/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, x3) -> f1#(rnd1, rnd2, rnd3) 
  f3#(I0, I1, I2) -> f3#(I3, I4, I5) [-1 <= I3 - 1 /\ 0 <= I0 - 1 /\ I3 + 1 <= I0]
  f2#(I6, I7, I8) -> f3#(I9, I10, I11) [-1 <= I9 - 1 /\ 0 <= I6 - 1 /\ I8 <= 0 /\ I9 + 1 <= I6]
  f2#(I12, I13, I14) -> f2#(I15, I13 - 1, I13) [3 <= I15 - 1 /\ 0 <= I12 - 1 /\ 0 <= I14 - 1 /\ I15 - 3 <= I12]
  f1#(I16, I17, I18) -> f2#(I19, 27, 28) [0 <= I19 - 1]
R = 
  init(x1, x2, x3) -> f1(rnd1, rnd2, rnd3) 
  f3(I0, I1, I2) -> f3(I3, I4, I5) [-1 <= I3 - 1 /\ 0 <= I0 - 1 /\ I3 + 1 <= I0]
  f2(I6, I7, I8) -> f3(I9, I10, I11) [-1 <= I9 - 1 /\ 0 <= I6 - 1 /\ I8 <= 0 /\ I9 + 1 <= I6]
  f2(I12, I13, I14) -> f2(I15, I13 - 1, I13) [3 <= I15 - 1 /\ 0 <= I12 - 1 /\ 0 <= I14 - 1 /\ I15 - 3 <= I12]
  f1(I16, I17, I18) -> f2(I19, 27, 28) [0 <= I19 - 1]

The dependency graph for this problem is:
  0 -> 4
  1 -> 1
  2 -> 1
  3 -> 2, 3
  4 -> 3
Where:
  0) init#(x1, x2, x3) -> f1#(rnd1, rnd2, rnd3) 
  1) f3#(I0, I1, I2) -> f3#(I3, I4, I5) [-1 <= I3 - 1 /\ 0 <= I0 - 1 /\ I3 + 1 <= I0]
  2) f2#(I6, I7, I8) -> f3#(I9, I10, I11) [-1 <= I9 - 1 /\ 0 <= I6 - 1 /\ I8 <= 0 /\ I9 + 1 <= I6]
  3) f2#(I12, I13, I14) -> f2#(I15, I13 - 1, I13) [3 <= I15 - 1 /\ 0 <= I12 - 1 /\ 0 <= I14 - 1 /\ I15 - 3 <= I12]
  4) f1#(I16, I17, I18) -> f2#(I19, 27, 28) [0 <= I19 - 1]

We have the following SCCs.
  { 3 }
  { 1 }

DP problem for innermost termination.
P =
  f3#(I0, I1, I2) -> f3#(I3, I4, I5) [-1 <= I3 - 1 /\ 0 <= I0 - 1 /\ I3 + 1 <= I0]
R = 
  init(x1, x2, x3) -> f1(rnd1, rnd2, rnd3) 
  f3(I0, I1, I2) -> f3(I3, I4, I5) [-1 <= I3 - 1 /\ 0 <= I0 - 1 /\ I3 + 1 <= I0]
  f2(I6, I7, I8) -> f3(I9, I10, I11) [-1 <= I9 - 1 /\ 0 <= I6 - 1 /\ I8 <= 0 /\ I9 + 1 <= I6]
  f2(I12, I13, I14) -> f2(I15, I13 - 1, I13) [3 <= I15 - 1 /\ 0 <= I12 - 1 /\ 0 <= I14 - 1 /\ I15 - 3 <= I12]
  f1(I16, I17, I18) -> f2(I19, 27, 28) [0 <= I19 - 1]

We use the basic value criterion with the projection function NU:
  NU[f3#(z1,z2,z3)] = z1

This gives the following inequalities:
  -1 <= I3 - 1 /\ 0 <= I0 - 1 /\ I3 + 1 <= I0  ==>  I0 >! I3

All dependency pairs are strictly oriented, so the entire dependency pair problem may be removed.

DP problem for innermost termination.
P =
  f2#(I12, I13, I14) -> f2#(I15, I13 - 1, I13) [3 <= I15 - 1 /\ 0 <= I12 - 1 /\ 0 <= I14 - 1 /\ I15 - 3 <= I12]
R = 
  init(x1, x2, x3) -> f1(rnd1, rnd2, rnd3) 
  f3(I0, I1, I2) -> f3(I3, I4, I5) [-1 <= I3 - 1 /\ 0 <= I0 - 1 /\ I3 + 1 <= I0]
  f2(I6, I7, I8) -> f3(I9, I10, I11) [-1 <= I9 - 1 /\ 0 <= I6 - 1 /\ I8 <= 0 /\ I9 + 1 <= I6]
  f2(I12, I13, I14) -> f2(I15, I13 - 1, I13) [3 <= I15 - 1 /\ 0 <= I12 - 1 /\ 0 <= I14 - 1 /\ I15 - 3 <= I12]
  f1(I16, I17, I18) -> f2(I19, 27, 28) [0 <= I19 - 1]