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

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

We have the following SCCs.
  { 1, 2, 3 }

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

We use the reverse value criterion with the projection function NU:
  NU[f2#(z1,z2,z3)] = z2 - 1 + -1 * z3
  NU[f3#(z1,z2,z3)] = z1 - 1 - 1 + -1 * z3

This gives the following inequalities:
  I2 <= I1 - 1  ==>  I0 - 1 - 1 + -1 * I2 >= I0 - 1 - 1 + -1 * I2
  I4 <= I5  ==>  I3 - 1 - 1 + -1 * I5 >= I3 - 1 - 1 + -1 * I5
  I8 <= I7 - 1  ==>  I7 - 1 + -1 * I8 > I7 - 1 - 1 + -1 * I8  with  I7 - 1 + -1 * I8 >= 0

We remove all the strictly oriented dependency pairs.

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

The dependency graph for this problem is:
  1 -> 1, 2
  2 -> 
Where:
  1) f3#(I0, I1, I2) -> f3#(I0, I1 - 1, I2) [I2 <= I1 - 1]
  2) f3#(I3, I4, I5) -> f2#(I4, I3 - 1, I5) [I4 <= I5]

We have the following SCCs.
  { 1 }

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

We use the reverse value criterion with the projection function NU:
  NU[f3#(z1,z2,z3)] = z2 - 1 + -1 * z3

This gives the following inequalities:
  I2 <= I1 - 1  ==>  I1 - 1 + -1 * I2 > I1 - 1 - 1 + -1 * I2  with  I1 - 1 + -1 * I2 >= 0

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