/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 =
  f6#(x1, x2) -> f5#(x1, x2) 
  f5#(I0, I1) -> f4#(I0, I1) 
  f4#(I4, I5) -> f1#(I4, -1 + I5) [0 <= -1 + I5]
  f2#(I9, I10) -> f1#(I9, I10) 
  f1#(I11, I12) -> f2#(I11, -1 + I12) [0 <= -1 + I12]
R = 
  f6(x1, x2) -> f5(x1, x2) 
  f5(I0, I1) -> f4(I0, I1) 
  f4(I2, I3) -> f3(rnd1, I3) [I3 <= 0 /\ y1 = y1 /\ rnd1 = rnd1]
  f4(I4, I5) -> f1(I4, -1 + I5) [0 <= -1 + I5]
  f1(I6, I7) -> f3(I8, I7) [I7 <= 0 /\ B0 = B0 /\ I8 = I8]
  f2(I9, I10) -> f1(I9, I10) 
  f1(I11, I12) -> f2(I11, -1 + I12) [0 <= -1 + I12]

The dependency graph for this problem is:
  0 -> 1
  1 -> 2
  2 -> 4
  3 -> 4
  4 -> 3
Where:
  0) f6#(x1, x2) -> f5#(x1, x2) 
  1) f5#(I0, I1) -> f4#(I0, I1) 
  2) f4#(I4, I5) -> f1#(I4, -1 + I5) [0 <= -1 + I5]
  3) f2#(I9, I10) -> f1#(I9, I10) 
  4) f1#(I11, I12) -> f2#(I11, -1 + I12) [0 <= -1 + I12]

We have the following SCCs.
  { 3, 4 }

DP problem for innermost termination.
P =
  f2#(I9, I10) -> f1#(I9, I10) 
  f1#(I11, I12) -> f2#(I11, -1 + I12) [0 <= -1 + I12]
R = 
  f6(x1, x2) -> f5(x1, x2) 
  f5(I0, I1) -> f4(I0, I1) 
  f4(I2, I3) -> f3(rnd1, I3) [I3 <= 0 /\ y1 = y1 /\ rnd1 = rnd1]
  f4(I4, I5) -> f1(I4, -1 + I5) [0 <= -1 + I5]
  f1(I6, I7) -> f3(I8, I7) [I7 <= 0 /\ B0 = B0 /\ I8 = I8]
  f2(I9, I10) -> f1(I9, I10) 
  f1(I11, I12) -> f2(I11, -1 + I12) [0 <= -1 + I12]

We use the basic value criterion with the projection function NU:
  NU[f1#(z1,z2)] = z2
  NU[f2#(z1,z2)] = z2

This gives the following inequalities:
    ==>  I10 (>! \union =) I10
  0 <= -1 + I12  ==>  I12 >! -1 + I12

We remove all the strictly oriented dependency pairs.

DP problem for innermost termination.
P =
  f2#(I9, I10) -> f1#(I9, I10) 
R = 
  f6(x1, x2) -> f5(x1, x2) 
  f5(I0, I1) -> f4(I0, I1) 
  f4(I2, I3) -> f3(rnd1, I3) [I3 <= 0 /\ y1 = y1 /\ rnd1 = rnd1]
  f4(I4, I5) -> f1(I4, -1 + I5) [0 <= -1 + I5]
  f1(I6, I7) -> f3(I8, I7) [I7 <= 0 /\ B0 = B0 /\ I8 = I8]
  f2(I9, I10) -> f1(I9, I10) 
  f1(I11, I12) -> f2(I11, -1 + I12) [0 <= -1 + I12]

The dependency graph for this problem is:
  3 -> 
Where:
  3) f2#(I9, I10) -> f1#(I9, I10) 

We have the following SCCs.