YES

DP problem for innermost termination.
P =
  f7#(x1, x2, x3, x4, x5, x6, x7, x8) -> f1#(x1, x2, x3, x4, x5, x6, x7, x8) 
  f6#(I0, I1, I2, I3, I4, I5, I6, I7) -> f3#(I0, I1, I2, I3, I4, I5, I6, I7) 
  f3#(I8, I9, I10, I11, I12, I13, I14, I15) -> f6#(I8, I9, -1 + I10, I11, rnd5, -1 + I13, I14, rnd8) [1 <= rnd8 /\ -1 + rnd8 <= -1 + I13 /\ -1 + I13 <= -1 + rnd8 /\ -1 + rnd5 <= -1 + I10 /\ -1 + I10 <= -1 + rnd5 /\ 1 <= I13 /\ rnd8 = rnd8 /\ rnd5 = rnd5]
  f3#(I16, I17, I18, I19, I20, I21, I22, I23) -> f2#(I16, I17, I18, I19, I20, I21, I22, I23) [I21 <= 0 /\ I21 <= 0]
  f5#(I24, I25, I26, I27, I28, I29, I30, I31) -> f3#(I24, I25, -1 + I26, rnd4, I28, -1 + I29, rnd7, I31) [1 <= rnd7 /\ 1 <= rnd4 /\ -1 + rnd7 <= -1 + I29 /\ -1 + I29 <= -1 + rnd7 /\ -1 + rnd4 <= -1 + I26 /\ -1 + I26 <= -1 + rnd4 /\ 1 <= I29 /\ rnd7 = rnd7 /\ rnd4 = rnd4]
  f2#(I40, I41, I42, I43, I44, I45, I46, I47) -> f3#(I40, I41, I42, I43, I44, rnd6, I46, I47) [1 <= I42 /\ 1 <= rnd6 /\ rnd6 = rnd6 /\ 1 <= I42]
  f1#(I48, I49, I50, I51, I52, I53, I54, I55) -> f2#(I48, I49, rnd3, I51, I52, I56, I54, I55) [I56 = I56 /\ rnd3 = rnd3]
R = 
  f7(x1, x2, x3, x4, x5, x6, x7, x8) -> f1(x1, x2, x3, x4, x5, x6, x7, x8) 
  f6(I0, I1, I2, I3, I4, I5, I6, I7) -> f3(I0, I1, I2, I3, I4, I5, I6, I7) 
  f3(I8, I9, I10, I11, I12, I13, I14, I15) -> f6(I8, I9, -1 + I10, I11, rnd5, -1 + I13, I14, rnd8) [1 <= rnd8 /\ -1 + rnd8 <= -1 + I13 /\ -1 + I13 <= -1 + rnd8 /\ -1 + rnd5 <= -1 + I10 /\ -1 + I10 <= -1 + rnd5 /\ 1 <= I13 /\ rnd8 = rnd8 /\ rnd5 = rnd5]
  f3(I16, I17, I18, I19, I20, I21, I22, I23) -> f2(I16, I17, I18, I19, I20, I21, I22, I23) [I21 <= 0 /\ I21 <= 0]
  f5(I24, I25, I26, I27, I28, I29, I30, I31) -> f3(I24, I25, -1 + I26, rnd4, I28, -1 + I29, rnd7, I31) [1 <= rnd7 /\ 1 <= rnd4 /\ -1 + rnd7 <= -1 + I29 /\ -1 + I29 <= -1 + rnd7 /\ -1 + rnd4 <= -1 + I26 /\ -1 + I26 <= -1 + rnd4 /\ 1 <= I29 /\ rnd7 = rnd7 /\ rnd4 = rnd4]
  f2(I32, I33, I34, I35, I36, I37, I38, I39) -> f4(I33, I33, I34, I35, I36, I37, I38, I39) [I34 <= 0]
  f2(I40, I41, I42, I43, I44, I45, I46, I47) -> f3(I40, I41, I42, I43, I44, rnd6, I46, I47) [1 <= I42 /\ 1 <= rnd6 /\ rnd6 = rnd6 /\ 1 <= I42]
  f1(I48, I49, I50, I51, I52, I53, I54, I55) -> f2(I48, I49, rnd3, I51, I52, I56, I54, I55) [I56 = I56 /\ rnd3 = rnd3]

