/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 =
  f5#(x1, x2, x3, x4, x5) -> f4#(x1, x2, x3, x4, x5) 
  f4#(I0, I1, I2, I3, I4) -> f1#(I0, I1, I2, 1000, I4) 
  f3#(I5, I6, I7, I8, I9) -> f1#(I5, I6, I7, I8, I9) 
  f1#(I10, I11, I12, I13, I14) -> f3#(I10, I11, I12, -1 + I13, I14) [0 <= -101 + I13 /\ 0 <= 9 - I14]
R = 
  f5(x1, x2, x3, x4, x5) -> f4(x1, x2, x3, x4, x5) 
  f4(I0, I1, I2, I3, I4) -> f1(I0, I1, I2, 1000, I4) 
  f3(I5, I6, I7, I8, I9) -> f1(I5, I6, I7, I8, I9) 
  f1(I10, I11, I12, I13, I14) -> f3(I10, I11, I12, -1 + I13, I14) [0 <= -101 + I13 /\ 0 <= 9 - I14]
  f1(I15, I16, I17, I18, I19) -> f2(rnd1, rnd2, 0, I18, I19) [rnd1 = rnd2 /\ rnd2 = 0 /\ -100 + I18 <= 0 /\ 0 <= 9 - I19]
  f1(I20, I21, I22, I23, I24) -> f2(I25, I26, 0, I23, I24) [I25 = I26 /\ I26 = 0 /\ 10 - I24 <= 0]

The dependency graph for this problem is:
  0 -> 1
  1 -> 3
  2 -> 3
  3 -> 2
Where:
  0) f5#(x1, x2, x3, x4, x5) -> f4#(x1, x2, x3, x4, x5) 
  1) f4#(I0, I1, I2, I3, I4) -> f1#(I0, I1, I2, 1000, I4) 
  2) f3#(I5, I6, I7, I8, I9) -> f1#(I5, I6, I7, I8, I9) 
  3) f1#(I10, I11, I12, I13, I14) -> f3#(I10, I11, I12, -1 + I13, I14) [0 <= -101 + I13 /\ 0 <= 9 - I14]

We have the following SCCs.
  { 2, 3 }

DP problem for innermost termination.
P =
  f3#(I5, I6, I7, I8, I9) -> f1#(I5, I6, I7, I8, I9) 
  f1#(I10, I11, I12, I13, I14) -> f3#(I10, I11, I12, -1 + I13, I14) [0 <= -101 + I13 /\ 0 <= 9 - I14]
R = 
  f5(x1, x2, x3, x4, x5) -> f4(x1, x2, x3, x4, x5) 
  f4(I0, I1, I2, I3, I4) -> f1(I0, I1, I2, 1000, I4) 
  f3(I5, I6, I7, I8, I9) -> f1(I5, I6, I7, I8, I9) 
  f1(I10, I11, I12, I13, I14) -> f3(I10, I11, I12, -1 + I13, I14) [0 <= -101 + I13 /\ 0 <= 9 - I14]
  f1(I15, I16, I17, I18, I19) -> f2(rnd1, rnd2, 0, I18, I19) [rnd1 = rnd2 /\ rnd2 = 0 /\ -100 + I18 <= 0 /\ 0 <= 9 - I19]
  f1(I20, I21, I22, I23, I24) -> f2(I25, I26, 0, I23, I24) [I25 = I26 /\ I26 = 0 /\ 10 - I24 <= 0]

We use the basic value criterion with the projection function NU:
  NU[f1#(z1,z2,z3,z4,z5)] = z4
  NU[f3#(z1,z2,z3,z4,z5)] = z4

This gives the following inequalities:
    ==>  I8 (>! \union =) I8
  0 <= -101 + I13 /\ 0 <= 9 - I14  ==>  I13 >! -1 + I13

We remove all the strictly oriented dependency pairs.

DP problem for innermost termination.
P =
  f3#(I5, I6, I7, I8, I9) -> f1#(I5, I6, I7, I8, I9) 
R = 
  f5(x1, x2, x3, x4, x5) -> f4(x1, x2, x3, x4, x5) 
  f4(I0, I1, I2, I3, I4) -> f1(I0, I1, I2, 1000, I4) 
  f3(I5, I6, I7, I8, I9) -> f1(I5, I6, I7, I8, I9) 
  f1(I10, I11, I12, I13, I14) -> f3(I10, I11, I12, -1 + I13, I14) [0 <= -101 + I13 /\ 0 <= 9 - I14]
  f1(I15, I16, I17, I18, I19) -> f2(rnd1, rnd2, 0, I18, I19) [rnd1 = rnd2 /\ rnd2 = 0 /\ -100 + I18 <= 0 /\ 0 <= 9 - I19]
  f1(I20, I21, I22, I23, I24) -> f2(I25, I26, 0, I23, I24) [I25 = I26 /\ I26 = 0 /\ 10 - I24 <= 0]

The dependency graph for this problem is:
  2 -> 
Where:
  2) f3#(I5, I6, I7, I8, I9) -> f1#(I5, I6, I7, I8, I9) 

We have the following SCCs.