10.50/10.39 YES 10.50/10.39 10.50/10.39 DP problem for innermost termination. 10.50/10.39 P = 10.50/10.39 f8#(x1, x2, x3, x4) -> f1#(x1, x2, x3, x4) 10.50/10.39 f7#(I0, I1, I2, I3) -> f2#(I0, I1, I2, I3) 10.50/10.39 f2#(I4, I5, I6, I7) -> f7#(I4, I5, 1 + I6, I7) [0 <= -1 - I6 + I7] 10.50/10.39 f6#(I8, I9, I10, I11) -> f2#(I8, I9, I10, I11) 10.50/10.39 f2#(I12, I13, I14, I15) -> f6#(I12, I13, 1 + I14, I15) [I15 <= I14 /\ I14 <= I15 /\ -1 * I14 + I15 <= 0 /\ -1 * I14 + I15 <= 0] 10.50/10.39 f3#(I20, I21, I22, I23) -> f4#(I20, I21, I22, I23) [I21 = I21] 10.50/10.39 f2#(I24, I25, I26, I27) -> f3#(I24, I25, I26, I27) [-1 * I26 + I27 <= 0 /\ -1 * I26 + I27 <= 0] 10.50/10.39 f1#(I28, I29, I30, I31) -> f2#(I28, I29, I30, I31) 10.50/10.39 R = 10.50/10.39 f8(x1, x2, x3, x4) -> f1(x1, x2, x3, x4) 10.50/10.39 f7(I0, I1, I2, I3) -> f2(I0, I1, I2, I3) 10.50/10.39 f2(I4, I5, I6, I7) -> f7(I4, I5, 1 + I6, I7) [0 <= -1 - I6 + I7] 10.50/10.39 f6(I8, I9, I10, I11) -> f2(I8, I9, I10, I11) 10.50/10.39 f2(I12, I13, I14, I15) -> f6(I12, I13, 1 + I14, I15) [I15 <= I14 /\ I14 <= I15 /\ -1 * I14 + I15 <= 0 /\ -1 * I14 + I15 <= 0] 10.50/10.39 f4(I16, I17, I18, I19) -> f5(rnd1, I17, I18, I19) [rnd1 = rnd1] 10.50/10.39 f3(I20, I21, I22, I23) -> f4(I20, I21, I22, I23) [I21 = I21] 10.50/10.39 f2(I24, I25, I26, I27) -> f3(I24, I25, I26, I27) [-1 * I26 + I27 <= 0 /\ -1 * I26 + I27 <= 0] 10.50/10.39 f1(I28, I29, I30, I31) -> f2(I28, I29, I30, I31) 10.50/10.39 10.50/10.39 The dependency graph for this problem is: 10.50/10.39 0 -> 7 10.50/10.39 1 -> 2, 4, 6 10.50/10.39 2 -> 1 10.50/10.39 3 -> 2, 4, 6 10.50/10.39 4 -> 3 10.50/10.39 5 -> 10.50/10.39 6 -> 5 10.50/10.39 7 -> 2, 4, 6 10.50/10.39 Where: 10.50/10.39 0) f8#(x1, x2, x3, x4) -> f1#(x1, x2, x3, x4) 10.50/10.39 1) f7#(I0, I1, I2, I3) -> f2#(I0, I1, I2, I3) 10.50/10.39 2) f2#(I4, I5, I6, I7) -> f7#(I4, I5, 1 + I6, I7) [0 <= -1 - I6 + I7] 10.50/10.39 3) f6#(I8, I9, I10, I11) -> f2#(I8, I9, I10, I11) 10.50/10.39 4) f2#(I12, I13, I14, I15) -> f6#(I12, I13, 1 + I14, I15) [I15 <= I14 /\ I14 <= I15 /\ -1 * I14 + I15 <= 0 /\ -1 * I14 + I15 <= 0] 10.50/10.39 5) f3#(I20, I21, I22, I23) -> f4#(I20, I21, I22, I23) [I21 = I21] 10.50/10.39 6) f2#(I24, I25, I26, I27) -> f3#(I24, I25, I26, I27) [-1 * I26 + I27 <= 0 /\ -1 * I26 + I27 <= 0] 10.50/10.39 7) f1#(I28, I29, I30, I31) -> f2#(I28, I29, I30, I31) 10.50/10.39 10.50/10.39 We have the following SCCs. 10.50/10.39 { 1, 2, 3, 4 } 10.50/10.39 10.50/10.39 DP problem for innermost termination. 