/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) -> f3#(rnd1, rnd2) 
  f4#(I0, I1) -> f4#(I0 - 1, I2) [0 <= I0 - 1]
  f3#(I3, I4) -> f4#(I5, I6) [0 <= I3 - 1 /\ 0 <= I4 - 1 /\ -1 <= I5 - 1]
  f2#(I7, I8) -> f2#(0, I9) [2 = I7 /\ 4 <= I9 - 1 /\ 0 <= I8 - 1 /\ I9 - 4 <= I8]
  f2#(I10, I11) -> f2#(2, I12) [1 = I10 /\ 2 <= I12 - 1 /\ 0 <= I11 - 1 /\ I12 - 2 <= I11]
  f2#(I13, I14) -> f2#(1, I15) [0 = I13 /\ -1 <= I15 - 1 /\ 6 <= I14 - 1 /\ I15 + 7 <= I14]
  f3#(I16, I17) -> f2#(0, I18) [-1 <= y1 - 1 /\ 0 <= I17 - 1 /\ 0 <= I16 - 1 /\ 0 <= I18 - 1]
  f1#(I19, I20) -> f2#(0, I21) [I20 + 2 <= I19 /\ 0 <= I21 - 1 /\ 0 <= I19 - 1 /\ I21 <= I19]
R = 
  init(x1, x2) -> f3(rnd1, rnd2) 
  f4(I0, I1) -> f4(I0 - 1, I2) [0 <= I0 - 1]
  f3(I3, I4) -> f4(I5, I6) [0 <= I3 - 1 /\ 0 <= I4 - 1 /\ -1 <= I5 - 1]
  f2(I7, I8) -> f2(0, I9) [2 = I7 /\ 4 <= I9 - 1 /\ 0 <= I8 - 1 /\ I9 - 4 <= I8]
  f2(I10, I11) -> f2(2, I12) [1 = I10 /\ 2 <= I12 - 1 /\ 0 <= I11 - 1 /\ I12 - 2 <= I11]
  f2(I13, I14) -> f2(1, I15) [0 = I13 /\ -1 <= I15 - 1 /\ 6 <= I14 - 1 /\ I15 + 7 <= I14]
  f3(I16, I17) -> f2(0, I18) [-1 <= y1 - 1 /\ 0 <= I17 - 1 /\ 0 <= I16 - 1 /\ 0 <= I18 - 1]
  f1(I19, I20) -> f2(0, I21) [I20 + 2 <= I19 /\ 0 <= I21 - 1 /\ 0 <= I19 - 1 /\ I21 <= I19]

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

We have the following SCCs.
  { 1 }
  { 3, 4, 5 }

DP problem for innermost termination.
P =
  f2#(I7, I8) -> f2#(0, I9) [2 = I7 /\ 4 <= I9 - 1 /\ 0 <= I8 - 1 /\ I9 - 4 <= I8]
  f2#(I10, I11) -> f2#(2, I12) [1 = I10 /\ 2 <= I12 - 1 /\ 0 <= I11 - 1 /\ I12 - 2 <= I11]
  f2#(I13, I14) -> f2#(1, I15) [0 = I13 /\ -1 <= I15 - 1 /\ 6 <= I14 - 1 /\ I15 + 7 <= I14]
R = 
  init(x1, x2) -> f3(rnd1, rnd2) 
  f4(I0, I1) -> f4(I0 - 1, I2) [0 <= I0 - 1]
  f3(I3, I4) -> f4(I5, I6) [0 <= I3 - 1 /\ 0 <= I4 - 1 /\ -1 <= I5 - 1]
  f2(I7, I8) -> f2(0, I9) [2 = I7 /\ 4 <= I9 - 1 /\ 0 <= I8 - 1 /\ I9 - 4 <= I8]
  f2(I10, I11) -> f2(2, I12) [1 = I10 /\ 2 <= I12 - 1 /\ 0 <= I11 - 1 /\ I12 - 2 <= I11]
  f2(I13, I14) -> f2(1, I15) [0 = I13 /\ -1 <= I15 - 1 /\ 6 <= I14 - 1 /\ I15 + 7 <= I14]
  f3(I16, I17) -> f2(0, I18) [-1 <= y1 - 1 /\ 0 <= I17 - 1 /\ 0 <= I16 - 1 /\ 0 <= I18 - 1]
  f1(I19, I20) -> f2(0, I21) [I20 + 2 <= I19 /\ 0 <= I21 - 1 /\ 0 <= I19 - 1 /\ I21 <= I19]