/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, x3, x4, x5) -> f1#(x1, x2, x3, x4, x5) 
  f6#(I0, I1, I2, I3, I4) -> f2#(I0, I1, I2, I3, I4) 
  f2#(I5, I6, I7, I8, I9) -> f6#(I5, 1 + I6, rnd3, I8, I9) [1 + rnd3 <= 10 /\ 1 + rnd3 <= 1 + I6 /\ 1 + I6 <= 1 + rnd3 /\ 1 + I6 <= 10 /\ rnd3 = rnd3]
  f4#(I15, I16, I17, I18, I19) -> f2#(I15, 1 + I16, I17, I18, I19) [1 + I16 <= 1 /\ 1 <= 1 + I16 /\ 1 + I16 <= 10]
  f3#(I20, I21, I22, I23, I24) -> f2#(rnd1, rnd2, I22, I23, I24) [1 + I21 <= 10 /\ y1 = 1 + I21 /\ rnd2 = rnd2 /\ 2 <= rnd2 /\ rnd2 <= 2 /\ rnd1 = rnd1 /\ rnd1 <= rnd2 /\ rnd2 <= rnd1]
  f1#(I25, I26, I27, I28, I29) -> f2#(I25, 0, I27, I28, I29) [0 <= 0 /\ 0 <= 0]
R = 
  f7(x1, x2, x3, x4, x5) -> f1(x1, x2, x3, x4, x5) 
  f6(I0, I1, I2, I3, I4) -> f2(I0, I1, I2, I3, I4) 
  f2(I5, I6, I7, I8, I9) -> f6(I5, 1 + I6, rnd3, I8, I9) [1 + rnd3 <= 10 /\ 1 + rnd3 <= 1 + I6 /\ 1 + I6 <= 1 + rnd3 /\ 1 + I6 <= 10 /\ rnd3 = rnd3]
  f2(I10, I11, I12, I13, I14) -> f5(I10, I11, I12, I14, I14) [10 <= I11]
  f4(I15, I16, I17, I18, I19) -> f2(I15, 1 + I16, I17, I18, I19) [1 + I16 <= 1 /\ 1 <= 1 + I16 /\ 1 + I16 <= 10]
  f3(I20, I21, I22, I23, I24) -> f2(rnd1, rnd2, I22, I23, I24) [1 + I21 <= 10 /\ y1 = 1 + I21 /\ rnd2 = rnd2 /\ 2 <= rnd2 /\ rnd2 <= 2 /\ rnd1 = rnd1 /\ rnd1 <= rnd2 /\ rnd2 <= rnd1]
  f1(I25, I26, I27, I28, I29) -> f2(I25, 0, I27, I28, I29) [0 <= 0 /\ 0 <= 0]

The dependency graph for this problem is:
  0 -> 5
  1 -> 2
  2 -> 1
  3 -> 2
  4 -> 2
  5 -> 2
Where:
  0) f7#(x1, x2, x3, x4, x5) -> f1#(x1, x2, x3, x4, x5) 
  1) f6#(I0, I1, I2, I3, I4) -> f2#(I0, I1, I2, I3, I4) 
  2) f2#(I5, I6, I7, I8, I9) -> f6#(I5, 1 + I6, rnd3, I8, I9) [1 + rnd3 <= 10 /\ 1 + rnd3 <= 1 + I6 /\ 1 + I6 <= 1 + rnd3 /\ 1 + I6 <= 10 /\ rnd3 = rnd3]
  3) f4#(I15, I16, I17, I18, I19) -> f2#(I15, 1 + I16, I17, I18, I19) [1 + I16 <= 1 /\ 1 <= 1 + I16 /\ 1 + I16 <= 10]
  4) f3#(I20, I21, I22, I23, I24) -> f2#(rnd1, rnd2, I22, I23, I24) [1 + I21 <= 10 /\ y1 = 1 + I21 /\ rnd2 = rnd2 /\ 2 <= rnd2 /\ rnd2 <= 2 /\ rnd1 = rnd1 /\ rnd1 <= rnd2 /\ rnd2 <= rnd1]
  5) f1#(I25, I26, I27, I28, I29) -> f2#(I25, 0, I27, I28, I29) [0 <= 0 /\ 0 <= 0]

