/export/starexec/sandbox2/solver/bin/starexec_run_Transition /export/starexec/sandbox2/benchmark/theBenchmark.smt2 /export/starexec/sandbox2/output/output_files


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YES

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

The dependency graph for this problem is:
  0 -> 6
  1 -> 4
  2 -> 
  3 -> 1, 3
  4 -> 3
  5 -> 1
  6 -> 2, 4, 5
Where:
  0) init#(x1, x2) -> f1#(rnd1, rnd2) 
  1) f3#(I0, I1) -> f2#(I1, I0) [I0 <= I1 - 1]
  2) f2#(I2, I3) -> f2#(I2, 0) [I2 = I3 /\ 0 <= I2 - 1]
  3) f3#(I4, I5) -> f3#(I4 - I5, I5) [0 <= I5 - 1 /\ 0 <= I4 - 1 /\ I5 <= I4 - 1]
  4) f2#(I6, I7) -> f3#(I6, I7) [0 <= I7 - 1 /\ -1 <= I6 - 1 /\ I7 <= I6 - 1]
  5) f2#(I8, I9) -> f3#(I8, I9) [0 <= I9 - 1 /\ -1 <= I8 - 1 /\ I8 <= I9 - 1]
  6) f1#(I10, I11) -> f2#(I12, I13) [0 <= I10 - 1 /\ -1 <= I12 - 1 /\ -1 <= I11 - 1 /\ -1 <= I13 - 1]

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

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

We use the basic value criterion with the projection function NU:
  NU[f2#(z1,z2)] = z1
  NU[f3#(z1,z2)] = z2

This gives the following inequalities:
  I0 <= I1 - 1  ==>  I1 (>! \union =) I1
  0 <= I5 - 1 /\ 0 <= I4 - 1 /\ I5 <= I4 - 1  ==>  I5 (>! \union =) I5
  0 <= I7 - 1 /\ -1 <= I6 - 1 /\ I7 <= I6 - 1  ==>  I6 >! I7

We remove all the strictly oriented dependency pairs.

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

The dependency graph for this problem is:
  1 -> 
  3 -> 1, 3
Where:
  1) f3#(I0, I1) -> f2#(I1, I0) [I0 <= I1 - 1]
  3) f3#(I4, I5) -> f3#(I4 - I5, I5) [0 <= I5 - 1 /\ 0 <= I4 - 1 /\ I5 <= I4 - 1]

We have the following SCCs.
  { 3 }

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

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

This gives the following inequalities:
  0 <= I5 - 1 /\ 0 <= I4 - 1 /\ I5 <= I4 - 1  ==>  I4 >! I4 - I5

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