/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 =
  f6#(x1, x2) -> f5#(x1, x2) 
  f5#(I0, I1) -> f2#(I0, rnd2) [y1 = I0 /\ rnd2 = rnd2]
  f2#(I2, I3) -> f3#(I2, I3) 
  f3#(I4, I5) -> f1#(I4, I5) [1 + I5 <= 4]
  f1#(I8, I9) -> f2#(I8, 1 + I9) [1 <= I9]
  f1#(I10, I11) -> f2#(I10, 1) [I11 <= 0]
R = 
  f6(x1, x2) -> f5(x1, x2) 
  f5(I0, I1) -> f2(I0, rnd2) [y1 = I0 /\ rnd2 = rnd2]
  f2(I2, I3) -> f3(I2, I3) 
  f3(I4, I5) -> f1(I4, I5) [1 + I5 <= 4]
  f3(I6, I7) -> f4(I6, I7) [4 <= I7]
  f1(I8, I9) -> f2(I8, 1 + I9) [1 <= I9]
  f1(I10, I11) -> f2(I10, 1) [I11 <= 0]

The dependency graph for this problem is:
  0 -> 1
  1 -> 2
  2 -> 3
  3 -> 4, 5
  4 -> 2
  5 -> 2
Where:
  0) f6#(x1, x2) -> f5#(x1, x2) 
  1) f5#(I0, I1) -> f2#(I0, rnd2) [y1 = I0 /\ rnd2 = rnd2]
  2) f2#(I2, I3) -> f3#(I2, I3) 
  3) f3#(I4, I5) -> f1#(I4, I5) [1 + I5 <= 4]
  4) f1#(I8, I9) -> f2#(I8, 1 + I9) [1 <= I9]
  5) f1#(I10, I11) -> f2#(I10, 1) [I11 <= 0]

We have the following SCCs.
  { 2, 3, 4, 5 }

DP problem for innermost termination.
P =
  f2#(I2, I3) -> f3#(I2, I3) 
  f3#(I4, I5) -> f1#(I4, I5) [1 + I5 <= 4]
  f1#(I8, I9) -> f2#(I8, 1 + I9) [1 <= I9]
  f1#(I10, I11) -> f2#(I10, 1) [I11 <= 0]
R = 
  f6(x1, x2) -> f5(x1, x2) 
  f5(I0, I1) -> f2(I0, rnd2) [y1 = I0 /\ rnd2 = rnd2]
  f2(I2, I3) -> f3(I2, I3) 
  f3(I4, I5) -> f1(I4, I5) [1 + I5 <= 4]
  f3(I6, I7) -> f4(I6, I7) [4 <= I7]
  f1(I8, I9) -> f2(I8, 1 + I9) [1 <= I9]
  f1(I10, I11) -> f2(I10, 1) [I11 <= 0]

We use the extended value criterion with the projection function NU:
  NU[f1#(x0,x1)] = -x1 + 2
  NU[f3#(x0,x1)] = -x1 + 3
  NU[f2#(x0,x1)] = -x1 + 3

This gives the following inequalities:
    ==>  -I3 + 3 >= -I3 + 3
  1 + I5 <= 4  ==>  -I5 + 3 > -I5 + 2  with  -I5 + 3 >= 0
  1 <= I9  ==>  -I9 + 2 >= -(1 + I9) + 3
  I11 <= 0  ==>  -I11 + 2 >= -1 + 3

We remove all the strictly oriented dependency pairs.

DP problem for innermost termination.
P =
  f2#(I2, I3) -> f3#(I2, I3) 
  f1#(I8, I9) -> f2#(I8, 1 + I9) [1 <= I9]
  f1#(I10, I11) -> f2#(I10, 1) [I11 <= 0]
R = 
  f6(x1, x2) -> f5(x1, x2) 
  f5(I0, I1) -> f2(I0, rnd2) [y1 = I0 /\ rnd2 = rnd2]
  f2(I2, I3) -> f3(I2, I3) 
  f3(I4, I5) -> f1(I4, I5) [1 + I5 <= 4]
  f3(I6, I7) -> f4(I6, I7) [4 <= I7]
  f1(I8, I9) -> f2(I8, 1 + I9) [1 <= I9]
  f1(I10, I11) -> f2(I10, 1) [I11 <= 0]

The dependency graph for this problem is:
  2 -> 
  4 -> 2
  5 -> 2
Where:
  2) f2#(I2, I3) -> f3#(I2, I3) 
  4) f1#(I8, I9) -> f2#(I8, 1 + I9) [1 <= I9]
  5) f1#(I10, I11) -> f2#(I10, 1) [I11 <= 0]

We have the following SCCs.