MAYBE

DP problem for innermost termination.
P =
  init#(x1, x2, x3, x4) -> f1#(rnd1, rnd2, rnd3, rnd4) 
  f3#(I0, I1, I2, I3) -> f2#(I1, I1, I1, I2) [1 <= I1 - 1 /\ 1 <= I0 - 1 /\ 0 <= I2 - 1]
  f2#(I4, I5, I6, I7) -> f3#(I4, I5 + 2, I7 + 1, I8) [I5 = I6 /\ -1 <= I7 - 1 /\ 0 <= I5 - 1 /\ I7 <= I5]
  f2#(I9, I10, I11, I12) -> f3#(I9, I10 + 4, I12, I13) [I10 = I11 /\ 0 <= I10 - 1 /\ I10 <= I12 - 1]
  f1#(I14, I15, I16, I17) -> f2#(2, 2, 2, 0) 
R = 
  init(x1, x2, x3, x4) -> f1(rnd1, rnd2, rnd3, rnd4) 
  f3(I0, I1, I2, I3) -> f2(I1, I1, I1, I2) [1 <= I1 - 1 /\ 1 <= I0 - 1 /\ 0 <= I2 - 1]
  f2(I4, I5, I6, I7) -> f3(I4, I5 + 2, I7 + 1, I8) [I5 = I6 /\ -1 <= I7 - 1 /\ 0 <= I5 - 1 /\ I7 <= I5]
  f2(I9, I10, I11, I12) -> f3(I9, I10 + 4, I12, I13) [I10 = I11 /\ 0 <= I10 - 1 /\ I10 <= I12 - 1]
  f1(I14, I15, I16, I17) -> f2(2, 2, 2, 0) 

The dependency graph for this problem is:
  0 -> 4
  1 -> 2, 3
  2 -> 1
  3 -> 1
  4 -> 2
Where:
  0) init#(x1, x2, x3, x4) -> f1#(rnd1, rnd2, rnd3, rnd4) 
  1) f3#(I0, I1, I2, I3) -> f2#(I1, I1, I1, I2) [1 <= I1 - 1 /\ 1 <= I0 - 1 /\ 0 <= I2 - 1]
  2) f2#(I4, I5, I6, I7) -> f3#(I4, I5 + 2, I7 + 1, I8) [I5 = I6 /\ -1 <= I7 - 1 /\ 0 <= I5 - 1 /\ I7 <= I5]
  3) f2#(I9, I10, I11, I12) -> f3#(I9, I10 + 4, I12, I13) [I10 = I11 /\ 0 <= I10 - 1 /\ I10 <= I12 - 1]
  4) f1#(I14, I15, I16, I17) -> f2#(2, 2, 2, 0) 

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

DP problem for innermost termination.
P =
  f3#(I0, I1, I2, I3) -> f2#(I1, I1, I1, I2) [1 <= I1 - 1 /\ 1 <= I0 - 1 /\ 0 <= I2 - 1]
  f2#(I4, I5, I6, I7) -> f3#(I4, I5 + 2, I7 + 1, I8) [I5 = I6 /\ -1 <= I7 - 1 /\ 0 <= I5 - 1 /\ I7 <= I5]
  f2#(I9, I10, I11, I12) -> f3#(I9, I10 + 4, I12, I13) [I10 = I11 /\ 0 <= I10 - 1 /\ I10 <= I12 - 1]
R = 
  init(x1, x2, x3, x4) -> f1(rnd1, rnd2, rnd3, rnd4) 
  f3(I0, I1, I2, I3) -> f2(I1, I1, I1, I2) [1 <= I1 - 1 /\ 1 <= I0 - 1 /\ 0 <= I2 - 1]
  f2(I4, I5, I6, I7) -> f3(I4, I5 + 2, I7 + 1, I8) [I5 = I6 /\ -1 <= I7 - 1 /\ 0 <= I5 - 1 /\ I7 <= I5]
  f2(I9, I10, I11, I12) -> f3(I9, I10 + 4, I12, I13) [I10 = I11 /\ 0 <= I10 - 1 /\ I10 <= I12 - 1]
  f1(I14, I15, I16, I17) -> f2(2, 2, 2, 0) 

We use the reverse value criterion with the projection function NU:
  NU[f2#(z1,z2,z3,z4)] = z4 - 1 + -1 * z2
  NU[f3#(z1,z2,z3,z4)] = z3 - 1 + -1 * z2

This gives the following inequalities:
  1 <= I1 - 1 /\ 1 <= I0 - 1 /\ 0 <= I2 - 1  ==>  I2 - 1 + -1 * I1 >= I2 - 1 + -1 * I1
  I5 = I6 /\ -1 <= I7 - 1 /\ 0 <= I5 - 1 /\ I7 <= I5  ==>  I7 - 1 + -1 * I5 >= I7 + 1 - 1 + -1 * (I5 + 2)
  I10 = I11 /\ 0 <= I10 - 1 /\ I10 <= I12 - 1  ==>  I12 - 1 + -1 * I10 > I12 - 1 + -1 * (I10 + 4)  with  I12 - 1 + -1 * I10 >= 0

We remove all the strictly oriented dependency pairs.

DP problem for innermost termination.
P =
  f3#(I0, I1, I2, I3) -> f2#(I1, I1, I1, I2) [1 <= I1 - 1 /\ 1 <= I0 - 1 /\ 0 <= I2 - 1]
  f2#(I4, I5, I6, I7) -> f3#(I4, I5 + 2, I7 + 1, I8) [I5 = I6 /\ -1 <= I7 - 1 /\ 0 <= I5 - 1 /\ I7 <= I5]
R = 
  init(x1, x2, x3, x4) -> f1(rnd1, rnd2, rnd3, rnd4) 
  f3(I0, I1, I2, I3) -> f2(I1, I1, I1, I2) [1 <= I1 - 1 /\ 1 <= I0 - 1 /\ 0 <= I2 - 1]
  f2(I4, I5, I6, I7) -> f3(I4, I5 + 2, I7 + 1, I8) [I5 = I6 /\ -1 <= I7 - 1 /\ 0 <= I5 - 1 /\ I7 <= I5]
  f2(I9, I10, I11, I12) -> f3(I9, I10 + 4, I12, I13) [I10 = I11 /\ 0 <= I10 - 1 /\ I10 <= I12 - 1]
  f1(I14, I15, I16, I17) -> f2(2, 2, 2, 0)