/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) 
  f3#(I0, I1) -> f3#(0, I2) [2 = I0 /\ 4 <= I2 - 1 /\ 0 <= I1 - 1 /\ I2 - 4 <= I1]
  f3#(I3, I4) -> f3#(2, I5) [1 = I3 /\ 2 <= I5 - 1 /\ 0 <= I4 - 1 /\ I5 - 2 <= I4]
  f3#(I6, I7) -> f3#(1, I8) [0 = I6 /\ -1 <= I8 - 1 /\ 6 <= I7 - 1 /\ I8 + 7 <= I7]
  f4#(I9, I10) -> f4#(I9 - 1, I11) [I9 - 1 <= I9 - 1 /\ 1 <= I9 - 1]
  f1#(I12, I13) -> f4#(I14, I15) [0 <= I12 - 1 /\ 0 <= I13 - 1 /\ -1 <= I14 - 1]
  f2#(I16, I17) -> f3#(0, I18) [I18 <= I17 /\ 0 <= y1 - 1 /\ 0 <= I16 - 1 /\ 0 <= I17 - 1 /\ 0 <= I18 - 1]
  f1#(I19, I20) -> f2#(I21, I22) [-1 <= I23 - 1 /\ 0 <= I20 - 1 /\ I21 <= I19 /\ I22 - 1 <= I19 /\ 0 <= I19 - 1 /\ 0 <= I21 - 1 /\ 1 <= I22 - 1]
  f1#(I24, I25) -> f2#(I26, I27) [-1 <= I28 - 1 /\ 0 <= I25 - 1 /\ I26 <= I24 /\ 0 <= I24 - 1 /\ 0 <= I26 - 1 /\ 2 <= I27 - 1]
R = 
  init(x1, x2) -> f1(rnd1, rnd2) 
  f3(I0, I1) -> f3(0, I2) [2 = I0 /\ 4 <= I2 - 1 /\ 0 <= I1 - 1 /\ I2 - 4 <= I1]
  f3(I3, I4) -> f3(2, I5) [1 = I3 /\ 2 <= I5 - 1 /\ 0 <= I4 - 1 /\ I5 - 2 <= I4]
  f3(I6, I7) -> f3(1, I8) [0 = I6 /\ -1 <= I8 - 1 /\ 6 <= I7 - 1 /\ I8 + 7 <= I7]
  f4(I9, I10) -> f4(I9 - 1, I11) [I9 - 1 <= I9 - 1 /\ 1 <= I9 - 1]
  f1(I12, I13) -> f4(I14, I15) [0 <= I12 - 1 /\ 0 <= I13 - 1 /\ -1 <= I14 - 1]
  f2(I16, I17) -> f3(0, I18) [I18 <= I17 /\ 0 <= y1 - 1 /\ 0 <= I16 - 1 /\ 0 <= I17 - 1 /\ 0 <= I18 - 1]
  f1(I19, I20) -> f2(I21, I22) [-1 <= I23 - 1 /\ 0 <= I20 - 1 /\ I21 <= I19 /\ I22 - 1 <= I19 /\ 0 <= I19 - 1 /\ 0 <= I21 - 1 /\ 1 <= I22 - 1]
  f1(I24, I25) -> f2(I26, I27) [-1 <= I28 - 1 /\ 0 <= I25 - 1 /\ I26 <= I24 /\ 0 <= I24 - 1 /\ 0 <= I26 - 1 /\ 2 <= I27 - 1]

