/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) 
  f4#(I0, I1) -> f4#(I0, I1 - 1) [0 <= I1 - 1]
  f4#(I2, I3) -> f2#(I2, 0) [0 = I3]
  f3#(I4, I5) -> f3#(I4, I5 - 1) [0 <= I5 - 1]
  f3#(I6, I7) -> f2#(0, I6) [0 = I7]
  f2#(I8, I9) -> f4#(I8, I9) [0 <= I8 - 1 /\ 0 <= I9 - 1 /\ I8 <= I9 - 1]
  f2#(I10, I11) -> f3#(I11, I10) [0 <= I10 - 1 /\ 0 <= I11 - 1 /\ I11 <= I10]
  f1#(I12, I13) -> f2#(I14, I15) [0 <= I12 - 1 /\ -1 <= I15 - 1 /\ -1 <= I13 - 1 /\ -1 <= I14 - 1]
R = 
  init(x1, x2) -> f1(rnd1, rnd2) 
  f4(I0, I1) -> f4(I0, I1 - 1) [0 <= I1 - 1]
  f4(I2, I3) -> f2(I2, 0) [0 = I3]
  f3(I4, I5) -> f3(I4, I5 - 1) [0 <= I5 - 1]
  f3(I6, I7) -> f2(0, I6) [0 = I7]
  f2(I8, I9) -> f4(I8, I9) [0 <= I8 - 1 /\ 0 <= I9 - 1 /\ I8 <= I9 - 1]
  f2(I10, I11) -> f3(I11, I10) [0 <= I10 - 1 /\ 0 <= I11 - 1 /\ I11 <= I10]
  f1(I12, I13) -> f2(I14, I15) [0 <= I12 - 1 /\ -1 <= I15 - 1 /\ -1 <= I13 - 1 /\ -1 <= I14 - 1]

The dependency graph for this problem is:
  0 -> 7
  1 -> 1, 2
  2 -> 
  3 -> 3, 4
  4 -> 
  5 -> 1
  6 -> 3
  7 -> 5, 6
Where:
  0) init#(x1, x2) -> f1#(rnd1, rnd2) 
  1) f4#(I0, I1) -> f4#(I0, I1 - 1) [0 <= I1 - 1]
  2) f4#(I2, I3) -> f2#(I2, 0) [0 = I3]
  3) f3#(I4, I5) -> f3#(I4, I5 - 1) [0 <= I5 - 1]
  4) f3#(I6, I7) -> f2#(0, I6) [0 = I7]
  5) f2#(I8, I9) -> f4#(I8, I9) [0 <= I8 - 1 /\ 0 <= I9 - 1 /\ I8 <= I9 - 1]
  6) f2#(I10, I11) -> f3#(I11, I10) [0 <= I10 - 1 /\ 0 <= I11 - 1 /\ I11 <= I10]
  7) f1#(I12, I13) -> f2#(I14, I15) [0 <= I12 - 1 /\ -1 <= I15 - 1 /\ -1 <= I13 - 1 /\ -1 <= I14 - 1]

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

DP problem for innermost termination.
P =
  f4#(I0, I1) -> f4#(I0, I1 - 1) [0 <= I1 - 1]
R = 
  init(x1, x2) -> f1(rnd1, rnd2) 
  f4(I0, I1) -> f4(I0, I1 - 1) [0 <= I1 - 1]
  f4(I2, I3) -> f2(I2, 0) [0 = I3]
  f3(I4, I5) -> f3(I4, I5 - 1) [0 <= I5 - 1]
  f3(I6, I7) -> f2(0, I6) [0 = I7]
  f2(I8, I9) -> f4(I8, I9) [0 <= I8 - 1 /\ 0 <= I9 - 1 /\ I8 <= I9 - 1]
  f2(I10, I11) -> f3(I11, I10) [0 <= I10 - 1 /\ 0 <= I11 - 1 /\ I11 <= I10]
  f1(I12, I13) -> f2(I14, I15) [0 <= I12 - 1 /\ -1 <= I15 - 1 /\ -1 <= I13 - 1 /\ -1 <= I14 - 1]

We use the basic value criterion with the projection function NU:
  NU[f4#(z1,z2)] = z2

This gives the following inequalities:
  0 <= I1 - 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 =
  f3#(I4, I5) -> f3#(I4, I5 - 1) [0 <= I5 - 1]
R = 
  init(x1, x2) -> f1(rnd1, rnd2) 
  f4(I0, I1) -> f4(I0, I1 - 1) [0 <= I1 - 1]
  f4(I2, I3) -> f2(I2, 0) [0 = I3]
  f3(I4, I5) -> f3(I4, I5 - 1) [0 <= I5 - 1]
  f3(I6, I7) -> f2(0, I6) [0 = I7]
  f2(I8, I9) -> f4(I8, I9) [0 <= I8 - 1 /\ 0 <= I9 - 1 /\ I8 <= I9 - 1]
  f2(I10, I11) -> f3(I11, I10) [0 <= I10 - 1 /\ 0 <= I11 - 1 /\ I11 <= I10]
  f1(I12, I13) -> f2(I14, I15) [0 <= I12 - 1 /\ -1 <= I15 - 1 /\ -1 <= I13 - 1 /\ -1 <= I14 - 1]

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

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

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