/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 =
  f6#(x1, x2) -> f1#(x1, x2) 
  f5#(I0, I1) -> f2#(I0, I1) 
  f2#(I2, I3) -> f5#(I2, rnd2) [0 <= -1 - I3 /\ y1 = y1 /\ rnd2 = -1 + y1]
  f4#(I4, I5) -> f2#(I4, I5) 
  f2#(I6, I7) -> f4#(I6, I8) [-1 * I7 <= 0 /\ 0 <= -1 + I7 /\ I9 = I9 /\ I8 = 1 + I9]
  f1#(I12, I13) -> f2#(I12, I13) 
R = 
  f6(x1, x2) -> f1(x1, x2) 
  f5(I0, I1) -> f2(I0, I1) 
  f2(I2, I3) -> f5(I2, rnd2) [0 <= -1 - I3 /\ y1 = y1 /\ rnd2 = -1 + y1]
  f4(I4, I5) -> f2(I4, I5) 
  f2(I6, I7) -> f4(I6, I8) [-1 * I7 <= 0 /\ 0 <= -1 + I7 /\ I9 = I9 /\ I8 = 1 + I9]
  f2(I10, I11) -> f3(rnd1, I11) [rnd1 = rnd1 /\ I11 <= 0 /\ -1 * I11 <= 0]
  f1(I12, I13) -> f2(I12, I13) 

The dependency graph for this problem is:
  0 -> 5
  1 -> 2, 4
  2 -> 1
  3 -> 2, 4
  4 -> 3
  5 -> 2, 4
Where:
  0) f6#(x1, x2) -> f1#(x1, x2) 
  1) f5#(I0, I1) -> f2#(I0, I1) 
  2) f2#(I2, I3) -> f5#(I2, rnd2) [0 <= -1 - I3 /\ y1 = y1 /\ rnd2 = -1 + y1]
  3) f4#(I4, I5) -> f2#(I4, I5) 
  4) f2#(I6, I7) -> f4#(I6, I8) [-1 * I7 <= 0 /\ 0 <= -1 + I7 /\ I9 = I9 /\ I8 = 1 + I9]
  5) f1#(I12, I13) -> f2#(I12, I13) 

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

DP problem for innermost termination.
P =
  f5#(I0, I1) -> f2#(I0, I1) 
  f2#(I2, I3) -> f5#(I2, rnd2) [0 <= -1 - I3 /\ y1 = y1 /\ rnd2 = -1 + y1]
  f4#(I4, I5) -> f2#(I4, I5) 
  f2#(I6, I7) -> f4#(I6, I8) [-1 * I7 <= 0 /\ 0 <= -1 + I7 /\ I9 = I9 /\ I8 = 1 + I9]
R = 
  f6(x1, x2) -> f1(x1, x2) 
  f5(I0, I1) -> f2(I0, I1) 
  f2(I2, I3) -> f5(I2, rnd2) [0 <= -1 - I3 /\ y1 = y1 /\ rnd2 = -1 + y1]
  f4(I4, I5) -> f2(I4, I5) 
  f2(I6, I7) -> f4(I6, I8) [-1 * I7 <= 0 /\ 0 <= -1 + I7 /\ I9 = I9 /\ I8 = 1 + I9]
  f2(I10, I11) -> f3(rnd1, I11) [rnd1 = rnd1 /\ I11 <= 0 /\ -1 * I11 <= 0]
  f1(I12, I13) -> f2(I12, I13)