/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, x4) -> f1#(rnd1, rnd2, rnd3, rnd4) 
  f2#(I0, I1, I2, I3) -> f2#(I0 - 1, I1 - 1, I2, I2) [I2 = I3 /\ 0 <= I1 - 1 /\ 0 <= I2 - 1 /\ 0 <= I0 - 1]
  f2#(I4, I5, I6, I7) -> f2#(I4, I6, I6, I6) [I6 = I7 /\ 0 = I5 /\ 0 <= I6 - 1]
  f1#(I8, I9, I10, I11) -> f2#(I12, I13, I14, I15) [I13 = I15 /\ I13 = I14 /\ 0 <= I8 - 1 /\ -1 <= I12 - 1 /\ -1 <= I9 - 1 /\ -1 <= I13 - 1]
R = 
  init(x1, x2, x3, x4) -> f1(rnd1, rnd2, rnd3, rnd4) 
  f2(I0, I1, I2, I3) -> f2(I0 - 1, I1 - 1, I2, I2) [I2 = I3 /\ 0 <= I1 - 1 /\ 0 <= I2 - 1 /\ 0 <= I0 - 1]
  f2(I4, I5, I6, I7) -> f2(I4, I6, I6, I6) [I6 = I7 /\ 0 = I5 /\ 0 <= I6 - 1]
  f1(I8, I9, I10, I11) -> f2(I12, I13, I14, I15) [I13 = I15 /\ I13 = I14 /\ 0 <= I8 - 1 /\ -1 <= I12 - 1 /\ -1 <= I9 - 1 /\ -1 <= I13 - 1]

The dependency graph for this problem is:
  0 -> 3
  1 -> 1, 2
  2 -> 1
  3 -> 1
Where:
  0) init#(x1, x2, x3, x4) -> f1#(rnd1, rnd2, rnd3, rnd4) 
  1) f2#(I0, I1, I2, I3) -> f2#(I0 - 1, I1 - 1, I2, I2) [I2 = I3 /\ 0 <= I1 - 1 /\ 0 <= I2 - 1 /\ 0 <= I0 - 1]
  2) f2#(I4, I5, I6, I7) -> f2#(I4, I6, I6, I6) [I6 = I7 /\ 0 = I5 /\ 0 <= I6 - 1]
  3) f1#(I8, I9, I10, I11) -> f2#(I12, I13, I14, I15) [I13 = I15 /\ I13 = I14 /\ 0 <= I8 - 1 /\ -1 <= I12 - 1 /\ -1 <= I9 - 1 /\ -1 <= I13 - 1]

We have the following SCCs.
  { 1, 2 }

DP problem for innermost termination.
P =
  f2#(I0, I1, I2, I3) -> f2#(I0 - 1, I1 - 1, I2, I2) [I2 = I3 /\ 0 <= I1 - 1 /\ 0 <= I2 - 1 /\ 0 <= I0 - 1]
  f2#(I4, I5, I6, I7) -> f2#(I4, I6, I6, I6) [I6 = I7 /\ 0 = I5 /\ 0 <= I6 - 1]
R = 
  init(x1, x2, x3, x4) -> f1(rnd1, rnd2, rnd3, rnd4) 
  f2(I0, I1, I2, I3) -> f2(I0 - 1, I1 - 1, I2, I2) [I2 = I3 /\ 0 <= I1 - 1 /\ 0 <= I2 - 1 /\ 0 <= I0 - 1]
  f2(I4, I5, I6, I7) -> f2(I4, I6, I6, I6) [I6 = I7 /\ 0 = I5 /\ 0 <= I6 - 1]
  f1(I8, I9, I10, I11) -> f2(I12, I13, I14, I15) [I13 = I15 /\ I13 = I14 /\ 0 <= I8 - 1 /\ -1 <= I12 - 1 /\ -1 <= I9 - 1 /\ -1 <= I13 - 1]

We use the basic value criterion with the projection function NU:
  NU[f2#(z1,z2,z3,z4)] = z1

This gives the following inequalities:
  I2 = I3 /\ 0 <= I1 - 1 /\ 0 <= I2 - 1 /\ 0 <= I0 - 1  ==>  I0 >! I0 - 1
  I6 = I7 /\ 0 = I5 /\ 0 <= I6 - 1  ==>  I4 (>! \union =) I4

We remove all the strictly oriented dependency pairs.

DP problem for innermost termination.
P =
  f2#(I4, I5, I6, I7) -> f2#(I4, I6, I6, I6) [I6 = I7 /\ 0 = I5 /\ 0 <= I6 - 1]
R = 
  init(x1, x2, x3, x4) -> f1(rnd1, rnd2, rnd3, rnd4) 
  f2(I0, I1, I2, I3) -> f2(I0 - 1, I1 - 1, I2, I2) [I2 = I3 /\ 0 <= I1 - 1 /\ 0 <= I2 - 1 /\ 0 <= I0 - 1]
  f2(I4, I5, I6, I7) -> f2(I4, I6, I6, I6) [I6 = I7 /\ 0 = I5 /\ 0 <= I6 - 1]
  f1(I8, I9, I10, I11) -> f2(I12, I13, I14, I15) [I13 = I15 /\ I13 = I14 /\ 0 <= I8 - 1 /\ -1 <= I12 - 1 /\ -1 <= I9 - 1 /\ -1 <= I13 - 1]

The dependency graph for this problem is:
  2 -> 
Where:
  2) f2#(I4, I5, I6, I7) -> f2#(I4, I6, I6, I6) [I6 = I7 /\ 0 = I5 /\ 0 <= I6 - 1]

We have the following SCCs.