MAYBE

DP problem for innermost termination.
P =
  init#(x1, x2) -> f1#(rnd1, rnd2) 
  f3#(I0, I1) -> f2#(I0 + 3, I2) [I0 - 2 * y1 = 1 /\ 0 <= I0 - 1 /\ I0 - 2 * y1 <= 1 /\ 0 <= I0 - 2 * y1]
  f2#(I3, I4) -> f3#(I3, I5) [0 <= I3 - 1 /\ I3 - 2 * I6 = 1]
  f3#(I7, I8) -> f2#(I7 - 1, I9) [0 <= I7 - 1 /\ I7 - 2 * I10 = 0 /\ I7 - 2 * I10 <= 1 /\ 0 <= I7 - 2 * I10]
  f2#(I11, I12) -> f3#(I11, I13) [I11 - 2 * I14 = 0 /\ 0 <= I11 - 1]
  f1#(I15, I16) -> f2#(I16, I17) [-1 <= I16 - 1 /\ 0 <= I15 - 1]
R = 
  init(x1, x2) -> f1(rnd1, rnd2) 
  f3(I0, I1) -> f2(I0 + 3, I2) [I0 - 2 * y1 = 1 /\ 0 <= I0 - 1 /\ I0 - 2 * y1 <= 1 /\ 0 <= I0 - 2 * y1]
  f2(I3, I4) -> f3(I3, I5) [0 <= I3 - 1 /\ I3 - 2 * I6 = 1]
  f3(I7, I8) -> f2(I7 - 1, I9) [0 <= I7 - 1 /\ I7 - 2 * I10 = 0 /\ I7 - 2 * I10 <= 1 /\ 0 <= I7 - 2 * I10]
  f2(I11, I12) -> f3(I11, I13) [I11 - 2 * I14 = 0 /\ 0 <= I11 - 1]
  f1(I15, I16) -> f2(I16, I17) [-1 <= I16 - 1 /\ 0 <= I15 - 1]

The dependency graph for this problem is:
  0 -> 5
  1 -> 4
  2 -> 1
  3 -> 2
  4 -> 3
  5 -> 2, 4
Where:
  0) init#(x1, x2) -> f1#(rnd1, rnd2) 
  1) f3#(I0, I1) -> f2#(I0 + 3, I2) [I0 - 2 * y1 = 1 /\ 0 <= I0 - 1 /\ I0 - 2 * y1 <= 1 /\ 0 <= I0 - 2 * y1]
  2) f2#(I3, I4) -> f3#(I3, I5) [0 <= I3 - 1 /\ I3 - 2 * I6 = 1]
  3) f3#(I7, I8) -> f2#(I7 - 1, I9) [0 <= I7 - 1 /\ I7 - 2 * I10 = 0 /\ I7 - 2 * I10 <= 1 /\ 0 <= I7 - 2 * I10]
  4) f2#(I11, I12) -> f3#(I11, I13) [I11 - 2 * I14 = 0 /\ 0 <= I11 - 1]
  5) f1#(I15, I16) -> f2#(I16, I17) [-1 <= I16 - 1 /\ 0 <= I15 - 1]

We have the following SCCs.
  { 1, 2, 3, 4 }

DP problem for innermost termination.
P =
  f3#(I0, I1) -> f2#(I0 + 3, I2) [I0 - 2 * y1 = 1 /\ 0 <= I0 - 1 /\ I0 - 2 * y1 <= 1 /\ 0 <= I0 - 2 * y1]
  f2#(I3, I4) -> f3#(I3, I5) [0 <= I3 - 1 /\ I3 - 2 * I6 = 1]
  f3#(I7, I8) -> f2#(I7 - 1, I9) [0 <= I7 - 1 /\ I7 - 2 * I10 = 0 /\ I7 - 2 * I10 <= 1 /\ 0 <= I7 - 2 * I10]
  f2#(I11, I12) -> f3#(I11, I13) [I11 - 2 * I14 = 0 /\ 0 <= I11 - 1]
R = 
  init(x1, x2) -> f1(rnd1, rnd2) 
  f3(I0, I1) -> f2(I0 + 3, I2) [I0 - 2 * y1 = 1 /\ 0 <= I0 - 1 /\ I0 - 2 * y1 <= 1 /\ 0 <= I0 - 2 * y1]
  f2(I3, I4) -> f3(I3, I5) [0 <= I3 - 1 /\ I3 - 2 * I6 = 1]
  f3(I7, I8) -> f2(I7 - 1, I9) [0 <= I7 - 1 /\ I7 - 2 * I10 = 0 /\ I7 - 2 * I10 <= 1 /\ 0 <= I7 - 2 * I10]
  f2(I11, I12) -> f3(I11, I13) [I11 - 2 * I14 = 0 /\ 0 <= I11 - 1]
  f1(I15, I16) -> f2(I16, I17) [-1 <= I16 - 1 /\ 0 <= I15 - 1]