The dependency graph for this problem is:
  0 -> 5, 7, 8
  1 -> 3
  2 -> 1
  3 -> 2
  4 -> 4
  5 -> 4
  6 -> 3
  7 -> 6
  8 -> 6
Where:
  0) init#(x1, x2) -> f1#(rnd1, rnd2) 
  1) f3#(I0, I1) -> f3#(0, I2) [2 = I0 /\ 4 <= I2 - 1 /\ 0 <= I1 - 1 /\ I2 - 4 <= I1]
  2) f3#(I3, I4) -> f3#(2, I5) [1 = I3 /\ 2 <= I5 - 1 /\ 0 <= I4 - 1 /\ I5 - 2 <= I4]
  3) f3#(I6, I7) -> f3#(1, I8) [0 = I6 /\ -1 <= I8 - 1 /\ 6 <= I7 - 1 /\ I8 + 7 <= I7]
  4) f4#(I9, I10) -> f4#(I9 - 1, I11) [I9 - 1 <= I9 - 1 /\ 1 <= I9 - 1]
  5) f1#(I12, I13) -> f4#(I14, I15) [0 <= I12 - 1 /\ 0 <= I13 - 1 /\ -1 <= I14 - 1]
  6) f2#(I16, I17) -> f3#(0, I18) [I18 <= I17 /\ 0 <= y1 - 1 /\ 0 <= I16 - 1 /\ 0 <= I17 - 1 /\ 0 <= I18 - 1]
  7) f1#(I19, I20) -> f2#(I21, I22) [-1 <= I23 - 1 /\ 0 <= I20 - 1 /\ I21 <= I19 /\ I22 - 1 <= I19 /\ 0 <= I19 - 1 /\ 0 <= I21 - 1 /\ 1 <= I22 - 1]
  8) f1#(I24, I25) -> f2#(I26, I27) [-1 <= I28 - 1 /\ 0 <= I25 - 1 /\ I26 <= I24 /\ 0 <= I24 - 1 /\ 0 <= I26 - 1 /\ 2 <= I27 - 1]

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

DP problem for innermost termination.
P =
  f3#(I0, I1) -> f3#(0, I2) [2 = I0 /\ 4 <= I2 - 1 /\ 0 <= I1 - 1 /\ I2 - 4 <= I1]
  f3#(I3, I4) -> f3#(2, I5) [1 = I3 /\ 2 <= I5 - 1 /\ 0 <= I4 - 1 /\ I5 - 2 <= I4]
  f3#(I6, I7) -> f3#(1, I8) [0 = I6 /\ -1 <= I8 - 1 /\ 6 <= I7 - 1 /\ I8 + 7 <= I7]
R = 
  init(x1, x2) -> f1(rnd1, rnd2) 
  f3(I0, I1) -> f3(0, I2) [2 = I0 /\ 4 <= I2 - 1 /\ 0 <= I1 - 1 /\ I2 - 4 <= I1]
  f3(I3, I4) -> f3(2, I5) [1 = I3 /\ 2 <= I5 - 1 /\ 0 <= I4 - 1 /\ I5 - 2 <= I4]
  f3(I6, I7) -> f3(1, I8) [0 = I6 /\ -1 <= I8 - 1 /\ 6 <= I7 - 1 /\ I8 + 7 <= I7]
  f4(I9, I10) -> f4(I9 - 1, I11) [I9 - 1 <= I9 - 1 /\ 1 <= I9 - 1]
  f1(I12, I13) -> f4(I14, I15) [0 <= I12 - 1 /\ 0 <= I13 - 1 /\ -1 <= I14 - 1]
  f2(I16, I17) -> f3(0, I18) [I18 <= I17 /\ 0 <= y1 - 1 /\ 0 <= I16 - 1 /\ 0 <= I17 - 1 /\ 0 <= I18 - 1]
  f1(I19, I20) -> f2(I21, I22) [-1 <= I23 - 1 /\ 0 <= I20 - 1 /\ I21 <= I19 /\ I22 - 1 <= I19 /\ 0 <= I19 - 1 /\ 0 <= I21 - 1 /\ 1 <= I22 - 1]
  f1(I24, I25) -> f2(I26, I27) [-1 <= I28 - 1 /\ 0 <= I25 - 1 /\ I26 <= I24 /\ 0 <= I24 - 1 /\ 0 <= I26 - 1 /\ 2 <= I27 - 1]