16.28/16.55 YES 16.28/16.55 16.28/16.55 DP problem for innermost termination. 16.28/16.55 P = 16.28/16.55 f7#(x1, x2, x3, x4, x5, x6, x7) -> f6#(x1, x2, x3, x4, x5, x6, x7) 16.28/16.55 f6#(I0, I1, I2, I3, I4, I5, I6) -> f1#(0, I1, I2, rnd4, rnd5, rnd6, I6) [rnd4 = rnd4 /\ rnd6 = rnd6 /\ rnd5 = rnd5] 16.28/16.55 f2#(I7, I8, I9, I10, I11, I12, I13) -> f1#(1 + I7, I8, I9, I10, I11, I12, I13) [1 + I7 <= I10] 16.28/16.55 f2#(I14, I15, I16, I17, I18, I19, I20) -> f5#(I14, 0, I18, I17, I18, I19, rnd7) [rnd7 = rnd7 /\ I17 <= I14] 16.28/16.55 f5#(I21, I22, I23, I24, I25, I26, I27) -> f3#(I21, I22, I23, I24, I25, I26, I27) 16.28/16.55 f3#(I28, I29, I30, I31, I32, I33, I34) -> f5#(I28, 1 + I29, I30, I31, I32, I33, I34) [1 + I29 <= I30] 16.28/16.55 f1#(I42, I43, I44, I45, I46, I47, I48) -> f2#(I42, I43, I44, I45, I46, I47, I48) 16.28/16.55 R = 16.28/16.55 f7(x1, x2, x3, x4, x5, x6, x7) -> f6(x1, x2, x3, x4, x5, x6, x7) 16.28/16.55 f6(I0, I1, I2, I3, I4, I5, I6) -> f1(0, I1, I2, rnd4, rnd5, rnd6, I6) [rnd4 = rnd4 /\ rnd6 = rnd6 /\ rnd5 = rnd5] 16.28/16.55 f2(I7, I8, I9, I10, I11, I12, I13) -> f1(1 + I7, I8, I9, I10, I11, I12, I13) [1 + I7 <= I10] 16.28/16.55 f2(I14, I15, I16, I17, I18, I19, I20) -> f5(I14, 0, I18, I17, I18, I19, rnd7) [rnd7 = rnd7 /\ I17 <= I14] 16.28/16.55 f5(I21, I22, I23, I24, I25, I26, I27) -> f3(I21, I22, I23, I24, I25, I26, I27) 16.28/16.55 f3(I28, I29, I30, I31, I32, I33, I34) -> f5(I28, 1 + I29, I30, I31, I32, I33, I34) [1 + I29 <= I30] 16.28/16.55 f3(I35, I36, I37, I38, I39, I40, I41) -> f4(I35, I36, I37, I38, I39, I40, I41) [I37 <= I36] 16.28/16.55 f1(I42, I43, I44, I45, I46, I47, I48) -> f2(I42, I43, I44, I45, I46, I47, I48) 16.28/16.55 16.28/16.55 The dependency graph for this problem is: 16.28/16.55 0 -> 1 16.28/16.55 1 -> 6 16.28/16.55 2 -> 6 16.28/16.55 3 -> 4 16.28/16.55 4 -> 5 16.28/16.55 5 -> 4 16.28/16.55 6 -> 2, 3 16.28/16.55 Where: 16.28/16.55 0) f7#(x1, x2, x3, x4, x5, x6, x7) -> f6#(x1, x2, x3, x4, x5, x6, x7) 16.28/16.55 1) f6#(I0, I1, I2, I3, I4, I5, I6) -> f1#(0, I1, I2, rnd4, rnd5, rnd6, I6) [rnd4 = rnd4 /\ rnd6 = rnd6 /\ rnd5 = rnd5] 16.28/16.55 2) f2#(I7, I8, I9, I10, I11, I12, I13) -> f1#(1 + I7, I8, I9, I10, I11, I12, I13) [1 + I7 <= I10] 16.28/16.55 3) f2#(I14, I15, I16, I17, I18, I19, I20) -> f5#(I14, 0, I18, I17, I18, I19, rnd7) [rnd7 = rnd7 /\ I17 <= I14] 16.28/16.55 4) f5#(I21, I22, I23, I24, I25, I26, I27) -> f3#(I21, I22, I23, I24, I25, I26, I27) 16.28/16.55 5) f3#(I28, I29, I30, I31, I32, I33, I34) -> f5#(I28, 1 + I29, I30, I31, I32, I33, I34) [1 + I29 <= I30] 16.28/16.55 6) f1#(I42, I43, I44, I45, I46, I47, I48) -> f2#(I42, I43, I44, I45, I46, I47, I48) 16.28/16.55 16.28/16.55 We have the following SCCs. 16.28/16.55 { 2, 6 } 16.28/16.55 { 4, 5 } 16.28/16.55 16.28/16.55 DP problem for innermost termination. 