10.50/10.39 P = 10.50/10.39 f7#(I0, I1, I2, I3) -> f2#(I0, I1, I2, I3) 10.50/10.39 f2#(I4, I5, I6, I7) -> f7#(I4, I5, 1 + I6, I7) [0 <= -1 - I6 + I7] 10.50/10.39 f6#(I8, I9, I10, I11) -> f2#(I8, I9, I10, I11) 10.50/10.39 f2#(I12, I13, I14, I15) -> f6#(I12, I13, 1 + I14, I15) [I15 <= I14 /\ I14 <= I15 /\ -1 * I14 + I15 <= 0 /\ -1 * I14 + I15 <= 0] 10.50/10.39 R = 10.50/10.39 f8(x1, x2, x3, x4) -> f1(x1, x2, x3, x4) 10.50/10.39 f7(I0, I1, I2, I3) -> f2(I0, I1, I2, I3) 10.50/10.39 f2(I4, I5, I6, I7) -> f7(I4, I5, 1 + I6, I7) [0 <= -1 - I6 + I7] 10.50/10.39 f6(I8, I9, I10, I11) -> f2(I8, I9, I10, I11) 10.50/10.39 f2(I12, I13, I14, I15) -> f6(I12, I13, 1 + I14, I15) [I15 <= I14 /\ I14 <= I15 /\ -1 * I14 + I15 <= 0 /\ -1 * I14 + I15 <= 0] 10.50/10.39 f4(I16, I17, I18, I19) -> f5(rnd1, I17, I18, I19) [rnd1 = rnd1] 10.50/10.39 f3(I20, I21, I22, I23) -> f4(I20, I21, I22, I23) [I21 = I21] 10.50/10.39 f2(I24, I25, I26, I27) -> f3(I24, I25, I26, I27) [-1 * I26 + I27 <= 0 /\ -1 * I26 + I27 <= 0] 10.50/10.39 f1(I28, I29, I30, I31) -> f2(I28, I29, I30, I31) 10.50/10.39 10.50/10.39 We use the reverse value criterion with the projection function NU: 10.50/10.39 NU[f6#(z1,z2,z3,z4)] = z4 + -1 * z3 10.50/10.39 NU[f2#(z1,z2,z3,z4)] = z4 + -1 * z3 10.50/10.39 NU[f7#(z1,z2,z3,z4)] = z4 + -1 * z3 10.50/10.39 10.50/10.39 This gives the following inequalities: 10.50/10.39 ==> I3 + -1 * I2 >= I3 + -1 * I2 10.50/10.39 0 <= -1 - I6 + I7 ==> I7 + -1 * I6 > I7 + -1 * (1 + I6) with I7 + -1 * I6 >= 0 10.50/10.39 ==> I11 + -1 * I10 >= I11 + -1 * I10 10.50/10.39 I15 <= I14 /\ I14 <= I15 /\ -1 * I14 + I15 <= 0 /\ -1 * I14 + I15 <= 0 ==> I15 + -1 * I14 > I15 + -1 * (1 + I14) with I15 + -1 * I14 >= 0 10.50/10.39 10.50/10.39 We remove all the strictly oriented dependency pairs. 10.50/10.39 10.50/10.39 DP problem for innermost termination. 10.50/10.39 P = 10.50/10.39 f7#(I0, I1, I2, I3) -> f2#(I0, I1, I2, I3) 10.50/10.39 f6#(I8, I9, I10, I11) -> f2#(I8, I9, I10, I11) 10.50/10.39 R = 10.50/10.39 f8(x1, x2, x3, x4) -> f1(x1, x2, x3, x4) 10.50/10.39 f7(I0, I1, I2, I3) -> f2(I0, I1, I2, I3) 10.50/10.39 f2(I4, I5, I6, I7) -> f7(I4, I5, 1 + I6, I7) [0 <= -1 - I6 + I7] 10.50/10.39 f6(I8, I9, I10, I11) -> f2(I8, I9, I10, I11) 10.50/10.39 f2(I12, I13, I14, I15) -> f6(I12, I13, 1 + I14, I15) [I15 <= I14 /\ I14 <= I15 /\ -1 * I14 + I15 <= 0 /\ -1 * I14 + I15 <= 0] 10.50/10.39 f4(I16, I17, I18, I19) -> f5(rnd1, I17, I18, I19) [rnd1 = rnd1] 10.50/10.39 f3(I20, I21, I22, I23) -> f4(I20, I21, I22, I23) [I21 = I21] 10.50/10.39 f2(I24, I25, I26, I27) -> f3(I24, I25, I26, I27) [-1 * I26 + I27 <= 0 /\ -1 * I26 + I27 <= 0] 10.50/10.39 f1(I28, I29, I30, I31) -> f2(I28, I29, I30, I31) 10.50/10.39 10.50/10.39 The dependency graph for this problem is: 10.50/10.39 1 -> 10.50/10.39 3 -> 10.50/10.39 Where: 10.50/10.39 1) f7#(I0, I1, I2, I3) -> f2#(I0, I1, I2, I3) 10.50/10.39 3) f6#(I8, I9, I10, I11) -> f2#(I8, I9, I10, I11) 10.50/10.39 10.50/10.39 We have the following SCCs. 10.50/10.39 10.50/13.36 EOF