MAYBE

DP problem for innermost termination.
P =
  f4#(x1, x2, x3, x4, x5) -> f3#(x1, x2, x3, x4, x5) 
  f3#(I0, I1, I2, I3, I4) -> f2#(I0, I1, I2, I3, I0 + I4) [1 + I0 <= I0 + I4]
  f3#(I5, I6, I7, I8, I9) -> f1#(I5, I6, I7, I8, I6 + I9) [1 + I5 <= I6 + I9]
  f2#(I10, I11, I12, I13, I14) -> f1#(I10, I11, I12, I13, I13 + I14) [1 + I10 <= I13 + I14]
  f1#(I15, I16, I17, I18, I19) -> f2#(I15, I16, I17, I18, I17 + I19) [1 + I15 <= I17 + I19]
R = 
  f4(x1, x2, x3, x4, x5) -> f3(x1, x2, x3, x4, x5) 
  f3(I0, I1, I2, I3, I4) -> f2(I0, I1, I2, I3, I0 + I4) [1 + I0 <= I0 + I4]
  f3(I5, I6, I7, I8, I9) -> f1(I5, I6, I7, I8, I6 + I9) [1 + I5 <= I6 + I9]
  f2(I10, I11, I12, I13, I14) -> f1(I10, I11, I12, I13, I13 + I14) [1 + I10 <= I13 + I14]
  f1(I15, I16, I17, I18, I19) -> f2(I15, I16, I17, I18, I17 + I19) [1 + I15 <= I17 + I19]

The dependency graph for this problem is:
  0 -> 1, 2
  1 -> 3
  2 -> 4
  3 -> 4
  4 -> 3
Where:
  0) f4#(x1, x2, x3, x4, x5) -> f3#(x1, x2, x3, x4, x5) 
  1) f3#(I0, I1, I2, I3, I4) -> f2#(I0, I1, I2, I3, I0 + I4) [1 + I0 <= I0 + I4]
  2) f3#(I5, I6, I7, I8, I9) -> f1#(I5, I6, I7, I8, I6 + I9) [1 + I5 <= I6 + I9]
  3) f2#(I10, I11, I12, I13, I14) -> f1#(I10, I11, I12, I13, I13 + I14) [1 + I10 <= I13 + I14]
  4) f1#(I15, I16, I17, I18, I19) -> f2#(I15, I16, I17, I18, I17 + I19) [1 + I15 <= I17 + I19]

We have the following SCCs.
  { 3, 4 }

DP problem for innermost termination.
P =
  f2#(I10, I11, I12, I13, I14) -> f1#(I10, I11, I12, I13, I13 + I14) [1 + I10 <= I13 + I14]
  f1#(I15, I16, I17, I18, I19) -> f2#(I15, I16, I17, I18, I17 + I19) [1 + I15 <= I17 + I19]
R = 
  f4(x1, x2, x3, x4, x5) -> f3(x1, x2, x3, x4, x5) 
  f3(I0, I1, I2, I3, I4) -> f2(I0, I1, I2, I3, I0 + I4) [1 + I0 <= I0 + I4]
  f3(I5, I6, I7, I8, I9) -> f1(I5, I6, I7, I8, I6 + I9) [1 + I5 <= I6 + I9]
  f2(I10, I11, I12, I13, I14) -> f1(I10, I11, I12, I13, I13 + I14) [1 + I10 <= I13 + I14]
  f1(I15, I16, I17, I18, I19) -> f2(I15, I16, I17, I18, I17 + I19) [1 + I15 <= I17 + I19]