/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) -> f1#(rnd1, rnd2) 
  f3#(I0, I1) -> f2#(3 * I0 + 1, I2) [I0 - 2 * y1 = 1 /\ 1 <= I0 - 1 /\ 0 <= 3 * I0 - 1 /\ I0 - 2 * y1 <= 1 /\ 0 <= I0 - 2 * y1]
  f2#(I3, I4) -> f3#(I3, I5) [I3 - 2 * I6 = 1 /\ 0 <= 3 * I3 - 1 /\ 1 <= I3 - 1]
  f3#(I7, I8) -> f2#(I9, I10) [1 <= I7 - 1 /\ I7 - 2 * I11 = 0 /\ I9 <= I7 - 1 /\ 0 <= I7 - 2 * I9 /\ I7 - 2 * I9 <= 1 /\ I7 - 2 * I11 <= 1 /\ 0 <= I7 - 2 * I11]
  f2#(I12, I13) -> f3#(I12, I14) [1 <= I12 - 1 /\ I15 <= I12 - 1 /\ I12 - 2 * y2 = 0]
  f1#(I16, I17) -> f2#(I17, I18) [-1 <= I17 - 1 /\ 0 <= I16 - 1]
R = 
  init(x1, x2) -> f1(rnd1, rnd2) 
  f3(I0, I1) -> f2(3 * I0 + 1, I2) [I0 - 2 * y1 = 1 /\ 1 <= I0 - 1 /\ 0 <= 3 * I0 - 1 /\ I0 - 2 * y1 <= 1 /\ 0 <= I0 - 2 * y1]
  f2(I3, I4) -> f3(I3, I5) [I3 - 2 * I6 = 1 /\ 0 <= 3 * I3 - 1 /\ 1 <= I3 - 1]
  f3(I7, I8) -> f2(I9, I10) [1 <= I7 - 1 /\ I7 - 2 * I11 = 0 /\ I9 <= I7 - 1 /\ 0 <= I7 - 2 * I9 /\ I7 - 2 * I9 <= 1 /\ I7 - 2 * I11 <= 1 /\ 0 <= I7 - 2 * I11]
  f2(I12, I13) -> f3(I12, I14) [1 <= I12 - 1 /\ I15 <= I12 - 1 /\ I12 - 2 * y2 = 0]
  f1(I16, I17) -> f2(I17, I18) [-1 <= I17 - 1 /\ 0 <= I16 - 1]

The dependency graph for this problem is:
  0 -> 5
  1 -> 4
  2 -> 1
  3 -> 2, 4
  4 -> 3
  5 -> 2, 4
Where:
  0) init#(x1, x2) -> f1#(rnd1, rnd2) 
  1) f3#(I0, I1) -> f2#(3 * I0 + 1, I2) [I0 - 2 * y1 = 1 /\ 1 <= I0 - 1 /\ 0 <= 3 * I0 - 1 /\ I0 - 2 * y1 <= 1 /\ 0 <= I0 - 2 * y1]
  2) f2#(I3, I4) -> f3#(I3, I5) [I3 - 2 * I6 = 1 /\ 0 <= 3 * I3 - 1 /\ 1 <= I3 - 1]
  3) f3#(I7, I8) -> f2#(I9, I10) [1 <= I7 - 1 /\ I7 - 2 * I11 = 0 /\ I9 <= I7 - 1 /\ 0 <= I7 - 2 * I9 /\ I7 - 2 * I9 <= 1 /\ I7 - 2 * I11 <= 1 /\ 0 <= I7 - 2 * I11]
  4) f2#(I12, I13) -> f3#(I12, I14) [1 <= I12 - 1 /\ I15 <= I12 - 1 /\ I12 - 2 * y2 = 0]
  5) f1#(I16, I17) -> f2#(I17, I18) [-1 <= I17 - 1 /\ 0 <= I16 - 1]

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

DP problem for innermost termination.
P =
  f3#(I0, I1) -> f2#(3 * I0 + 1, I2) [I0 - 2 * y1 = 1 /\ 1 <= I0 - 1 /\ 0 <= 3 * I0 - 1 /\ I0 - 2 * y1 <= 1 /\ 0 <= I0 - 2 * y1]
  f2#(I3, I4) -> f3#(I3, I5) [I3 - 2 * I6 = 1 /\ 0 <= 3 * I3 - 1 /\ 1 <= I3 - 1]
  f3#(I7, I8) -> f2#(I9, I10) [1 <= I7 - 1 /\ I7 - 2 * I11 = 0 /\ I9 <= I7 - 1 /\ 0 <= I7 - 2 * I9 /\ I7 - 2 * I9 <= 1 /\ I7 - 2 * I11 <= 1 /\ 0 <= I7 - 2 * I11]
  f2#(I12, I13) -> f3#(I12, I14) [1 <= I12 - 1 /\ I15 <= I12 - 1 /\ I12 - 2 * y2 = 0]
R = 
  init(x1, x2) -> f1(rnd1, rnd2) 
  f3(I0, I1) -> f2(3 * I0 + 1, I2) [I0 - 2 * y1 = 1 /\ 1 <= I0 - 1 /\ 0 <= 3 * I0 - 1 /\ I0 - 2 * y1 <= 1 /\ 0 <= I0 - 2 * y1]
  f2(I3, I4) -> f3(I3, I5) [I3 - 2 * I6 = 1 /\ 0 <= 3 * I3 - 1 /\ 1 <= I3 - 1]
  f3(I7, I8) -> f2(I9, I10) [1 <= I7 - 1 /\ I7 - 2 * I11 = 0 /\ I9 <= I7 - 1 /\ 0 <= I7 - 2 * I9 /\ I7 - 2 * I9 <= 1 /\ I7 - 2 * I11 <= 1 /\ 0 <= I7 - 2 * I11]
  f2(I12, I13) -> f3(I12, I14) [1 <= I12 - 1 /\ I15 <= I12 - 1 /\ I12 - 2 * y2 = 0]
  f1(I16, I17) -> f2(I17, I18) [-1 <= I17 - 1 /\ 0 <= I16 - 1]