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

The dependency graph for this problem is:
  0 -> 1
  1 -> 3, 4
  2 -> 3, 4
  3 -> 2
  4 -> 6
  5 -> 6
  6 -> 5
Where:
  0) f6#(x1, x2) -> f5#(x1, x2) 
  1) f5#(I0, I1) -> f3#(rnd1, 1) [rnd1 = rnd1]
  2) f4#(I2, I3) -> f3#(I2, I3) 
  3) f3#(I4, I5) -> f4#(I4, I5) [1 <= I4]
  4) f3#(I6, I7) -> f1#(I6, 1) [I6 <= 0 /\ y1 = 0]
  5) f2#(I8, I9) -> f1#(I8, I9) 
  6) f1#(I10, I11) -> f2#(I10, I11) 

We have the following SCCs.
  { 2, 3 }
  { 5, 6 }

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