/export/starexec/sandbox/solver/bin/starexec_run_Transition /export/starexec/sandbox/benchmark/theBenchmark.smt2 /export/starexec/sandbox/output/output_files


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YES

DP problem for innermost termination.
P =
  f5#(x1, x2, x3) -> f4#(x1, x2, x3) 
  f4#(I0, I1, I2) -> f3#(I0, 0, rnd3) [y1 = 0 /\ rnd3 = rnd3]
  f3#(I3, I4, I5) -> f1#(I3, I4, I5) 
  f1#(I6, I7, I8) -> f3#(I6, 1 + I7, I8) [1 + I7 <= I6]
R = 
  f5(x1, x2, x3) -> f4(x1, x2, x3) 
  f4(I0, I1, I2) -> f3(I0, 0, rnd3) [y1 = 0 /\ rnd3 = rnd3]
  f3(I3, I4, I5) -> f1(I3, I4, I5) 
  f1(I6, I7, I8) -> f3(I6, 1 + I7, I8) [1 + I7 <= I6]
  f1(I9, I10, I11) -> f2(I9, I10, I11) [I9 <= I10]

The dependency graph for this problem is:
  0 -> 1
  1 -> 2
  2 -> 3
  3 -> 2
Where:
  0) f5#(x1, x2, x3) -> f4#(x1, x2, x3) 
  1) f4#(I0, I1, I2) -> f3#(I0, 0, rnd3) [y1 = 0 /\ rnd3 = rnd3]
  2) f3#(I3, I4, I5) -> f1#(I3, I4, I5) 
  3) f1#(I6, I7, I8) -> f3#(I6, 1 + I7, I8) [1 + I7 <= I6]

We have the following SCCs.
  { 2, 3 }

DP problem for innermost termination.
P =
  f3#(I3, I4, I5) -> f1#(I3, I4, I5) 
  f1#(I6, I7, I8) -> f3#(I6, 1 + I7, I8) [1 + I7 <= I6]
R = 
  f5(x1, x2, x3) -> f4(x1, x2, x3) 
  f4(I0, I1, I2) -> f3(I0, 0, rnd3) [y1 = 0 /\ rnd3 = rnd3]
  f3(I3, I4, I5) -> f1(I3, I4, I5) 
  f1(I6, I7, I8) -> f3(I6, 1 + I7, I8) [1 + I7 <= I6]
  f1(I9, I10, I11) -> f2(I9, I10, I11) [I9 <= I10]

We use the reverse value criterion with the projection function NU:
  NU[f1#(z1,z2,z3)] = z1 + -1 * (1 + z2)
  NU[f3#(z1,z2,z3)] = z1 + -1 * (1 + z2)

This gives the following inequalities:
    ==>  I3 + -1 * (1 + I4) >= I3 + -1 * (1 + I4)
  1 + I7 <= I6  ==>  I6 + -1 * (1 + I7) > I6 + -1 * (1 + (1 + I7))  with  I6 + -1 * (1 + I7) >= 0

We remove all the strictly oriented dependency pairs.

DP problem for innermost termination.
P =
  f3#(I3, I4, I5) -> f1#(I3, I4, I5) 
R = 
  f5(x1, x2, x3) -> f4(x1, x2, x3) 
  f4(I0, I1, I2) -> f3(I0, 0, rnd3) [y1 = 0 /\ rnd3 = rnd3]
  f3(I3, I4, I5) -> f1(I3, I4, I5) 
  f1(I6, I7, I8) -> f3(I6, 1 + I7, I8) [1 + I7 <= I6]
  f1(I9, I10, I11) -> f2(I9, I10, I11) [I9 <= I10]

The dependency graph for this problem is:
  2 -> 
Where:
  2) f3#(I3, I4, I5) -> f1#(I3, I4, I5) 

We have the following SCCs.