/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, x4, x5) -> f4#(x1, x2, x3, x4, x5) 
  f4#(I0, I1, I2, I3, I4) -> f3#(0, 1, rnd3, rnd4, rnd5) [rnd5 = rnd5 /\ rnd4 = rnd4 /\ rnd3 = rnd3]
  f3#(I5, I6, I7, I8, I9) -> f1#(I5, I6, I7, I8, I9) 
  f3#(I10, I11, I12, I13, I14) -> f1#(I10, I11, I12, I13, I14) 
R = 
  f5(x1, x2, x3, x4, x5) -> f4(x1, x2, x3, x4, x5) 
  f4(I0, I1, I2, I3, I4) -> f3(0, 1, rnd3, rnd4, rnd5) [rnd5 = rnd5 /\ rnd4 = rnd4 /\ rnd3 = rnd3]
  f3(I5, I6, I7, I8, I9) -> f1(I5, I6, I7, I8, I9) 
  f3(I10, I11, I12, I13, I14) -> f1(I10, I11, I12, I13, I14) 
  f1(I15, I16, I17, I18, I19) -> f2(I15, I16, I17, I18, I19) 

The dependency graph for this problem is:
  0 -> 1
  1 -> 2, 3
  2 -> 
  3 -> 
Where:
  0) f5#(x1, x2, x3, x4, x5) -> f4#(x1, x2, x3, x4, x5) 
  1) f4#(I0, I1, I2, I3, I4) -> f3#(0, 1, rnd3, rnd4, rnd5) [rnd5 = rnd5 /\ rnd4 = rnd4 /\ rnd3 = rnd3]
  2) f3#(I5, I6, I7, I8, I9) -> f1#(I5, I6, I7, I8, I9) 
  3) f3#(I10, I11, I12, I13, I14) -> f1#(I10, I11, I12, I13, I14) 

We have the following SCCs.