YES

DP problem for innermost termination.
P =
  f9#(x1, x2, x3, x4, x5, x6, x7, x8) -> f8#(x1, x2, x3, x4, x5, x6, x7, x8) 
  f8#(I0, I1, I2, I3, I4, I5, I6, I7) -> f4#(I0, I1, I2, I3, 4, 0, 0, I7) 
  f2#(I8, I9, I10, I11, I12, I13, I14, I15) -> f4#(I8, I9, I10, I11, I12, 1 + I13, I14, I15) 
  f4#(I16, I17, I18, I19, I20, I21, I22, I23) -> f2#(rnd1, I23, I18, I19, I20, I21, rnd7, I23) [I22 <= 0 /\ 0 <= I22 /\ 0 <= -1 + I20 - I21 /\ y1 = I23 /\ rnd7 = y1 /\ rnd1 = rnd1]
  f4#(I42, I43, I44, I45, I46, I47, I48, I49) -> f5#(I42, I43, I44, I45, I46, I47, I48, I49) [I45 = I45]
  f3#(I50, I51, I52, I53, I54, I55, I56, I57) -> f4#(I58, I57, I52, I53, I54, 1 + I55, I59, I57) [I60 = I57 /\ I59 = I60 /\ I58 = I58]
  f1#(I61, I62, I63, I64, I65, I66, I67, I68) -> f2#(I69, I68, I63, I64, I65, I66, I70, I68) [I71 = I68 /\ I70 = I71 /\ I69 = I69]
R = 
  f9(x1, x2, x3, x4, x5, x6, x7, x8) -> f8(x1, x2, x3, x4, x5, x6, x7, x8) 
  f8(I0, I1, I2, I3, I4, I5, I6, I7) -> f4(I0, I1, I2, I3, 4, 0, 0, I7) 
  f2(I8, I9, I10, I11, I12, I13, I14, I15) -> f4(I8, I9, I10, I11, I12, 1 + I13, I14, I15) 
  f4(I16, I17, I18, I19, I20, I21, I22, I23) -> f2(rnd1, I23, I18, I19, I20, I21, rnd7, I23) [I22 <= 0 /\ 0 <= I22 /\ 0 <= -1 + I20 - I21 /\ y1 = I23 /\ rnd7 = y1 /\ rnd1 = rnd1]
  f4(I24, I25, I26, I27, I28, I29, I30, I31) -> f7(I32, I25, I30, I27, I28, I29, I30, I31) [I32 = I30 /\ I28 - I29 <= 0 /\ 0 <= I30 /\ I30 <= 0]
  f5(I33, I34, I35, I36, I37, I38, I39, I40) -> f6(I41, I34, I39, I36, I37, I38, I39, I40) [I41 = I39]
  f4(I42, I43, I44, I45, I46, I47, I48, I49) -> f5(I42, I43, I44, I45, I46, I47, I48, I49) [I45 = I45]
  f3(I50, I51, I52, I53, I54, I55, I56, I57) -> f4(I58, I57, I52, I53, I54, 1 + I55, I59, I57) [I60 = I57 /\ I59 = I60 /\ I58 = I58]
  f1(I61, I62, I63, I64, I65, I66, I67, I68) -> f2(I69, I68, I63, I64, I65, I66, I70, I68) [I71 = I68 /\ I70 = I71 /\ I69 = I69]

The dependency graph for this problem is:
  0 -> 1
  1 -> 3, 4
  2 -> 3, 4
  3 -> 2
  4 -> 
  5 -> 3, 4
  6 -> 2
Where:
  0) f9#(x1, x2, x3, x4, x5, x6, x7, x8) -> f8#(x1, x2, x3, x4, x5, x6, x7, x8) 
  1) f8#(I0, I1, I2, I3, I4, I5, I6, I7) -> f4#(I0, I1, I2, I3, 4, 0, 0, I7) 
  2) f2#(I8, I9, I10, I11, I12, I13, I14, I15) -> f4#(I8, I9, I10, I11, I12, 1 + I13, I14, I15) 
  3) f4#(I16, I17, I18, I19, I20, I21, I22, I23) -> f2#(rnd1, I23, I18, I19, I20, I21, rnd7, I23) [I22 <= 0 /\ 0 <= I22 /\ 0 <= -1 + I20 - I21 /\ y1 = I23 /\ rnd7 = y1 /\ rnd1 = rnd1]
  4) f4#(I42, I43, I44, I45, I46, I47, I48, I49) -> f5#(I42, I43, I44, I45, I46, I47, I48, I49) [I45 = I45]
  5) f3#(I50, I51, I52, I53, I54, I55, I56, I57) -> f4#(I58, I57, I52, I53, I54, 1 + I55, I59, I57) [I60 = I57 /\ I59 = I60 /\ I58 = I58]
  6) f1#(I61, I62, I63, I64, I65, I66, I67, I68) -> f2#(I69, I68, I63, I64, I65, I66, I70, I68) [I71 = I68 /\ I70 = I71 /\ I69 = I69]