The dependency graph for this problem is:
  0 -> 6
  1 -> 2, 3
  2 -> 1
  3 -> 5
  4 -> 2, 3
  5 -> 2
  6 -> 5
Where:
  0) f7#(x1, x2, x3, x4, x5, x6, x7, x8) -> f1#(x1, x2, x3, x4, x5, x6, x7, x8) 
  1) f6#(I0, I1, I2, I3, I4, I5, I6, I7) -> f3#(I0, I1, I2, I3, I4, I5, I6, I7) 
  2) f3#(I8, I9, I10, I11, I12, I13, I14, I15) -> f6#(I8, I9, -1 + I10, I11, rnd5, -1 + I13, I14, rnd8) [1 <= rnd8 /\ -1 + rnd8 <= -1 + I13 /\ -1 + I13 <= -1 + rnd8 /\ -1 + rnd5 <= -1 + I10 /\ -1 + I10 <= -1 + rnd5 /\ 1 <= I13 /\ rnd8 = rnd8 /\ rnd5 = rnd5]
  3) f3#(I16, I17, I18, I19, I20, I21, I22, I23) -> f2#(I16, I17, I18, I19, I20, I21, I22, I23) [I21 <= 0 /\ I21 <= 0]
  4) f5#(I24, I25, I26, I27, I28, I29, I30, I31) -> f3#(I24, I25, -1 + I26, rnd4, I28, -1 + I29, rnd7, I31) [1 <= rnd7 /\ 1 <= rnd4 /\ -1 + rnd7 <= -1 + I29 /\ -1 + I29 <= -1 + rnd7 /\ -1 + rnd4 <= -1 + I26 /\ -1 + I26 <= -1 + rnd4 /\ 1 <= I29 /\ rnd7 = rnd7 /\ rnd4 = rnd4]
  5) f2#(I40, I41, I42, I43, I44, I45, I46, I47) -> f3#(I40, I41, I42, I43, I44, rnd6, I46, I47) [1 <= I42 /\ 1 <= rnd6 /\ rnd6 = rnd6 /\ 1 <= I42]
  6) f1#(I48, I49, I50, I51, I52, I53, I54, I55) -> f2#(I48, I49, rnd3, I51, I52, I56, I54, I55) [I56 = I56 /\ rnd3 = rnd3]

We have the following SCCs.
  { 1, 2, 3, 5 }

DP problem for innermost termination.
P =
  f6#(I0, I1, I2, I3, I4, I5, I6, I7) -> f3#(I0, I1, I2, I3, I4, I5, I6, I7) 
  f3#(I8, I9, I10, I11, I12, I13, I14, I15) -> f6#(I8, I9, -1 + I10, I11, rnd5, -1 + I13, I14, rnd8) [1 <= rnd8 /\ -1 + rnd8 <= -1 + I13 /\ -1 + I13 <= -1 + rnd8 /\ -1 + rnd5 <= -1 + I10 /\ -1 + I10 <= -1 + rnd5 /\ 1 <= I13 /\ rnd8 = rnd8 /\ rnd5 = rnd5]
  f3#(I16, I17, I18, I19, I20, I21, I22, I23) -> f2#(I16, I17, I18, I19, I20, I21, I22, I23) [I21 <= 0 /\ I21 <= 0]
  f2#(I40, I41, I42, I43, I44, I45, I46, I47) -> f3#(I40, I41, I42, I43, I44, rnd6, I46, I47) [1 <= I42 /\ 1 <= rnd6 /\ rnd6 = rnd6 /\ 1 <= I42]
R = 
  f7(x1, x2, x3, x4, x5, x6, x7, x8) -> f1(x1, x2, x3, x4, x5, x6, x7, x8) 
  f6(I0, I1, I2, I3, I4, I5, I6, I7) -> f3(I0, I1, I2, I3, I4, I5, I6, I7) 
  f3(I8, I9, I10, I11, I12, I13, I14, I15) -> f6(I8, I9, -1 + I10, I11, rnd5, -1 + I13, I14, rnd8) [1 <= rnd8 /\ -1 + rnd8 <= -1 + I13 /\ -1 + I13 <= -1 + rnd8 /\ -1 + rnd5 <= -1 + I10 /\ -1 + I10 <= -1 + rnd5 /\ 1 <= I13 /\ rnd8 = rnd8 /\ rnd5 = rnd5]
  f3(I16, I17, I18, I19, I20, I21, I22, I23) -> f2(I16, I17, I18, I19, I20, I21, I22, I23) [I21 <= 0 /\ I21 <= 0]
  f5(I24, I25, I26, I27, I28, I29, I30, I31) -> f3(I24, I25, -1 + I26, rnd4, I28, -1 + I29, rnd7, I31) [1 <= rnd7 /\ 1 <= rnd4 /\ -1 + rnd7 <= -1 + I29 /\ -1 + I29 <= -1 + rnd7 /\ -1 + rnd4 <= -1 + I26 /\ -1 + I26 <= -1 + rnd4 /\ 1 <= I29 /\ rnd7 = rnd7 /\ rnd4 = rnd4]
  f2(I32, I33, I34, I35, I36, I37, I38, I39) -> f4(I33, I33, I34, I35, I36, I37, I38, I39) [I34 <= 0]
  f2(I40, I41, I42, I43, I44, I45, I46, I47) -> f3(I40, I41, I42, I43, I44, rnd6, I46, I47) [1 <= I42 /\ 1 <= rnd6 /\ rnd6 = rnd6 /\ 1 <= I42]
  f1(I48, I49, I50, I51, I52, I53, I54, I55) -> f2(I48, I49, rnd3, I51, I52, I56, I54, I55) [I56 = I56 /\ rnd3 = rnd3]

