/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) 
  f4#(I0, I1) -> f4#(I0 - 2, I2) [I0 <= 2 /\ I0 <= -1]
  f3#(I3, I4) -> f3#(I3 - 2, I5) [1 <= I3 - 1]
  f2#(I6, I7) -> f4#(I8, I9) [-1 <= y1 - 1 /\ 1 <= I7 - 1 /\ -1 <= y2 - 1 /\ y2 - 2 * y3 = 0 /\ 0 <= I6 - 1 /\ y2 - 2 * y3 <= 1 /\ 0 <= y2 - 2 * y3 /\ 0 - y1 = I8]
  f1#(I10, I11) -> f2#(I10, I11) [-1 <= I12 - 1 /\ 1 <= I11 - 1 /\ -1 <= I13 - 1 /\ I13 - 2 * I14 = 0 /\ 0 <= I10 - 1]
  f3#(I15, I16) -> f4#(0, I17) [0 = I15]
  f2#(I18, I19) -> f3#(I20, I21) [-1 <= I20 - 1 /\ 1 <= I19 - 1 /\ I22 - 2 * I23 = 1 /\ -1 <= I22 - 1 /\ 0 <= I18 - 1 /\ I22 - 2 * I23 <= 1 /\ 0 <= I22 - 2 * I23]
  f1#(I24, I25) -> f2#(I24, I25) [-1 <= I26 - 1 /\ 1 <= I25 - 1 /\ I27 - 2 * I28 = 1 /\ -1 <= I27 - 1 /\ 0 <= I24 - 1]
R = 
  init(x1, x2) -> f1(rnd1, rnd2) 
  f4(I0, I1) -> f4(I0 - 2, I2) [I0 <= 2 /\ I0 <= -1]
  f3(I3, I4) -> f3(I3 - 2, I5) [1 <= I3 - 1]
  f2(I6, I7) -> f4(I8, I9) [-1 <= y1 - 1 /\ 1 <= I7 - 1 /\ -1 <= y2 - 1 /\ y2 - 2 * y3 = 0 /\ 0 <= I6 - 1 /\ y2 - 2 * y3 <= 1 /\ 0 <= y2 - 2 * y3 /\ 0 - y1 = I8]
  f1(I10, I11) -> f2(I10, I11) [-1 <= I12 - 1 /\ 1 <= I11 - 1 /\ -1 <= I13 - 1 /\ I13 - 2 * I14 = 0 /\ 0 <= I10 - 1]
  f3(I15, I16) -> f4(0, I17) [0 = I15]
  f2(I18, I19) -> f3(I20, I21) [-1 <= I20 - 1 /\ 1 <= I19 - 1 /\ I22 - 2 * I23 = 1 /\ -1 <= I22 - 1 /\ 0 <= I18 - 1 /\ I22 - 2 * I23 <= 1 /\ 0 <= I22 - 2 * I23]
  f1(I24, I25) -> f2(I24, I25) [-1 <= I26 - 1 /\ 1 <= I25 - 1 /\ I27 - 2 * I28 = 1 /\ -1 <= I27 - 1 /\ 0 <= I24 - 1]

The dependency graph for this problem is:
  0 -> 4, 7
  1 -> 1
  2 -> 2, 5
  3 -> 1
  4 -> 3, 6
  5 -> 
  6 -> 2, 5
  7 -> 3, 6
Where:
  0) init#(x1, x2) -> f1#(rnd1, rnd2) 
  1) f4#(I0, I1) -> f4#(I0 - 2, I2) [I0 <= 2 /\ I0 <= -1]
  2) f3#(I3, I4) -> f3#(I3 - 2, I5) [1 <= I3 - 1]
  3) f2#(I6, I7) -> f4#(I8, I9) [-1 <= y1 - 1 /\ 1 <= I7 - 1 /\ -1 <= y2 - 1 /\ y2 - 2 * y3 = 0 /\ 0 <= I6 - 1 /\ y2 - 2 * y3 <= 1 /\ 0 <= y2 - 2 * y3 /\ 0 - y1 = I8]
  4) f1#(I10, I11) -> f2#(I10, I11) [-1 <= I12 - 1 /\ 1 <= I11 - 1 /\ -1 <= I13 - 1 /\ I13 - 2 * I14 = 0 /\ 0 <= I10 - 1]
  5) f3#(I15, I16) -> f4#(0, I17) [0 = I15]
  6) f2#(I18, I19) -> f3#(I20, I21) [-1 <= I20 - 1 /\ 1 <= I19 - 1 /\ I22 - 2 * I23 = 1 /\ -1 <= I22 - 1 /\ 0 <= I18 - 1 /\ I22 - 2 * I23 <= 1 /\ 0 <= I22 - 2 * I23]
  7) f1#(I24, I25) -> f2#(I24, I25) [-1 <= I26 - 1 /\ 1 <= I25 - 1 /\ I27 - 2 * I28 = 1 /\ -1 <= I27 - 1 /\ 0 <= I24 - 1]

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

DP problem for innermost termination.
P =
  f4#(I0, I1) -> f4#(I0 - 2, I2) [I0 <= 2 /\ I0 <= -1]
R = 
  init(x1, x2) -> f1(rnd1, rnd2) 
  f4(I0, I1) -> f4(I0 - 2, I2) [I0 <= 2 /\ I0 <= -1]
  f3(I3, I4) -> f3(I3 - 2, I5) [1 <= I3 - 1]
  f2(I6, I7) -> f4(I8, I9) [-1 <= y1 - 1 /\ 1 <= I7 - 1 /\ -1 <= y2 - 1 /\ y2 - 2 * y3 = 0 /\ 0 <= I6 - 1 /\ y2 - 2 * y3 <= 1 /\ 0 <= y2 - 2 * y3 /\ 0 - y1 = I8]
  f1(I10, I11) -> f2(I10, I11) [-1 <= I12 - 1 /\ 1 <= I11 - 1 /\ -1 <= I13 - 1 /\ I13 - 2 * I14 = 0 /\ 0 <= I10 - 1]
  f3(I15, I16) -> f4(0, I17) [0 = I15]
  f2(I18, I19) -> f3(I20, I21) [-1 <= I20 - 1 /\ 1 <= I19 - 1 /\ I22 - 2 * I23 = 1 /\ -1 <= I22 - 1 /\ 0 <= I18 - 1 /\ I22 - 2 * I23 <= 1 /\ 0 <= I22 - 2 * I23]
  f1(I24, I25) -> f2(I24, I25) [-1 <= I26 - 1 /\ 1 <= I25 - 1 /\ I27 - 2 * I28 = 1 /\ -1 <= I27 - 1 /\ 0 <= I24 - 1]