/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 =
  f5#(x1, x2, x3) -> f4#(x1, x2, x3) 
  f4#(I0, I1, I2) -> f1#(2, I1, I2) [I1 <= 3 /\ 0 <= I1 /\ I2 <= 3 /\ 0 <= I2]
  f3#(I3, I4, I5) -> f1#(I3, I4, I5) 
  f1#(I6, I7, I8) -> f3#(I6, 1 + I7, I8) [1 + 2 * I7 <= I6 + I8]
  f2#(I9, I10, I11) -> f1#(I9, I10, I11) 
  f1#(I12, I13, I14) -> f2#(I12, -1 + I13, I14) [1 + I12 + I14 <= -1 + 2 * I13]
R = 
  f5(x1, x2, x3) -> f4(x1, x2, x3) 
  f4(I0, I1, I2) -> f1(2, I1, I2) [I1 <= 3 /\ 0 <= I1 /\ I2 <= 3 /\ 0 <= I2]
  f3(I3, I4, I5) -> f1(I3, I4, I5) 
  f1(I6, I7, I8) -> f3(I6, 1 + I7, I8) [1 + 2 * I7 <= I6 + I8]
  f2(I9, I10, I11) -> f1(I9, I10, I11) 
  f1(I12, I13, I14) -> f2(I12, -1 + I13, I14) [1 + I12 + I14 <= -1 + 2 * I13]

The dependency graph for this problem is:
  0 -> 1
  1 -> 3, 5
  2 -> 3, 5
  3 -> 2
  4 -> 3, 5
  5 -> 4
Where:
  0) f5#(x1, x2, x3) -> f4#(x1, x2, x3) 
  1) f4#(I0, I1, I2) -> f1#(2, I1, I2) [I1 <= 3 /\ 0 <= I1 /\ I2 <= 3 /\ 0 <= I2]
  2) f3#(I3, I4, I5) -> f1#(I3, I4, I5) 
  3) f1#(I6, I7, I8) -> f3#(I6, 1 + I7, I8) [1 + 2 * I7 <= I6 + I8]
  4) f2#(I9, I10, I11) -> f1#(I9, I10, I11) 
  5) f1#(I12, I13, I14) -> f2#(I12, -1 + I13, I14) [1 + I12 + I14 <= -1 + 2 * I13]

We have the following SCCs.
  { 2, 3, 4, 5 }

DP problem for innermost termination.
P =
  f3#(I3, I4, I5) -> f1#(I3, I4, I5) 
  f1#(I6, I7, I8) -> f3#(I6, 1 + I7, I8) [1 + 2 * I7 <= I6 + I8]
  f2#(I9, I10, I11) -> f1#(I9, I10, I11) 
  f1#(I12, I13, I14) -> f2#(I12, -1 + I13, I14) [1 + I12 + I14 <= -1 + 2 * I13]
R = 
  f5(x1, x2, x3) -> f4(x1, x2, x3) 
  f4(I0, I1, I2) -> f1(2, I1, I2) [I1 <= 3 /\ 0 <= I1 /\ I2 <= 3 /\ 0 <= I2]
  f3(I3, I4, I5) -> f1(I3, I4, I5) 
  f1(I6, I7, I8) -> f3(I6, 1 + I7, I8) [1 + 2 * I7 <= I6 + I8]
  f2(I9, I10, I11) -> f1(I9, I10, I11) 
  f1(I12, I13, I14) -> f2(I12, -1 + I13, I14) [1 + I12 + I14 <= -1 + 2 * I13]