16.28/16.55 P = 16.28/16.55 f5#(I21, I22, I23, I24, I25, I26, I27) -> f3#(I21, I22, I23, I24, I25, I26, I27) 16.28/16.55 f3#(I28, I29, I30, I31, I32, I33, I34) -> f5#(I28, 1 + I29, I30, I31, I32, I33, I34) [1 + I29 <= I30] 16.28/16.55 R = 16.28/16.55 f7(x1, x2, x3, x4, x5, x6, x7) -> f6(x1, x2, x3, x4, x5, x6, x7) 16.28/16.55 f6(I0, I1, I2, I3, I4, I5, I6) -> f1(0, I1, I2, rnd4, rnd5, rnd6, I6) [rnd4 = rnd4 /\ rnd6 = rnd6 /\ rnd5 = rnd5] 16.28/16.55 f2(I7, I8, I9, I10, I11, I12, I13) -> f1(1 + I7, I8, I9, I10, I11, I12, I13) [1 + I7 <= I10] 16.28/16.55 f2(I14, I15, I16, I17, I18, I19, I20) -> f5(I14, 0, I18, I17, I18, I19, rnd7) [rnd7 = rnd7 /\ I17 <= I14] 16.28/16.55 f5(I21, I22, I23, I24, I25, I26, I27) -> f3(I21, I22, I23, I24, I25, I26, I27) 16.28/16.55 f3(I28, I29, I30, I31, I32, I33, I34) -> f5(I28, 1 + I29, I30, I31, I32, I33, I34) [1 + I29 <= I30] 16.28/16.55 f3(I35, I36, I37, I38, I39, I40, I41) -> f4(I35, I36, I37, I38, I39, I40, I41) [I37 <= I36] 16.28/16.55 f1(I42, I43, I44, I45, I46, I47, I48) -> f2(I42, I43, I44, I45, I46, I47, I48) 16.28/16.55 16.28/16.55 We use the reverse value criterion with the projection function NU: 16.28/16.55 NU[f3#(z1,z2,z3,z4,z5,z6,z7)] = z3 + -1 * (1 + z2) 16.28/16.55 NU[f5#(z1,z2,z3,z4,z5,z6,z7)] = z3 + -1 * (1 + z2) 16.28/16.55 16.28/16.55 This gives the following inequalities: 16.28/16.55 ==> I23 + -1 * (1 + I22) >= I23 + -1 * (1 + I22) 16.28/16.55 1 + I29 <= I30 ==> I30 + -1 * (1 + I29) > I30 + -1 * (1 + (1 + I29)) with I30 + -1 * (1 + I29) >= 0 16.28/16.55 16.28/16.55 We remove all the strictly oriented dependency pairs. 16.28/16.55 16.28/16.55 DP problem for innermost termination. 16.28/16.55 P = 16.28/16.55 f5#(I21, I22, I23, I24, I25, I26, I27) -> f3#(I21, I22, I23, I24, I25, I26, I27) 16.28/16.55 R = 16.28/16.55 f7(x1, x2, x3, x4, x5, x6, x7) -> f6(x1, x2, x3, x4, x5, x6, x7) 16.28/16.55 f6(I0, I1, I2, I3, I4, I5, I6) -> f1(0, I1, I2, rnd4, rnd5, rnd6, I6) [rnd4 = rnd4 /\ rnd6 = rnd6 /\ rnd5 = rnd5] 16.28/16.55 f2(I7, I8, I9, I10, I11, I12, I13) -> f1(1 + I7, I8, I9, I10, I11, I12, I13) [1 + I7 <= I10] 16.28/16.55 f2(I14, I15, I16, I17, I18, I19, I20) -> f5(I14, 0, I18, I17, I18, I19, rnd7) [rnd7 = rnd7 /\ I17 <= I14] 16.28/16.55 f5(I21, I22, I23, I24, I25, I26, I27) -> f3(I21, I22, I23, I24, I25, I26, I27) 16.28/16.55 f3(I28, I29, I30, I31, I32, I33, I34) -> f5(I28, 1 + I29, I30, I31, I32, I33, I34) [1 + I29 <= I30] 16.28/16.55 f3(I35, I36, I37, I38, I39, I40, I41) -> f4(I35, I36, I37, I38, I39, I40, I41) [I37 <= I36] 16.28/16.55 f1(I42, I43, I44, I45, I46, I47, I48) -> f2(I42, I43, I44, I45, I46, I47, I48) 16.28/16.55 16.28/16.55 The dependency graph for this problem is: 16.28/16.55 4 -> 16.28/16.55 Where: 16.28/16.55 4) f5#(I21, I22, I23, I24, I25, I26, I27) -> f3#(I21, I22, I23, I24, I25, I26, I27) 16.28/16.55 16.28/16.55 We have the following SCCs. 16.28/16.55 16.28/16.55 16.28/16.55 DP problem for innermost termination. 