/export/starexec/sandbox2/solver/bin/starexec_run_Transition /export/starexec/sandbox2/benchmark/theBenchmark.smt2 /export/starexec/sandbox2/output/output_files


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MAYBE

DP problem for innermost termination.
P =
  f4#(x1) -> f3#(x1) 
  f3#(I0) -> f1#(rnd1) [2 <= rnd1 /\ rnd1 = rnd1]
  f2#(I1) -> f1#(I1) 
  f1#(I2) -> f2#(I3) [y1 = y1 /\ 1 <= y1 /\ 2 <= y1 /\ I3 = I3 /\ -1 <= I3 /\ 1 + I3 <= 2]
R = 
  f4(x1) -> f3(x1) 
  f3(I0) -> f1(rnd1) [2 <= rnd1 /\ rnd1 = rnd1]
  f2(I1) -> f1(I1) 
  f1(I2) -> f2(I3) [y1 = y1 /\ 1 <= y1 /\ 2 <= y1 /\ I3 = I3 /\ -1 <= I3 /\ 1 + I3 <= 2]

The dependency graph for this problem is:
  0 -> 1
  1 -> 3
  2 -> 3
  3 -> 2
Where:
  0) f4#(x1) -> f3#(x1) 
  1) f3#(I0) -> f1#(rnd1) [2 <= rnd1 /\ rnd1 = rnd1]
  2) f2#(I1) -> f1#(I1) 
  3) f1#(I2) -> f2#(I3) [y1 = y1 /\ 1 <= y1 /\ 2 <= y1 /\ I3 = I3 /\ -1 <= I3 /\ 1 + I3 <= 2]

We have the following SCCs.
  { 2, 3 }

DP problem for innermost termination.
P =
  f2#(I1) -> f1#(I1) 
  f1#(I2) -> f2#(I3) [y1 = y1 /\ 1 <= y1 /\ 2 <= y1 /\ I3 = I3 /\ -1 <= I3 /\ 1 + I3 <= 2]
R = 
  f4(x1) -> f3(x1) 
  f3(I0) -> f1(rnd1) [2 <= rnd1 /\ rnd1 = rnd1]
  f2(I1) -> f1(I1) 
  f1(I2) -> f2(I3) [y1 = y1 /\ 1 <= y1 /\ 2 <= y1 /\ I3 = I3 /\ -1 <= I3 /\ 1 + I3 <= 2]