/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 =
  init#(x1, x2, x3) -> f1#(rnd1, rnd2, rnd3) 
  f2#(I0, I1, I2) -> f2#(I0, 1, 1) [0 = I2 /\ I0 <= 1]
  f2#(I3, I4, I5) -> f2#(I3, 1, I5 + 1) [1 = I4 /\ I3 <= 1 /\ I5 <= 9 /\ 0 <= I5 - 1 /\ I5 <= 10]
  f2#(I6, I7, I8) -> f2#(I6, 0, 9) [10 = I8 /\ I6 <= 1]
  f2#(I9, I10, I11) -> f2#(I9, 0, I11 - 1) [0 = I10 /\ 0 <= I11 - 1 /\ I9 <= 1 /\ I11 <= 10 /\ I11 <= 9]
  f2#(I12, I13, I14) -> f2#(1, 1, 1) [0 = I14 /\ I12 <= 1]
  f2#(I15, I16, I17) -> f2#(1, 1, I17 + 1) [1 = I16 /\ I15 <= 1 /\ I17 <= 9 /\ 0 <= I17 - 1 /\ I17 <= 10]
  f2#(I18, I19, I20) -> f2#(0, 0, 9) [10 = I20 /\ I18 <= 1]
  f2#(I21, I22, I23) -> f2#(0, 0, I23 - 1) [0 = I22 /\ 0 <= I23 - 1 /\ I21 <= 1 /\ I23 <= 10 /\ I23 <= 9]
  f1#(I24, I25, I26) -> f2#(0, 0, I25) [-1 <= I25 - 1 /\ 0 <= I24 - 1]
R = 
  init(x1, x2, x3) -> f1(rnd1, rnd2, rnd3) 
  f2(I0, I1, I2) -> f2(I0, 1, 1) [0 = I2 /\ I0 <= 1]
  f2(I3, I4, I5) -> f2(I3, 1, I5 + 1) [1 = I4 /\ I3 <= 1 /\ I5 <= 9 /\ 0 <= I5 - 1 /\ I5 <= 10]
  f2(I6, I7, I8) -> f2(I6, 0, 9) [10 = I8 /\ I6 <= 1]
  f2(I9, I10, I11) -> f2(I9, 0, I11 - 1) [0 = I10 /\ 0 <= I11 - 1 /\ I9 <= 1 /\ I11 <= 10 /\ I11 <= 9]
  f2(I12, I13, I14) -> f2(1, 1, 1) [0 = I14 /\ I12 <= 1]
  f2(I15, I16, I17) -> f2(1, 1, I17 + 1) [1 = I16 /\ I15 <= 1 /\ I17 <= 9 /\ 0 <= I17 - 1 /\ I17 <= 10]
  f2(I18, I19, I20) -> f2(0, 0, 9) [10 = I20 /\ I18 <= 1]
  f2(I21, I22, I23) -> f2(0, 0, I23 - 1) [0 = I22 /\ 0 <= I23 - 1 /\ I21 <= 1 /\ I23 <= 10 /\ I23 <= 9]
  f1(I24, I25, I26) -> f2(0, 0, I25) [-1 <= I25 - 1 /\ 0 <= I24 - 1]