16.28/16.55 P = 16.28/16.55 f2#(I7, I8, I9, I10, I11, I12, I13) -> f1#(1 + I7, I8, I9, I10, I11, I12, I13) [1 + I7 <= I10] 16.28/16.55 f1#(I42, I43, I44, I45, I46, I47, I48) -> f2#(I42, I43, I44, I45, I46, I47, I48) 16.28/16.55 R = 16.28/16.55 f7(x1, x2, x3, x4, x5, x6, x7) -> f6(x1, x2, x3, x4, x5, x6, x7) 16.28/16.55 f6(I0, I1, I2, I3, I4, I5, I6) -> f1(0, I1, I2, rnd4, rnd5, rnd6, I6) [rnd4 = rnd4 /\ rnd6 = rnd6 /\ rnd5 = rnd5] 16.28/16.55 f2(I7, I8, I9, I10, I11, I12, I13) -> f1(1 + I7, I8, I9, I10, I11, I12, I13) [1 + I7 <= I10] 16.28/16.55 f2(I14, I15, I16, I17, I18, I19, I20) -> f5(I14, 0, I18, I17, I18, I19, rnd7) [rnd7 = rnd7 /\ I17 <= I14] 16.28/16.55 f5(I21, I22, I23, I24, I25, I26, I27) -> f3(I21, I22, I23, I24, I25, I26, I27) 16.28/16.55 f3(I28, I29, I30, I31, I32, I33, I34) -> f5(I28, 1 + I29, I30, I31, I32, I33, I34) [1 + I29 <= I30] 16.28/16.55 f3(I35, I36, I37, I38, I39, I40, I41) -> f4(I35, I36, I37, I38, I39, I40, I41) [I37 <= I36] 16.28/16.55 f1(I42, I43, I44, I45, I46, I47, I48) -> f2(I42, I43, I44, I45, I46, I47, I48) 16.28/16.55 16.28/16.55 We use the reverse value criterion with the projection function NU: 16.28/16.55 NU[f1#(z1,z2,z3,z4,z5,z6,z7)] = z4 + -1 * (1 + z1) 16.28/16.55 NU[f2#(z1,z2,z3,z4,z5,z6,z7)] = z4 + -1 * (1 + z1) 16.28/16.55 16.28/16.55 This gives the following inequalities: 16.28/16.55 1 + I7 <= I10 ==> I10 + -1 * (1 + I7) > I10 + -1 * (1 + (1 + I7)) with I10 + -1 * (1 + I7) >= 0 16.28/16.55 ==> I45 + -1 * (1 + I42) >= I45 + -1 * (1 + I42) 16.28/16.55 16.28/16.55 We remove all the strictly oriented dependency pairs. 16.28/16.55 16.28/16.55 DP problem for innermost termination. 16.28/16.55 P = 16.28/16.55 f1#(I42, I43, I44, I45, I46, I47, I48) -> f2#(I42, I43, I44, I45, I46, I47, I48) 16.28/16.55 R = 16.28/16.55 f7(x1, x2, x3, x4, x5, x6, x7) -> f6(x1, x2, x3, x4, x5, x6, x7) 16.28/16.55 f6(I0, I1, I2, I3, I4, I5, I6) -> f1(0, I1, I2, rnd4, rnd5, rnd6, I6) [rnd4 = rnd4 /\ rnd6 = rnd6 /\ rnd5 = rnd5] 16.28/16.55 f2(I7, I8, I9, I10, I11, I12, I13) -> f1(1 + I7, I8, I9, I10, I11, I12, I13) [1 + I7 <= I10] 16.28/16.55 f2(I14, I15, I16, I17, I18, I19, I20) -> f5(I14, 0, I18, I17, I18, I19, rnd7) [rnd7 = rnd7 /\ I17 <= I14] 16.28/16.55 f5(I21, I22, I23, I24, I25, I26, I27) -> f3(I21, I22, I23, I24, I25, I26, I27) 16.28/16.55 f3(I28, I29, I30, I31, I32, I33, I34) -> f5(I28, 1 + I29, I30, I31, I32, I33, I34) [1 + I29 <= I30] 16.28/16.55 f3(I35, I36, I37, I38, I39, I40, I41) -> f4(I35, I36, I37, I38, I39, I40, I41) [I37 <= I36] 16.28/16.55 f1(I42, I43, I44, I45, I46, I47, I48) -> f2(I42, I43, I44, I45, I46, I47, I48) 16.28/16.55 16.28/16.55 The dependency graph for this problem is: 16.28/16.55 6 -> 16.28/16.55 Where: 16.28/16.55 6) f1#(I42, I43, I44, I45, I46, I47, I48) -> f2#(I42, I43, I44, I45, I46, I47, I48) 16.28/16.55 16.28/16.55 We have the following SCCs. 16.28/16.55 16.28/19.53 EOF