We have the following SCCs.
  { 2, 3 }

DP problem for innermost termination.
P =
  f2#(I8, I9, I10, I11, I12, I13, I14, I15) -> f4#(I8, I9, I10, I11, I12, 1 + I13, I14, I15) 
  f4#(I16, I17, I18, I19, I20, I21, I22, I23) -> f2#(rnd1, I23, I18, I19, I20, I21, rnd7, I23) [I22 <= 0 /\ 0 <= I22 /\ 0 <= -1 + I20 - I21 /\ y1 = I23 /\ rnd7 = y1 /\ rnd1 = rnd1]
R = 
  f9(x1, x2, x3, x4, x5, x6, x7, x8) -> f8(x1, x2, x3, x4, x5, x6, x7, x8) 
  f8(I0, I1, I2, I3, I4, I5, I6, I7) -> f4(I0, I1, I2, I3, 4, 0, 0, I7) 
  f2(I8, I9, I10, I11, I12, I13, I14, I15) -> f4(I8, I9, I10, I11, I12, 1 + I13, I14, I15) 
  f4(I16, I17, I18, I19, I20, I21, I22, I23) -> f2(rnd1, I23, I18, I19, I20, I21, rnd7, I23) [I22 <= 0 /\ 0 <= I22 /\ 0 <= -1 + I20 - I21 /\ y1 = I23 /\ rnd7 = y1 /\ rnd1 = rnd1]
  f4(I24, I25, I26, I27, I28, I29, I30, I31) -> f7(I32, I25, I30, I27, I28, I29, I30, I31) [I32 = I30 /\ I28 - I29 <= 0 /\ 0 <= I30 /\ I30 <= 0]
  f5(I33, I34, I35, I36, I37, I38, I39, I40) -> f6(I41, I34, I39, I36, I37, I38, I39, I40) [I41 = I39]
  f4(I42, I43, I44, I45, I46, I47, I48, I49) -> f5(I42, I43, I44, I45, I46, I47, I48, I49) [I45 = I45]
  f3(I50, I51, I52, I53, I54, I55, I56, I57) -> f4(I58, I57, I52, I53, I54, 1 + I55, I59, I57) [I60 = I57 /\ I59 = I60 /\ I58 = I58]
  f1(I61, I62, I63, I64, I65, I66, I67, I68) -> f2(I69, I68, I63, I64, I65, I66, I70, I68) [I71 = I68 /\ I70 = I71 /\ I69 = I69]

We use the reverse value criterion with the projection function NU:
  NU[f4#(z1,z2,z3,z4,z5,z6,z7,z8)] = -1 + z5 - z6 + -1 * 0
  NU[f2#(z1,z2,z3,z4,z5,z6,z7,z8)] = -1 + z5 - (1 + z6) + -1 * 0

This gives the following inequalities:
    ==>  -1 + I12 - (1 + I13) + -1 * 0 >= -1 + I12 - (1 + I13) + -1 * 0
  I22 <= 0 /\ 0 <= I22 /\ 0 <= -1 + I20 - I21 /\ y1 = I23 /\ rnd7 = y1 /\ rnd1 = rnd1  ==>  -1 + I20 - I21 + -1 * 0 > -1 + I20 - (1 + I21) + -1 * 0  with  -1 + I20 - I21 + -1 * 0 >= 0

We remove all the strictly oriented dependency pairs.

DP problem for innermost termination.
P =
  f2#(I8, I9, I10, I11, I12, I13, I14, I15) -> f4#(I8, I9, I10, I11, I12, 1 + I13, I14, I15) 
R = 
  f9(x1, x2, x3, x4, x5, x6, x7, x8) -> f8(x1, x2, x3, x4, x5, x6, x7, x8) 
  f8(I0, I1, I2, I3, I4, I5, I6, I7) -> f4(I0, I1, I2, I3, 4, 0, 0, I7) 
  f2(I8, I9, I10, I11, I12, I13, I14, I15) -> f4(I8, I9, I10, I11, I12, 1 + I13, I14, I15) 
  f4(I16, I17, I18, I19, I20, I21, I22, I23) -> f2(rnd1, I23, I18, I19, I20, I21, rnd7, I23) [I22 <= 0 /\ 0 <= I22 /\ 0 <= -1 + I20 - I21 /\ y1 = I23 /\ rnd7 = y1 /\ rnd1 = rnd1]
  f4(I24, I25, I26, I27, I28, I29, I30, I31) -> f7(I32, I25, I30, I27, I28, I29, I30, I31) [I32 = I30 /\ I28 - I29 <= 0 /\ 0 <= I30 /\ I30 <= 0]
  f5(I33, I34, I35, I36, I37, I38, I39, I40) -> f6(I41, I34, I39, I36, I37, I38, I39, I40) [I41 = I39]
  f4(I42, I43, I44, I45, I46, I47, I48, I49) -> f5(I42, I43, I44, I45, I46, I47, I48, I49) [I45 = I45]
  f3(I50, I51, I52, I53, I54, I55, I56, I57) -> f4(I58, I57, I52, I53, I54, 1 + I55, I59, I57) [I60 = I57 /\ I59 = I60 /\ I58 = I58]
  f1(I61, I62, I63, I64, I65, I66, I67, I68) -> f2(I69, I68, I63, I64, I65, I66, I70, I68) [I71 = I68 /\ I70 = I71 /\ I69 = I69]

The dependency graph for this problem is:
  2 -> 
Where:
  2) f2#(I8, I9, I10, I11, I12, I13, I14, I15) -> f4#(I8, I9, I10, I11, I12, 1 + I13, I14, I15) 

We have the following SCCs.