/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, I3) [0 <= I0 - 1 /\ 0 <= I1 - 1]
  f3#(I4, I5, I6) -> f3#(I4, I5 - 1, I6 - 1) [0 <= I4 - 1 /\ 0 <= I5 - 1]
  f3#(I7, I8, I9) -> f2#(I7 - I8, I9 - 1, I10) [0 <= I7 - 1 /\ 0 <= I8 - 1]
  f2#(I11, I12, I13) -> f3#(I11, 2, I12) [0 <= I11 - 1]
  f1#(I14, I15, I16) -> f2#(I15, I15, I17) [-1 <= I15 - 1 /\ 0 <= I14 - 1]
R = 
  init(x1, x2, x3) -> f1(rnd1, rnd2, rnd3) 
  f3(I0, I1, I2) -> f3(I0, I1 - 1, I3) [0 <= I0 - 1 /\ 0 <= I1 - 1]
  f3(I4, I5, I6) -> f3(I4, I5 - 1, I6 - 1) [0 <= I4 - 1 /\ 0 <= I5 - 1]
  f3(I7, I8, I9) -> f2(I7 - I8, I9 - 1, I10) [0 <= I7 - 1 /\ 0 <= I8 - 1]
  f2(I11, I12, I13) -> f3(I11, 2, I12) [0 <= I11 - 1]
  f1(I14, I15, I16) -> f2(I15, I15, I17) [-1 <= I15 - 1 /\ 0 <= I14 - 1]

The dependency graph for this problem is:
  0 -> 5
  1 -> 1, 2, 3
  2 -> 1, 2, 3
  3 -> 4
  4 -> 1, 2, 3
  5 -> 4
Where:
  0) init#(x1, x2, x3) -> f1#(rnd1, rnd2, rnd3) 
  1) f3#(I0, I1, I2) -> f3#(I0, I1 - 1, I3) [0 <= I0 - 1 /\ 0 <= I1 - 1]
  2) f3#(I4, I5, I6) -> f3#(I4, I5 - 1, I6 - 1) [0 <= I4 - 1 /\ 0 <= I5 - 1]
  3) f3#(I7, I8, I9) -> f2#(I7 - I8, I9 - 1, I10) [0 <= I7 - 1 /\ 0 <= I8 - 1]
  4) f2#(I11, I12, I13) -> f3#(I11, 2, I12) [0 <= I11 - 1]
  5) f1#(I14, I15, I16) -> f2#(I15, I15, I17) [-1 <= I15 - 1 /\ 0 <= I14 - 1]

We have the following SCCs.
  { 1, 2, 3, 4 }

DP problem for innermost termination.
P =
  f3#(I0, I1, I2) -> f3#(I0, I1 - 1, I3) [0 <= I0 - 1 /\ 0 <= I1 - 1]
  f3#(I4, I5, I6) -> f3#(I4, I5 - 1, I6 - 1) [0 <= I4 - 1 /\ 0 <= I5 - 1]
  f3#(I7, I8, I9) -> f2#(I7 - I8, I9 - 1, I10) [0 <= I7 - 1 /\ 0 <= I8 - 1]
  f2#(I11, I12, I13) -> f3#(I11, 2, I12) [0 <= I11 - 1]
R = 
  init(x1, x2, x3) -> f1(rnd1, rnd2, rnd3) 
  f3(I0, I1, I2) -> f3(I0, I1 - 1, I3) [0 <= I0 - 1 /\ 0 <= I1 - 1]
  f3(I4, I5, I6) -> f3(I4, I5 - 1, I6 - 1) [0 <= I4 - 1 /\ 0 <= I5 - 1]
  f3(I7, I8, I9) -> f2(I7 - I8, I9 - 1, I10) [0 <= I7 - 1 /\ 0 <= I8 - 1]
  f2(I11, I12, I13) -> f3(I11, 2, I12) [0 <= I11 - 1]
  f1(I14, I15, I16) -> f2(I15, I15, I17) [-1 <= I15 - 1 /\ 0 <= I14 - 1]