We use the extended value criterion with the projection function NU:
  NU[f2#(x0,x1,x2,x3,x4,x5,x6,x7)] = x2 - 1
  NU[f3#(x0,x1,x2,x3,x4,x5,x6,x7)] = x2 - x5 - 1
  NU[f6#(x0,x1,x2,x3,x4,x5,x6,x7)] = x2 - x5 - 1

This gives the following inequalities:
    ==>  I2 - I5 - 1 >= I2 - I5 - 1
  1 <= rnd8 /\ -1 + rnd8 <= -1 + I13 /\ -1 + I13 <= -1 + rnd8 /\ -1 + rnd5 <= -1 + I10 /\ -1 + I10 <= -1 + rnd5 /\ 1 <= I13 /\ rnd8 = rnd8 /\ rnd5 = rnd5  ==>  I10 - I13 - 1 >= (-1 + I10) - (-1 + I13) - 1
  I21 <= 0 /\ I21 <= 0  ==>  I18 - I21 - 1 >= I18 - 1
  1 <= I42 /\ 1 <= rnd6 /\ rnd6 = rnd6 /\ 1 <= I42  ==>  I42 - 1 > I42 - rnd6 - 1  with  I42 - 1 >= 0

We remove all the strictly oriented dependency pairs.

DP problem for innermost termination.
P =
  f6#(I0, I1, I2, I3, I4, I5, I6, I7) -> f3#(I0, I1, I2, I3, I4, I5, I6, I7) 
  f3#(I8, I9, I10, I11, I12, I13, I14, I15) -> f6#(I8, I9, -1 + I10, I11, rnd5, -1 + I13, I14, rnd8) [1 <= rnd8 /\ -1 + rnd8 <= -1 + I13 /\ -1 + I13 <= -1 + rnd8 /\ -1 + rnd5 <= -1 + I10 /\ -1 + I10 <= -1 + rnd5 /\ 1 <= I13 /\ rnd8 = rnd8 /\ rnd5 = rnd5]
  f3#(I16, I17, I18, I19, I20, I21, I22, I23) -> f2#(I16, I17, I18, I19, I20, I21, I22, I23) [I21 <= 0 /\ I21 <= 0]
R = 
  f7(x1, x2, x3, x4, x5, x6, x7, x8) -> f1(x1, x2, x3, x4, x5, x6, x7, x8) 
  f6(I0, I1, I2, I3, I4, I5, I6, I7) -> f3(I0, I1, I2, I3, I4, I5, I6, I7) 
  f3(I8, I9, I10, I11, I12, I13, I14, I15) -> f6(I8, I9, -1 + I10, I11, rnd5, -1 + I13, I14, rnd8) [1 <= rnd8 /\ -1 + rnd8 <= -1 + I13 /\ -1 + I13 <= -1 + rnd8 /\ -1 + rnd5 <= -1 + I10 /\ -1 + I10 <= -1 + rnd5 /\ 1 <= I13 /\ rnd8 = rnd8 /\ rnd5 = rnd5]
  f3(I16, I17, I18, I19, I20, I21, I22, I23) -> f2(I16, I17, I18, I19, I20, I21, I22, I23) [I21 <= 0 /\ I21 <= 0]
  f5(I24, I25, I26, I27, I28, I29, I30, I31) -> f3(I24, I25, -1 + I26, rnd4, I28, -1 + I29, rnd7, I31) [1 <= rnd7 /\ 1 <= rnd4 /\ -1 + rnd7 <= -1 + I29 /\ -1 + I29 <= -1 + rnd7 /\ -1 + rnd4 <= -1 + I26 /\ -1 + I26 <= -1 + rnd4 /\ 1 <= I29 /\ rnd7 = rnd7 /\ rnd4 = rnd4]
  f2(I32, I33, I34, I35, I36, I37, I38, I39) -> f4(I33, I33, I34, I35, I36, I37, I38, I39) [I34 <= 0]
  f2(I40, I41, I42, I43, I44, I45, I46, I47) -> f3(I40, I41, I42, I43, I44, rnd6, I46, I47) [1 <= I42 /\ 1 <= rnd6 /\ rnd6 = rnd6 /\ 1 <= I42]
  f1(I48, I49, I50, I51, I52, I53, I54, I55) -> f2(I48, I49, rnd3, I51, I52, I56, I54, I55) [I56 = I56 /\ rnd3 = rnd3]

