/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 =
  f7#(x1, x2) -> f6#(x1, x2) 
  f6#(I0, I1) -> f5#(I0, rnd2) [y1 = 0 /\ rnd2 = rnd2]
  f4#(I2, I3) -> f3#(I2, I3) 
  f5#(I4, I5) -> f4#(I4, I5) [1 <= I5]
  f5#(I6, I7) -> f1#(I6, I7) [I7 <= 0]
  f3#(I8, I9) -> f4#(I8, 1 + I9) [1 + I9 <= I8]
  f3#(I10, I11) -> f1#(I10, I11) [I10 <= I11]
R = 
  f7(x1, x2) -> f6(x1, x2) 
  f6(I0, I1) -> f5(I0, rnd2) [y1 = 0 /\ rnd2 = rnd2]
  f4(I2, I3) -> f3(I2, I3) 
  f5(I4, I5) -> f4(I4, I5) [1 <= I5]
  f5(I6, I7) -> f1(I6, I7) [I7 <= 0]
  f3(I8, I9) -> f4(I8, 1 + I9) [1 + I9 <= I8]
  f3(I10, I11) -> f1(I10, I11) [I10 <= I11]
  f1(I12, I13) -> f2(I12, I13) 

The dependency graph for this problem is:
  0 -> 1
  1 -> 3, 4
  2 -> 5, 6
  3 -> 2
  4 -> 
  5 -> 2
  6 -> 
Where:
  0) f7#(x1, x2) -> f6#(x1, x2) 
  1) f6#(I0, I1) -> f5#(I0, rnd2) [y1 = 0 /\ rnd2 = rnd2]
  2) f4#(I2, I3) -> f3#(I2, I3) 
  3) f5#(I4, I5) -> f4#(I4, I5) [1 <= I5]
  4) f5#(I6, I7) -> f1#(I6, I7) [I7 <= 0]
  5) f3#(I8, I9) -> f4#(I8, 1 + I9) [1 + I9 <= I8]
  6) f3#(I10, I11) -> f1#(I10, I11) [I10 <= I11]

We have the following SCCs.
  { 2, 5 }

DP problem for innermost termination.
P =
  f4#(I2, I3) -> f3#(I2, I3) 
  f3#(I8, I9) -> f4#(I8, 1 + I9) [1 + I9 <= I8]
R = 
  f7(x1, x2) -> f6(x1, x2) 
  f6(I0, I1) -> f5(I0, rnd2) [y1 = 0 /\ rnd2 = rnd2]
  f4(I2, I3) -> f3(I2, I3) 
  f5(I4, I5) -> f4(I4, I5) [1 <= I5]
  f5(I6, I7) -> f1(I6, I7) [I7 <= 0]
  f3(I8, I9) -> f4(I8, 1 + I9) [1 + I9 <= I8]
  f3(I10, I11) -> f1(I10, I11) [I10 <= I11]
  f1(I12, I13) -> f2(I12, I13) 

We use the reverse value criterion with the projection function NU:
  NU[f3#(z1,z2)] = z1 + -1 * (1 + z2)
  NU[f4#(z1,z2)] = z1 + -1 * (1 + z2)

This gives the following inequalities:
    ==>  I2 + -1 * (1 + I3) >= I2 + -1 * (1 + I3)
  1 + I9 <= I8  ==>  I8 + -1 * (1 + I9) > I8 + -1 * (1 + (1 + I9))  with  I8 + -1 * (1 + I9) >= 0

We remove all the strictly oriented dependency pairs.

DP problem for innermost termination.
P =
  f4#(I2, I3) -> f3#(I2, I3) 
R = 
  f7(x1, x2) -> f6(x1, x2) 
  f6(I0, I1) -> f5(I0, rnd2) [y1 = 0 /\ rnd2 = rnd2]
  f4(I2, I3) -> f3(I2, I3) 
  f5(I4, I5) -> f4(I4, I5) [1 <= I5]
  f5(I6, I7) -> f1(I6, I7) [I7 <= 0]
  f3(I8, I9) -> f4(I8, 1 + I9) [1 + I9 <= I8]
  f3(I10, I11) -> f1(I10, I11) [I10 <= I11]
  f1(I12, I13) -> f2(I12, I13) 

The dependency graph for this problem is:
  2 -> 
Where:
  2) f4#(I2, I3) -> f3#(I2, I3) 

We have the following SCCs.