The dependency graph for this problem is:
  0 -> 9
  1 -> 2, 6
  2 -> 2, 3, 6, 7
  3 -> 4, 8
  4 -> 1, 4, 5, 8
  5 -> 2, 6
  6 -> 2, 3, 6, 7
  7 -> 4, 8
  8 -> 1, 4, 5, 8
  9 -> 1, 3, 4, 5, 7, 8
Where:
  0) init#(x1, x2, x3) -> f1#(rnd1, rnd2, rnd3) 
  1) f2#(I0, I1, I2) -> f2#(I0, 1, 1) [0 = I2 /\ I0 <= 1]
  2) f2#(I3, I4, I5) -> f2#(I3, 1, I5 + 1) [1 = I4 /\ I3 <= 1 /\ I5 <= 9 /\ 0 <= I5 - 1 /\ I5 <= 10]
  3) f2#(I6, I7, I8) -> f2#(I6, 0, 9) [10 = I8 /\ I6 <= 1]
  4) f2#(I9, I10, I11) -> f2#(I9, 0, I11 - 1) [0 = I10 /\ 0 <= I11 - 1 /\ I9 <= 1 /\ I11 <= 10 /\ I11 <= 9]
  5) f2#(I12, I13, I14) -> f2#(1, 1, 1) [0 = I14 /\ I12 <= 1]
  6) f2#(I15, I16, I17) -> f2#(1, 1, I17 + 1) [1 = I16 /\ I15 <= 1 /\ I17 <= 9 /\ 0 <= I17 - 1 /\ I17 <= 10]
  7) f2#(I18, I19, I20) -> f2#(0, 0, 9) [10 = I20 /\ I18 <= 1]
  8) f2#(I21, I22, I23) -> f2#(0, 0, I23 - 1) [0 = I22 /\ 0 <= I23 - 1 /\ I21 <= 1 /\ I23 <= 10 /\ I23 <= 9]
  9) f1#(I24, I25, I26) -> f2#(0, 0, I25) [-1 <= I25 - 1 /\ 0 <= I24 - 1]

We have the following SCCs.
  { 1, 2, 3, 4, 5, 6, 7, 8 }

DP problem for innermost termination.
P =
  f2#(I0, I1, I2) -> f2#(I0, 1, 1) [0 = I2 /\ I0 <= 1]
  f2#(I3, I4, I5) -> f2#(I3, 1, I5 + 1) [1 = I4 /\ I3 <= 1 /\ I5 <= 9 /\ 0 <= I5 - 1 /\ I5 <= 10]
  f2#(I6, I7, I8) -> f2#(I6, 0, 9) [10 = I8 /\ I6 <= 1]
  f2#(I9, I10, I11) -> f2#(I9, 0, I11 - 1) [0 = I10 /\ 0 <= I11 - 1 /\ I9 <= 1 /\ I11 <= 10 /\ I11 <= 9]
  f2#(I12, I13, I14) -> f2#(1, 1, 1) [0 = I14 /\ I12 <= 1]
  f2#(I15, I16, I17) -> f2#(1, 1, I17 + 1) [1 = I16 /\ I15 <= 1 /\ I17 <= 9 /\ 0 <= I17 - 1 /\ I17 <= 10]
  f2#(I18, I19, I20) -> f2#(0, 0, 9) [10 = I20 /\ I18 <= 1]
  f2#(I21, I22, I23) -> f2#(0, 0, I23 - 1) [0 = I22 /\ 0 <= I23 - 1 /\ I21 <= 1 /\ I23 <= 10 /\ I23 <= 9]
R = 
  init(x1, x2, x3) -> f1(rnd1, rnd2, rnd3) 
  f2(I0, I1, I2) -> f2(I0, 1, 1) [0 = I2 /\ I0 <= 1]
  f2(I3, I4, I5) -> f2(I3, 1, I5 + 1) [1 = I4 /\ I3 <= 1 /\ I5 <= 9 /\ 0 <= I5 - 1 /\ I5 <= 10]
  f2(I6, I7, I8) -> f2(I6, 0, 9) [10 = I8 /\ I6 <= 1]
  f2(I9, I10, I11) -> f2(I9, 0, I11 - 1) [0 = I10 /\ 0 <= I11 - 1 /\ I9 <= 1 /\ I11 <= 10 /\ I11 <= 9]
  f2(I12, I13, I14) -> f2(1, 1, 1) [0 = I14 /\ I12 <= 1]
  f2(I15, I16, I17) -> f2(1, 1, I17 + 1) [1 = I16 /\ I15 <= 1 /\ I17 <= 9 /\ 0 <= I17 - 1 /\ I17 <= 10]
  f2(I18, I19, I20) -> f2(0, 0, 9) [10 = I20 /\ I18 <= 1]
  f2(I21, I22, I23) -> f2(0, 0, I23 - 1) [0 = I22 /\ 0 <= I23 - 1 /\ I21 <= 1 /\ I23 <= 10 /\ I23 <= 9]
  f1(I24, I25, I26) -> f2(0, 0, I25) [-1 <= I25 - 1 /\ 0 <= I24 - 1]