The dependency graph for this problem is:
  1 -> 2, 3
  2 -> 1
  3 -> 
Where:
  1) f6#(I0, I1, I2, I3, I4, I5, I6, I7) -> f3#(I0, I1, I2, I3, I4, I5, I6, I7) 
  2) f3#(I8, I9, I10, I11, I12, I13, I14, I15) -> f6#(I8, I9, -1 + I10, I11, rnd5, -1 + I13, I14, rnd8) [1 <= rnd8 /\ -1 + rnd8 <= -1 + I13 /\ -1 + I13 <= -1 + rnd8 /\ -1 + rnd5 <= -1 + I10 /\ -1 + I10 <= -1 + rnd5 /\ 1 <= I13 /\ rnd8 = rnd8 /\ rnd5 = rnd5]
  3) f3#(I16, I17, I18, I19, I20, I21, I22, I23) -> f2#(I16, I17, I18, I19, I20, I21, I22, I23) [I21 <= 0 /\ I21 <= 0]

We have the following SCCs.
  { 1, 2 }

DP problem for innermost termination.
P =
  f6#(I0, I1, I2, I3, I4, I5, I6, I7) -> f3#(I0, I1, I2, I3, I4, I5, I6, I7) 
  f3#(I8, I9, I10, I11, I12, I13, I14, I15) -> f6#(I8, I9, -1 + I10, I11, rnd5, -1 + I13, I14, rnd8) [1 <= rnd8 /\ -1 + rnd8 <= -1 + I13 /\ -1 + I13 <= -1 + rnd8 /\ -1 + rnd5 <= -1 + I10 /\ -1 + I10 <= -1 + rnd5 /\ 1 <= I13 /\ rnd8 = rnd8 /\ rnd5 = rnd5]
R = 
  f7(x1, x2, x3, x4, x5, x6, x7, x8) -> f1(x1, x2, x3, x4, x5, x6, x7, x8) 
  f6(I0, I1, I2, I3, I4, I5, I6, I7) -> f3(I0, I1, I2, I3, I4, I5, I6, I7) 
  f3(I8, I9, I10, I11, I12, I13, I14, I15) -> f6(I8, I9, -1 + I10, I11, rnd5, -1 + I13, I14, rnd8) [1 <= rnd8 /\ -1 + rnd8 <= -1 + I13 /\ -1 + I13 <= -1 + rnd8 /\ -1 + rnd5 <= -1 + I10 /\ -1 + I10 <= -1 + rnd5 /\ 1 <= I13 /\ rnd8 = rnd8 /\ rnd5 = rnd5]
  f3(I16, I17, I18, I19, I20, I21, I22, I23) -> f2(I16, I17, I18, I19, I20, I21, I22, I23) [I21 <= 0 /\ I21 <= 0]
  f5(I24, I25, I26, I27, I28, I29, I30, I31) -> f3(I24, I25, -1 + I26, rnd4, I28, -1 + I29, rnd7, I31) [1 <= rnd7 /\ 1 <= rnd4 /\ -1 + rnd7 <= -1 + I29 /\ -1 + I29 <= -1 + rnd7 /\ -1 + rnd4 <= -1 + I26 /\ -1 + I26 <= -1 + rnd4 /\ 1 <= I29 /\ rnd7 = rnd7 /\ rnd4 = rnd4]
  f2(I32, I33, I34, I35, I36, I37, I38, I39) -> f4(I33, I33, I34, I35, I36, I37, I38, I39) [I34 <= 0]
  f2(I40, I41, I42, I43, I44, I45, I46, I47) -> f3(I40, I41, I42, I43, I44, rnd6, I46, I47) [1 <= I42 /\ 1 <= rnd6 /\ rnd6 = rnd6 /\ 1 <= I42]
  f1(I48, I49, I50, I51, I52, I53, I54, I55) -> f2(I48, I49, rnd3, I51, I52, I56, I54, I55) [I56 = I56 /\ rnd3 = rnd3]

