/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 =
  f5#(x1, x2, x3) -> f4#(x1, x2, x3) 
  f4#(I0, I1, I2) -> f1#(I0, I1, I2) 
  f3#(I3, I4, I5) -> f1#(I3, I4, I5) 
  f1#(I6, I7, I8) -> f3#(I6, -2 + I8, 1 + -2 + I8) [0 <= -1 + I7 + I8 /\ 0 <= -1 + I8 /\ 0 <= -1 + I7]
R = 
  f5(x1, x2, x3) -> f4(x1, x2, x3) 
  f4(I0, I1, I2) -> f1(I0, I1, I2) 
  f3(I3, I4, I5) -> f1(I3, I4, I5) 
  f1(I6, I7, I8) -> f3(I6, -2 + I8, 1 + -2 + I8) [0 <= -1 + I7 + I8 /\ 0 <= -1 + I8 /\ 0 <= -1 + I7]
  f1(I9, I10, I11) -> f2(rnd1, I10, I11) [rnd1 = rnd1 /\ I10 + I11 <= 0 /\ 0 <= -1 + I11 /\ 0 <= -1 + I10]
  f1(I12, I13, I14) -> f2(I15, I13, I14) [I15 = I15 /\ I14 <= 0 /\ 0 <= -1 + I13]
  f1(I16, I17, I18) -> f2(I19, I17, I18) [I19 = I19 /\ I17 <= 0]

The dependency graph for this problem is:
  0 -> 1
  1 -> 3
  2 -> 3
  3 -> 2
Where:
  0) f5#(x1, x2, x3) -> f4#(x1, x2, x3) 
  1) f4#(I0, I1, I2) -> f1#(I0, I1, I2) 
  2) f3#(I3, I4, I5) -> f1#(I3, I4, I5) 
  3) f1#(I6, I7, I8) -> f3#(I6, -2 + I8, 1 + -2 + I8) [0 <= -1 + I7 + I8 /\ 0 <= -1 + I8 /\ 0 <= -1 + I7]

We have the following SCCs.
  { 2, 3 }

DP problem for innermost termination.
P =
  f3#(I3, I4, I5) -> f1#(I3, I4, I5) 
  f1#(I6, I7, I8) -> f3#(I6, -2 + I8, 1 + -2 + I8) [0 <= -1 + I7 + I8 /\ 0 <= -1 + I8 /\ 0 <= -1 + I7]
R = 
  f5(x1, x2, x3) -> f4(x1, x2, x3) 
  f4(I0, I1, I2) -> f1(I0, I1, I2) 
  f3(I3, I4, I5) -> f1(I3, I4, I5) 
  f1(I6, I7, I8) -> f3(I6, -2 + I8, 1 + -2 + I8) [0 <= -1 + I7 + I8 /\ 0 <= -1 + I8 /\ 0 <= -1 + I7]
  f1(I9, I10, I11) -> f2(rnd1, I10, I11) [rnd1 = rnd1 /\ I10 + I11 <= 0 /\ 0 <= -1 + I11 /\ 0 <= -1 + I10]
  f1(I12, I13, I14) -> f2(I15, I13, I14) [I15 = I15 /\ I14 <= 0 /\ 0 <= -1 + I13]
  f1(I16, I17, I18) -> f2(I19, I17, I18) [I19 = I19 /\ I17 <= 0]

We use the basic value criterion with the projection function NU:
  NU[f1#(z1,z2,z3)] = z3
  NU[f3#(z1,z2,z3)] = z3

This gives the following inequalities:
    ==>  I5 (>! \union =) I5
  0 <= -1 + I7 + I8 /\ 0 <= -1 + I8 /\ 0 <= -1 + I7  ==>  I8 >! 1 + -2 + I8

We remove all the strictly oriented dependency pairs.

DP problem for innermost termination.
P =
  f3#(I3, I4, I5) -> f1#(I3, I4, I5) 
R = 
  f5(x1, x2, x3) -> f4(x1, x2, x3) 
  f4(I0, I1, I2) -> f1(I0, I1, I2) 
  f3(I3, I4, I5) -> f1(I3, I4, I5) 
  f1(I6, I7, I8) -> f3(I6, -2 + I8, 1 + -2 + I8) [0 <= -1 + I7 + I8 /\ 0 <= -1 + I8 /\ 0 <= -1 + I7]
  f1(I9, I10, I11) -> f2(rnd1, I10, I11) [rnd1 = rnd1 /\ I10 + I11 <= 0 /\ 0 <= -1 + I11 /\ 0 <= -1 + I10]
  f1(I12, I13, I14) -> f2(I15, I13, I14) [I15 = I15 /\ I14 <= 0 /\ 0 <= -1 + I13]
  f1(I16, I17, I18) -> f2(I19, I17, I18) [I19 = I19 /\ I17 <= 0]

The dependency graph for this problem is:
  2 -> 
Where:
  2) f3#(I3, I4, I5) -> f1#(I3, I4, I5) 

We have the following SCCs.