We use the basic value criterion with the projection function NU:
  NU[f2#(z1,z2,z3)] = z1
  NU[f3#(z1,z2,z3)] = z1

This gives the following inequalities:
  0 <= I0 - 1 /\ 0 <= I1 - 1  ==>  I0 (>! \union =) I0
  0 <= I4 - 1 /\ 0 <= I5 - 1  ==>  I4 (>! \union =) I4
  0 <= I7 - 1 /\ 0 <= I8 - 1  ==>  I7 >! I7 - I8
  0 <= I11 - 1  ==>  I11 (>! \union =) I11

We remove all the strictly oriented dependency pairs.

DP problem for innermost termination.
P =
  f3#(I0, I1, I2) -> f3#(I0, I1 - 1, I3) [0 <= I0 - 1 /\ 0 <= I1 - 1]
  f3#(I4, I5, I6) -> f3#(I4, I5 - 1, I6 - 1) [0 <= I4 - 1 /\ 0 <= I5 - 1]
  f2#(I11, I12, I13) -> f3#(I11, 2, I12) [0 <= I11 - 1]
R = 
  init(x1, x2, x3) -> f1(rnd1, rnd2, rnd3) 
  f3(I0, I1, I2) -> f3(I0, I1 - 1, I3) [0 <= I0 - 1 /\ 0 <= I1 - 1]
  f3(I4, I5, I6) -> f3(I4, I5 - 1, I6 - 1) [0 <= I4 - 1 /\ 0 <= I5 - 1]
  f3(I7, I8, I9) -> f2(I7 - I8, I9 - 1, I10) [0 <= I7 - 1 /\ 0 <= I8 - 1]
  f2(I11, I12, I13) -> f3(I11, 2, I12) [0 <= I11 - 1]
  f1(I14, I15, I16) -> f2(I15, I15, I17) [-1 <= I15 - 1 /\ 0 <= I14 - 1]

The dependency graph for this problem is:
  1 -> 1, 2
  2 -> 1, 2
  4 -> 1, 2
Where:
  1) f3#(I0, I1, I2) -> f3#(I0, I1 - 1, I3) [0 <= I0 - 1 /\ 0 <= I1 - 1]
  2) f3#(I4, I5, I6) -> f3#(I4, I5 - 1, I6 - 1) [0 <= I4 - 1 /\ 0 <= I5 - 1]
  4) f2#(I11, I12, I13) -> f3#(I11, 2, I12) [0 <= I11 - 1]

We have the following SCCs.
  { 1, 2 }

DP problem for innermost termination.
P =
  f3#(I0, I1, I2) -> f3#(I0, I1 - 1, I3) [0 <= I0 - 1 /\ 0 <= I1 - 1]
  f3#(I4, I5, I6) -> f3#(I4, I5 - 1, I6 - 1) [0 <= I4 - 1 /\ 0 <= I5 - 1]
R = 
  init(x1, x2, x3) -> f1(rnd1, rnd2, rnd3) 
  f3(I0, I1, I2) -> f3(I0, I1 - 1, I3) [0 <= I0 - 1 /\ 0 <= I1 - 1]
  f3(I4, I5, I6) -> f3(I4, I5 - 1, I6 - 1) [0 <= I4 - 1 /\ 0 <= I5 - 1]
  f3(I7, I8, I9) -> f2(I7 - I8, I9 - 1, I10) [0 <= I7 - 1 /\ 0 <= I8 - 1]
  f2(I11, I12, I13) -> f3(I11, 2, I12) [0 <= I11 - 1]
  f1(I14, I15, I16) -> f2(I15, I15, I17) [-1 <= I15 - 1 /\ 0 <= I14 - 1]

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

This gives the following inequalities:
  0 <= I0 - 1 /\ 0 <= I1 - 1  ==>  I1 >! I1 - 1
  0 <= I4 - 1 /\ 0 <= I5 - 1  ==>  I5 >! I5 - 1

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