We use the basic value criterion with the projection function NU:
  NU[f3#(z1,z2,z3,z4,z5,z6,z7,z8)] = z6
  NU[f6#(z1,z2,z3,z4,z5,z6,z7,z8)] = z6

This gives the following inequalities:
    ==>  I5 (>! \union =) I5
  1 <= rnd8 /\ -1 + rnd8 <= -1 + I13 /\ -1 + I13 <= -1 + rnd8 /\ -1 + rnd5 <= -1 + I10 /\ -1 + I10 <= -1 + rnd5 /\ 1 <= I13 /\ rnd8 = rnd8 /\ rnd5 = rnd5  ==>  I13 >! -1 + I13

We remove all the strictly oriented dependency pairs.

DP problem for innermost termination.
P =
  f6#(I0, I1, I2, I3, I4, I5, I6, I7) -> f3#(I0, I1, I2, I3, I4, I5, I6, I7) 
R = 
  f7(x1, x2, x3, x4, x5, x6, x7, x8) -> f1(x1, x2, x3, x4, x5, x6, x7, x8) 
  f6(I0, I1, I2, I3, I4, I5, I6, I7) -> f3(I0, I1, I2, I3, I4, I5, I6, I7) 
  f3(I8, I9, I10, I11, I12, I13, I14, I15) -> f6(I8, I9, -1 + I10, I11, rnd5, -1 + I13, I14, rnd8) [1 <= rnd8 /\ -1 + rnd8 <= -1 + I13 /\ -1 + I13 <= -1 + rnd8 /\ -1 + rnd5 <= -1 + I10 /\ -1 + I10 <= -1 + rnd5 /\ 1 <= I13 /\ rnd8 = rnd8 /\ rnd5 = rnd5]
  f3(I16, I17, I18, I19, I20, I21, I22, I23) -> f2(I16, I17, I18, I19, I20, I21, I22, I23) [I21 <= 0 /\ I21 <= 0]
  f5(I24, I25, I26, I27, I28, I29, I30, I31) -> f3(I24, I25, -1 + I26, rnd4, I28, -1 + I29, rnd7, I31) [1 <= rnd7 /\ 1 <= rnd4 /\ -1 + rnd7 <= -1 + I29 /\ -1 + I29 <= -1 + rnd7 /\ -1 + rnd4 <= -1 + I26 /\ -1 + I26 <= -1 + rnd4 /\ 1 <= I29 /\ rnd7 = rnd7 /\ rnd4 = rnd4]
  f2(I32, I33, I34, I35, I36, I37, I38, I39) -> f4(I33, I33, I34, I35, I36, I37, I38, I39) [I34 <= 0]
  f2(I40, I41, I42, I43, I44, I45, I46, I47) -> f3(I40, I41, I42, I43, I44, rnd6, I46, I47) [1 <= I42 /\ 1 <= rnd6 /\ rnd6 = rnd6 /\ 1 <= I42]
  f1(I48, I49, I50, I51, I52, I53, I54, I55) -> f2(I48, I49, rnd3, I51, I52, I56, I54, I55) [I56 = I56 /\ rnd3 = rnd3]

The dependency graph for this problem is:
  1 -> 
Where:
  1) f6#(I0, I1, I2, I3, I4, I5, I6, I7) -> f3#(I0, I1, I2, I3, I4, I5, I6, I7) 

We have the following SCCs.