/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 =
  f6#(x1, x2, x3) -> f5#(x1, x2, x3) 
  f5#(I0, I1, I2) -> f4#(0, I1, I2) 
  f3#(I3, I4, I5) -> f1#(I3, I4, I5) 
  f4#(I9, I10, I11) -> f3#(I9, I10, rnd3) [rnd3 = rnd3 /\ I10 <= 127]
  f1#(I12, I13, I14) -> f3#(1 + I12, I13, 1 + I14) [1 + I12 <= 36]
R = 
  f6(x1, x2, x3) -> f5(x1, x2, x3) 
  f5(I0, I1, I2) -> f4(0, I1, I2) 
  f3(I3, I4, I5) -> f1(I3, I4, I5) 
  f4(I6, I7, I8) -> f2(I6, I7, I8) [128 <= I7]
  f4(I9, I10, I11) -> f3(I9, I10, rnd3) [rnd3 = rnd3 /\ I10 <= 127]
  f1(I12, I13, I14) -> f3(1 + I12, I13, 1 + I14) [1 + I12 <= 36]
  f1(I15, I16, I17) -> f2(I15, I16, I17) [36 <= I15]

The dependency graph for this problem is:
  0 -> 1
  1 -> 3
  2 -> 4
  3 -> 2
  4 -> 2
Where:
  0) f6#(x1, x2, x3) -> f5#(x1, x2, x3) 
  1) f5#(I0, I1, I2) -> f4#(0, I1, I2) 
  2) f3#(I3, I4, I5) -> f1#(I3, I4, I5) 
  3) f4#(I9, I10, I11) -> f3#(I9, I10, rnd3) [rnd3 = rnd3 /\ I10 <= 127]
  4) f1#(I12, I13, I14) -> f3#(1 + I12, I13, 1 + I14) [1 + I12 <= 36]

We have the following SCCs.
  { 2, 4 }

DP problem for innermost termination.
P =
  f3#(I3, I4, I5) -> f1#(I3, I4, I5) 
  f1#(I12, I13, I14) -> f3#(1 + I12, I13, 1 + I14) [1 + I12 <= 36]
R = 
  f6(x1, x2, x3) -> f5(x1, x2, x3) 
  f5(I0, I1, I2) -> f4(0, I1, I2) 
  f3(I3, I4, I5) -> f1(I3, I4, I5) 
  f4(I6, I7, I8) -> f2(I6, I7, I8) [128 <= I7]
  f4(I9, I10, I11) -> f3(I9, I10, rnd3) [rnd3 = rnd3 /\ I10 <= 127]
  f1(I12, I13, I14) -> f3(1 + I12, I13, 1 + I14) [1 + I12 <= 36]
  f1(I15, I16, I17) -> f2(I15, I16, I17) [36 <= I15]

We use the reverse value criterion with the projection function NU:
  NU[f1#(z1,z2,z3)] = 36 + -1 * (1 + z1)
  NU[f3#(z1,z2,z3)] = 36 + -1 * (1 + z1)

This gives the following inequalities:
    ==>  36 + -1 * (1 + I3) >= 36 + -1 * (1 + I3)
  1 + I12 <= 36  ==>  36 + -1 * (1 + I12) > 36 + -1 * (1 + (1 + I12))  with  36 + -1 * (1 + I12) >= 0

We remove all the strictly oriented dependency pairs.

DP problem for innermost termination.
P =
  f3#(I3, I4, I5) -> f1#(I3, I4, I5) 
R = 
  f6(x1, x2, x3) -> f5(x1, x2, x3) 
  f5(I0, I1, I2) -> f4(0, I1, I2) 
  f3(I3, I4, I5) -> f1(I3, I4, I5) 
  f4(I6, I7, I8) -> f2(I6, I7, I8) [128 <= I7]
  f4(I9, I10, I11) -> f3(I9, I10, rnd3) [rnd3 = rnd3 /\ I10 <= 127]
  f1(I12, I13, I14) -> f3(1 + I12, I13, 1 + I14) [1 + I12 <= 36]
  f1(I15, I16, I17) -> f2(I15, I16, I17) [36 <= I15]

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

We have the following SCCs.