We have the following SCCs.
  { 1, 2 }

DP problem for innermost termination.
P =
  f6#(I0, I1, I2, I3, I4) -> f2#(I0, I1, I2, I3, I4) 
  f2#(I5, I6, I7, I8, I9) -> f6#(I5, 1 + I6, rnd3, I8, I9) [1 + rnd3 <= 10 /\ 1 + rnd3 <= 1 + I6 /\ 1 + I6 <= 1 + rnd3 /\ 1 + I6 <= 10 /\ rnd3 = rnd3]
R = 
  f7(x1, x2, x3, x4, x5) -> f1(x1, x2, x3, x4, x5) 
  f6(I0, I1, I2, I3, I4) -> f2(I0, I1, I2, I3, I4) 
  f2(I5, I6, I7, I8, I9) -> f6(I5, 1 + I6, rnd3, I8, I9) [1 + rnd3 <= 10 /\ 1 + rnd3 <= 1 + I6 /\ 1 + I6 <= 1 + rnd3 /\ 1 + I6 <= 10 /\ rnd3 = rnd3]
  f2(I10, I11, I12, I13, I14) -> f5(I10, I11, I12, I14, I14) [10 <= I11]
  f4(I15, I16, I17, I18, I19) -> f2(I15, 1 + I16, I17, I18, I19) [1 + I16 <= 1 /\ 1 <= 1 + I16 /\ 1 + I16 <= 10]
  f3(I20, I21, I22, I23, I24) -> f2(rnd1, rnd2, I22, I23, I24) [1 + I21 <= 10 /\ y1 = 1 + I21 /\ rnd2 = rnd2 /\ 2 <= rnd2 /\ rnd2 <= 2 /\ rnd1 = rnd1 /\ rnd1 <= rnd2 /\ rnd2 <= rnd1]
  f1(I25, I26, I27, I28, I29) -> f2(I25, 0, I27, I28, I29) [0 <= 0 /\ 0 <= 0]

We use the reverse value criterion with the projection function NU:
  NU[f2#(z1,z2,z3,z4,z5)] = 10 + -1 * (1 + z2)
  NU[f6#(z1,z2,z3,z4,z5)] = 10 + -1 * (1 + z2)

This gives the following inequalities:
    ==>  10 + -1 * (1 + I1) >= 10 + -1 * (1 + I1)
  1 + rnd3 <= 10 /\ 1 + rnd3 <= 1 + I6 /\ 1 + I6 <= 1 + rnd3 /\ 1 + I6 <= 10 /\ rnd3 = rnd3  ==>  10 + -1 * (1 + I6) > 10 + -1 * (1 + (1 + I6))  with  10 + -1 * (1 + I6) >= 0

We remove all the strictly oriented dependency pairs.

DP problem for innermost termination.
P =
  f6#(I0, I1, I2, I3, I4) -> f2#(I0, I1, I2, I3, I4) 
R = 
  f7(x1, x2, x3, x4, x5) -> f1(x1, x2, x3, x4, x5) 
  f6(I0, I1, I2, I3, I4) -> f2(I0, I1, I2, I3, I4) 
  f2(I5, I6, I7, I8, I9) -> f6(I5, 1 + I6, rnd3, I8, I9) [1 + rnd3 <= 10 /\ 1 + rnd3 <= 1 + I6 /\ 1 + I6 <= 1 + rnd3 /\ 1 + I6 <= 10 /\ rnd3 = rnd3]
  f2(I10, I11, I12, I13, I14) -> f5(I10, I11, I12, I14, I14) [10 <= I11]
  f4(I15, I16, I17, I18, I19) -> f2(I15, 1 + I16, I17, I18, I19) [1 + I16 <= 1 /\ 1 <= 1 + I16 /\ 1 + I16 <= 10]
  f3(I20, I21, I22, I23, I24) -> f2(rnd1, rnd2, I22, I23, I24) [1 + I21 <= 10 /\ y1 = 1 + I21 /\ rnd2 = rnd2 /\ 2 <= rnd2 /\ rnd2 <= 2 /\ rnd1 = rnd1 /\ rnd1 <= rnd2 /\ rnd2 <= rnd1]
  f1(I25, I26, I27, I28, I29) -> f2(I25, 0, I27, I28, I29) [0 <= 0 /\ 0 <= 0]

The dependency graph for this problem is:
  1 -> 
Where:
  1) f6#(I0, I1, I2, I3, I4) -> f2#(I0, I1, I2, I3, I4) 

We have the following SCCs.