/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, x3, x4) -> f1#(rnd1, rnd2, rnd3, rnd4) 
  f2#(I0, I1, I2, I3) -> f2#(I4, I5, I2, I3 + 1) [0 <= I1 - 1 /\ 0 <= I2 - 1 /\ -1 <= I3 - 1 /\ I3 <= I2 - 1 /\ I3 + 1 <= I2 /\ -1 <= y1 - 1 /\ I0 - 1 - y1 = I4 /\ I0 - 1 - y1 = I5]
  f2#(I6, I7, I8, I9) -> f2#(I6 - 1, I6 - 1, I8, I9) [0 <= I7 - 1 /\ I8 <= I9 /\ I6 - 1 <= I6 - 1 /\ -1 <= I8 - 1]
  f1#(I10, I11, I12, I13) -> f2#(I14, I15, I11, 2) [0 <= I10 - 1 /\ -1 <= I14 - 1 /\ 1 <= I11 - 1 /\ -1 <= I15 - 1]
  f1#(I16, I17, I18, I19) -> f2#(I20, 0, 1, 1) [1 = I17 /\ -1 <= I20 - 1 /\ 0 <= I16 - 1]
  f1#(I21, I22, I23, I24) -> f2#(0, 0, 0, 0) [0 = I22 /\ 0 <= I21 - 1]
R = 
  init(x1, x2, x3, x4) -> f1(rnd1, rnd2, rnd3, rnd4) 
  f2(I0, I1, I2, I3) -> f2(I4, I5, I2, I3 + 1) [0 <= I1 - 1 /\ 0 <= I2 - 1 /\ -1 <= I3 - 1 /\ I3 <= I2 - 1 /\ I3 + 1 <= I2 /\ -1 <= y1 - 1 /\ I0 - 1 - y1 = I4 /\ I0 - 1 - y1 = I5]
  f2(I6, I7, I8, I9) -> f2(I6 - 1, I6 - 1, I8, I9) [0 <= I7 - 1 /\ I8 <= I9 /\ I6 - 1 <= I6 - 1 /\ -1 <= I8 - 1]
  f1(I10, I11, I12, I13) -> f2(I14, I15, I11, 2) [0 <= I10 - 1 /\ -1 <= I14 - 1 /\ 1 <= I11 - 1 /\ -1 <= I15 - 1]
  f1(I16, I17, I18, I19) -> f2(I20, 0, 1, 1) [1 = I17 /\ -1 <= I20 - 1 /\ 0 <= I16 - 1]
  f1(I21, I22, I23, I24) -> f2(0, 0, 0, 0) [0 = I22 /\ 0 <= I21 - 1]

The dependency graph for this problem is:
  0 -> 3, 4, 5
  1 -> 1, 2
  2 -> 2
  3 -> 1, 2
  4 -> 
  5 -> 
Where:
  0) init#(x1, x2, x3, x4) -> f1#(rnd1, rnd2, rnd3, rnd4) 
  1) f2#(I0, I1, I2, I3) -> f2#(I4, I5, I2, I3 + 1) [0 <= I1 - 1 /\ 0 <= I2 - 1 /\ -1 <= I3 - 1 /\ I3 <= I2 - 1 /\ I3 + 1 <= I2 /\ -1 <= y1 - 1 /\ I0 - 1 - y1 = I4 /\ I0 - 1 - y1 = I5]
  2) f2#(I6, I7, I8, I9) -> f2#(I6 - 1, I6 - 1, I8, I9) [0 <= I7 - 1 /\ I8 <= I9 /\ I6 - 1 <= I6 - 1 /\ -1 <= I8 - 1]
  3) f1#(I10, I11, I12, I13) -> f2#(I14, I15, I11, 2) [0 <= I10 - 1 /\ -1 <= I14 - 1 /\ 1 <= I11 - 1 /\ -1 <= I15 - 1]
  4) f1#(I16, I17, I18, I19) -> f2#(I20, 0, 1, 1) [1 = I17 /\ -1 <= I20 - 1 /\ 0 <= I16 - 1]
  5) f1#(I21, I22, I23, I24) -> f2#(0, 0, 0, 0) [0 = I22 /\ 0 <= I21 - 1]

We have the following SCCs.
  { 1 }
  { 2 }

DP problem for innermost termination.
P =
  f2#(I6, I7, I8, I9) -> f2#(I6 - 1, I6 - 1, I8, I9) [0 <= I7 - 1 /\ I8 <= I9 /\ I6 - 1 <= I6 - 1 /\ -1 <= I8 - 1]
R = 
  init(x1, x2, x3, x4) -> f1(rnd1, rnd2, rnd3, rnd4) 
  f2(I0, I1, I2, I3) -> f2(I4, I5, I2, I3 + 1) [0 <= I1 - 1 /\ 0 <= I2 - 1 /\ -1 <= I3 - 1 /\ I3 <= I2 - 1 /\ I3 + 1 <= I2 /\ -1 <= y1 - 1 /\ I0 - 1 - y1 = I4 /\ I0 - 1 - y1 = I5]
  f2(I6, I7, I8, I9) -> f2(I6 - 1, I6 - 1, I8, I9) [0 <= I7 - 1 /\ I8 <= I9 /\ I6 - 1 <= I6 - 1 /\ -1 <= I8 - 1]
  f1(I10, I11, I12, I13) -> f2(I14, I15, I11, 2) [0 <= I10 - 1 /\ -1 <= I14 - 1 /\ 1 <= I11 - 1 /\ -1 <= I15 - 1]
  f1(I16, I17, I18, I19) -> f2(I20, 0, 1, 1) [1 = I17 /\ -1 <= I20 - 1 /\ 0 <= I16 - 1]
  f1(I21, I22, I23, I24) -> f2(0, 0, 0, 0) [0 = I22 /\ 0 <= I21 - 1]