20, I21, I22, I23, I24) -> f3#(I20, I21, I22, I23, I24) [I20 = I20] 12.64/12.58 f1#(I25, I26, I27, I28, I29) -> f2#(I25, I26, I27, I29, I29) [I28 <= 2 * I29 /\ 2 * I29 <= I28] 12.64/12.58 f1#(I30, I31, I32, I33, I34) -> f2#(I30, I31, I32, 1 + 3 * I33, I34) [I33 <= 1 + 2 * I34 /\ 1 + 2 * I34 <= I33] 12.64/12.58 R = 12.64/12.58 f7(x1, x2, x3, x4, x5) -> f6(x1, x2, x3, x4, x5) 12.64/12.58 f6(I0, I1, I2, I3, I4) -> f2(I0, I1, I2, rnd4, I4) [1 <= rnd4 /\ rnd4 = rnd4] 12.64/12.58 f5(I5, I6, I7, I8, I9) -> f1(I5, I6, I7, I8, rnd5) [rnd5 = rnd5] 12.64/12.58 f4(I10, I11, I12, I13, I14) -> f5(I10, I11, I12, I13, I14) [I12 = I12] 12.64/12.58 f3(I15, I16, I17, I18, I19) -> f4(I15, I16, I17, I18, I19) [I16 = I16] 12.64/12.58 f2(I20, I21, I22, I23, I24) -> f3(I20, I21, I22, I23, I24) [I20 = I20] 12.64/12.58 f1(I25, I26, I27, I28, I29) -> f2(I25, I26, I27, I29, I29) [I28 <= 2 * I29 /\ 2 * I29 <= I28] 12.64/12.58 f1(I30, I31, I32, I33, I34) -> f2(I30, I31, I32, 1 + 3 * I33, I34) [I33 <= 1 + 2 * I34 /\ 1 + 2 * I34 <= I33] 12.64/12.58 12.64/12.58 The dependency graph for this problem is: 12.64/12.58 0 -> 1 12.64/12.58 1 -> 5 12.64/12.58 2 -> 6, 7 12.64/12.58 3 -> 2 12.64/12.58 4 -> 3 12.64/12.58 5 -> 4 12.64/12.58 6 -> 5 12.64/12.58 7 -> 5 12.64/12.58 Where: 12.64/12.58 0) f7#(x1, x2, x3, x4, x5) -> f6#(x1, x2, x3, x4, x5) 12.64/12.58 1) f6#(I0, I1, I2, I3, I4) -> f2#(I0, I1, I2, rnd4, I4) [1 <= rnd4 /\ rnd4 = rnd4] 12.64/12.58 2) f5#(I5, I6, I7, I8, I9) -> f1#(I5, I6, I7, I8, rnd5) [rnd5 = rnd5] 12.64/12.58 3) f4#(I10, I11, I12, I13, I14) -> f5#(I10, I11, I12, I13, I14) [I12 = I12] 12.64/12.58 4) f3#(I15, I16, I17, I18, I19) -> f4#(I15, I16, I17, I18, I19) [I16 = I16] 12.64/12.58 5) f2#(I20, I21, I22, I23, I24) -> f3#(I20, I21, I22, I23, I24) [I20 = I20] 12.64/12.58 6) f1#(I25, I26, I27, I28, I29) -> f2#(I25, I26, I27, I29, I29) [I28 <= 2 * I29 /\ 2 * I29 <= I28] 12.64/12.58 7) f1#(I30, I31, I32, I33, I34) -> f2#(I30, I31, I32, 1 + 3 * I33, I34) [I33 <= 1 + 2 * I34 /\ 1 + 2 * I34 <= I33] 12.64/12.58 12.64/12.58 We have the following SCCs. 12.64/12.58 { 2, 3, 4, 5, 6, 7 } 12.64/12.58 12.64/12.58 DP problem for innermost termination. 12.64/12.58 P = 12.64/12.58 f5#(I5, I6, I7, I8, I9) -> f1#(I5, I6, I7, I8, rnd5) [rnd5 = rnd5] 12.64/12.58 f4#(I10, I11, I12, I13, I14) -> f5#(I10, I11, I12, I13, I14) [I12 = I12] 12.64/12.58 f3#(I15, I16, I17, I18, I19) -> f4#(I15, I16, I17, I18, I19) [I16 = I16] 12.64/12.58 f2#(I20, I21, I22, I23, I24) -> f3#(I20, I21, I22, I23, I24) [I20 = I20] 12.64/12.58 f1#(I25, I26, I27, I28, I29) -> f2#(I25, I26, I27, I29, I29) [I28 <= 2 * I29 /\ 2 * I29 <= I28] 12.64/12.58 f1#(I30, I31, I32, I33, I34) -> f2#(I30, I31, I32, 1 + 3 * I33, I34) [I33 <= 1 + 2 * I34 /\ 1 + 2 * I34 <= I33] 12.64/12.58 R = 12.64/12.58 f7(x1, x2, x3, x4, x5) -> f6(x1, x2, x3, x4, x5) 12.64/12.58 f6(I0, I1, I2, I3, I4) -> f2(I0, I1, I2, rnd4, I4) [1 <= rnd4 /\ rnd4 = rnd4] 12.64/12.58 f5(I5, I6, I7, I8, I9) -> f1(I5, I6, I7, I8, rnd5) [rnd5 = rnd5] 12.64/12.58 f4(I10, I11, I12, I13, I14) -> f5(I10, I11, I12, I13, I14) [I12 = I12] 12.64/12.58 f3(I15, I16, I17, I18, I19) -> f4(I15, I16, I17, I18, I19) [I16 = I16] 12.64/12.58 f2(I20, I21, I22, I23, I24) -> f3(I20, I21, I22, I23, I24) [I20 = I20] 12.64/12.58 f1(I25, I26, I27, I28, I29) -> f2(I25, I26, I27, I29, I29) [I28 <= 2 * I29 /\ 2 * I29 <= I28] 12.64/12.58 f1(I30, I31, I32, I33, I34) -> f2(I30, I31, I32, 1 + 3 * I33, I34) [I33 <= 1 + 2 * I34 /\ 1 + 2 * I34 <= I33] 12.64/12.58 12.64/15.56 EOF