/export/starexec/sandbox/solver/bin/starexec_run_Transition /export/starexec/sandbox/benchmark/theBenchmark.smt2 /export/starexec/sandbox/output/output_files


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YES

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

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

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

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

We use the basic value criterion with the projection function NU:
  NU[f2#(z1,z2,z3)] = z2

This gives the following inequalities:
  0 = I0 /\ 0 <= I1 - 1 /\ 0 <= I2 - 1  ==>  I1 >! I1 - 1

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

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

We use the basic value criterion with the projection function NU:
  NU[f2#(z1,z2,z3)] = z1

This gives the following inequalities:
  -1 <= I7 - 1 /\ 0 <= I8 - 1 /\ 0 <= I6 - 1  ==>  I6 >! I6 - 1

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