/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, x4, x5) -> f1#(x1, x2, x3, x4, x5) 
  f4#(I0, I1, I2, I3, I4) -> f2#(I0, I1, I2, I3, I4) 
  f2#(I5, I6, I7, I8, I9) -> f4#(I5, 1 + I6, I7, I8, I9) [1 + I7 <= I5 /\ 1 + I7 <= 1 + I6 /\ 1 + I6 <= 1 + I7 /\ 1 + I6 <= I5]
  f1#(I15, I16, I17, I18, I19) -> f2#(I15, rnd2, I17, I18, I19) [y1 = 0 /\ 0 <= y1 /\ y1 <= 0 /\ 0 <= y1 /\ y1 <= 0 /\ 1 + y1 <= I15 /\ y2 = 1 + y1 /\ 1 <= y2 /\ y2 <= 1 /\ 1 <= y2 /\ y2 <= 1 /\ 1 + y2 <= I15 /\ rnd2 = 1 + y2 /\ 2 <= rnd2 /\ rnd2 <= 2]
R = 
  f5(x1, x2, x3, x4, x5) -> f1(x1, x2, x3, x4, x5) 
  f4(I0, I1, I2, I3, I4) -> f2(I0, I1, I2, I3, I4) 
  f2(I5, I6, I7, I8, I9) -> f4(I5, 1 + I6, I7, I8, I9) [1 + I7 <= I5 /\ 1 + I7 <= 1 + I6 /\ 1 + I6 <= 1 + I7 /\ 1 + I6 <= I5]
  f2(I10, I11, I12, I13, I14) -> f3(I10, I11, I12, I14, I14) [I10 <= I11]
  f1(I15, I16, I17, I18, I19) -> f2(I15, rnd2, I17, I18, I19) [y1 = 0 /\ 0 <= y1 /\ y1 <= 0 /\ 0 <= y1 /\ y1 <= 0 /\ 1 + y1 <= I15 /\ y2 = 1 + y1 /\ 1 <= y2 /\ y2 <= 1 /\ 1 <= y2 /\ y2 <= 1 /\ 1 + y2 <= I15 /\ rnd2 = 1 + y2 /\ 2 <= rnd2 /\ rnd2 <= 2]

The dependency graph for this problem is:
  0 -> 3
  1 -> 2
  2 -> 1
  3 -> 2
Where:
  0) f5#(x1, x2, x3, x4, x5) -> f1#(x1, x2, x3, x4, x5) 
  1) f4#(I0, I1, I2, I3, I4) -> f2#(I0, I1, I2, I3, I4) 
  2) f2#(I5, I6, I7, I8, I9) -> f4#(I5, 1 + I6, I7, I8, I9) [1 + I7 <= I5 /\ 1 + I7 <= 1 + I6 /\ 1 + I6 <= 1 + I7 /\ 1 + I6 <= I5]
  3) f1#(I15, I16, I17, I18, I19) -> f2#(I15, rnd2, I17, I18, I19) [y1 = 0 /\ 0 <= y1 /\ y1 <= 0 /\ 0 <= y1 /\ y1 <= 0 /\ 1 + y1 <= I15 /\ y2 = 1 + y1 /\ 1 <= y2 /\ y2 <= 1 /\ 1 <= y2 /\ y2 <= 1 /\ 1 + y2 <= I15 /\ rnd2 = 1 + y2 /\ 2 <= rnd2 /\ rnd2 <= 2]

We have the following SCCs.
  { 1, 2 }

DP problem for innermost termination.
P =
  f4#(I0, I1, I2, I3, I4) -> f2#(I0, I1, I2, I3, I4) 
  f2#(I5, I6, I7, I8, I9) -> f4#(I5, 1 + I6, I7, I8, I9) [1 + I7 <= I5 /\ 1 + I7 <= 1 + I6 /\ 1 + I6 <= 1 + I7 /\ 1 + I6 <= I5]
R = 
  f5(x1, x2, x3, x4, x5) -> f1(x1, x2, x3, x4, x5) 
  f4(I0, I1, I2, I3, I4) -> f2(I0, I1, I2, I3, I4) 
  f2(I5, I6, I7, I8, I9) -> f4(I5, 1 + I6, I7, I8, I9) [1 + I7 <= I5 /\ 1 + I7 <= 1 + I6 /\ 1 + I6 <= 1 + I7 /\ 1 + I6 <= I5]
  f2(I10, I11, I12, I13, I14) -> f3(I10, I11, I12, I14, I14) [I10 <= I11]
  f1(I15, I16, I17, I18, I19) -> f2(I15, rnd2, I17, I18, I19) [y1 = 0 /\ 0 <= y1 /\ y1 <= 0 /\ 0 <= y1 /\ y1 <= 0 /\ 1 + y1 <= I15 /\ y2 = 1 + y1 /\ 1 <= y2 /\ y2 <= 1 /\ 1 <= y2 /\ y2 <= 1 /\ 1 + y2 <= I15 /\ rnd2 = 1 + y2 /\ 2 <= rnd2 /\ rnd2 <= 2]

We use the reverse value criterion with the projection function NU:
  NU[f2#(z1,z2,z3,z4,z5)] = 1 + z3 + -1 * (1 + z2)
  NU[f4#(z1,z2,z3,z4,z5)] = 1 + z3 + -1 * (1 + z2)

This gives the following inequalities:
    ==>  1 + I2 + -1 * (1 + I1) >= 1 + I2 + -1 * (1 + I1)
  1 + I7 <= I5 /\ 1 + I7 <= 1 + I6 /\ 1 + I6 <= 1 + I7 /\ 1 + I6 <= I5  ==>  1 + I7 + -1 * (1 + I6) > 1 + I7 + -1 * (1 + (1 + I6))  with  1 + I7 + -1 * (1 + I6) >= 0

We remove all the strictly oriented dependency pairs.

DP problem for innermost termination.
P =
  f4#(I0, I1, I2, I3, I4) -> f2#(I0, I1, I2, I3, I4) 
R = 
  f5(x1, x2, x3, x4, x5) -> f1(x1, x2, x3, x4, x5) 
  f4(I0, I1, I2, I3, I4) -> f2(I0, I1, I2, I3, I4) 
  f2(I5, I6, I7, I8, I9) -> f4(I5, 1 + I6, I7, I8, I9) [1 + I7 <= I5 /\ 1 + I7 <= 1 + I6 /\ 1 + I6 <= 1 + I7 /\ 1 + I6 <= I5]
  f2(I10, I11, I12, I13, I14) -> f3(I10, I11, I12, I14, I14) [I10 <= I11]
  f1(I15, I16, I17, I18, I19) -> f2(I15, rnd2, I17, I18, I19) [y1 = 0 /\ 0 <= y1 /\ y1 <= 0 /\ 0 <= y1 /\ y1 <= 0 /\ 1 + y1 <= I15 /\ y2 = 1 + y1 /\ 1 <= y2 /\ y2 <= 1 /\ 1 <= y2 /\ y2 <= 1 /\ 1 + y2 <= I15 /\ rnd2 = 1 + y2 /\ 2 <= rnd2 /\ rnd2 <= 2]

The dependency graph for this problem is:
  1 -> 
Where:
  1) f4#(I0, I1, I2, I3, I4) -> f2#(I0, I1, I2, I3, I4) 

We have the following SCCs.