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


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MAYBE

DP problem for innermost termination.
P =
  init#(x1, x2) -> f1#(rnd1, rnd2) 
  f2#(I0, I1) -> f2#(1, I1) [0 = I0 /\ 0 <= I1 - 1]
  f2#(I2, I3) -> f2#(0, I2 - 1) [I2 = I3 /\ 0 <= I2 - 1]
  f2#(I4, I5) -> f2#(I4 + 1, I5) [0 <= I4 - 1 /\ 0 <= I5 - 1 /\ I4 <= I5 - 1]
  f2#(I6, I7) -> f2#(0, 0) [0 = I7 /\ 0 = I6]
  f1#(I8, I9) -> f2#(I9, 20) [-1 <= I9 - 1 /\ 0 <= I8 - 1]
R = 
  init(x1, x2) -> f1(rnd1, rnd2) 
  f2(I0, I1) -> f2(1, I1) [0 = I0 /\ 0 <= I1 - 1]
  f2(I2, I3) -> f2(0, I2 - 1) [I2 = I3 /\ 0 <= I2 - 1]
  f2(I4, I5) -> f2(I4 + 1, I5) [0 <= I4 - 1 /\ 0 <= I5 - 1 /\ I4 <= I5 - 1]
  f2(I6, I7) -> f2(0, 0) [0 = I7 /\ 0 = I6]
  f1(I8, I9) -> f2(I9, 20) [-1 <= I9 - 1 /\ 0 <= I8 - 1]

The dependency graph for this problem is:
  0 -> 5
  1 -> 2, 3
  2 -> 1, 4
  3 -> 2, 3
  4 -> 4
  5 -> 1, 2, 3
Where:
  0) init#(x1, x2) -> f1#(rnd1, rnd2) 
  1) f2#(I0, I1) -> f2#(1, I1) [0 = I0 /\ 0 <= I1 - 1]
  2) f2#(I2, I3) -> f2#(0, I2 - 1) [I2 = I3 /\ 0 <= I2 - 1]
  3) f2#(I4, I5) -> f2#(I4 + 1, I5) [0 <= I4 - 1 /\ 0 <= I5 - 1 /\ I4 <= I5 - 1]
  4) f2#(I6, I7) -> f2#(0, 0) [0 = I7 /\ 0 = I6]
  5) f1#(I8, I9) -> f2#(I9, 20) [-1 <= I9 - 1 /\ 0 <= I8 - 1]

We have the following SCCs.
  { 1, 2, 3 }
  { 4 }

DP problem for innermost termination.
P =
  f2#(I6, I7) -> f2#(0, 0) [0 = I7 /\ 0 = I6]
R = 
  init(x1, x2) -> f1(rnd1, rnd2) 
  f2(I0, I1) -> f2(1, I1) [0 = I0 /\ 0 <= I1 - 1]
  f2(I2, I3) -> f2(0, I2 - 1) [I2 = I3 /\ 0 <= I2 - 1]
  f2(I4, I5) -> f2(I4 + 1, I5) [0 <= I4 - 1 /\ 0 <= I5 - 1 /\ I4 <= I5 - 1]
  f2(I6, I7) -> f2(0, 0) [0 = I7 /\ 0 = I6]
  f1(I8, I9) -> f2(I9, 20) [-1 <= I9 - 1 /\ 0 <= I8 - 1]