/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) -> f1#(x1, x2, x3, x4) 
  f4#(I0, I1, I2, I3) -> f2#(I0, I1, I2, I3) 
  f2#(I4, I5, I6, I7) -> f4#(1 + I4, I5, I6, I7) [1 + I5 <= 10 /\ 1 + I5 <= 1 + I4 /\ 1 + I4 <= 1 + I5 /\ 1 + I4 <= 10]
  f1#(I12, I13, I14, I15) -> f2#(rnd1, I13, I14, I15) [y1 = 0 /\ 0 <= y1 /\ y1 <= 0 /\ 0 <= y1 /\ y1 <= 0 /\ 1 + y1 <= 10 /\ y2 = 1 + y1 /\ 1 <= y2 /\ y2 <= 1 /\ 1 <= y2 /\ y2 <= 1 /\ 1 + y2 <= 10 /\ rnd1 = 1 + y2 /\ 2 <= rnd1 /\ rnd1 <= 2]
R = 
  f5(x1, x2, x3, x4) -> f1(x1, x2, x3, x4) 
  f4(I0, I1, I2, I3) -> f2(I0, I1, I2, I3) 
  f2(I4, I5, I6, I7) -> f4(1 + I4, I5, I6, I7) [1 + I5 <= 10 /\ 1 + I5 <= 1 + I4 /\ 1 + I4 <= 1 + I5 /\ 1 + I4 <= 10]
  f2(I8, I9, I10, I11) -> f3(I8, I9, I11, I11) [10 <= I8]
  f1(I12, I13, I14, I15) -> f2(rnd1, I13, I14, I15) [y1 = 0 /\ 0 <= y1 /\ y1 <= 0 /\ 0 <= y1 /\ y1 <= 0 /\ 1 + y1 <= 10 /\ y2 = 1 + y1 /\ 1 <= y2 /\ y2 <= 1 /\ 1 <= y2 /\ y2 <= 1 /\ 1 + y2 <= 10 /\ rnd1 = 1 + y2 /\ 2 <= rnd1 /\ rnd1 <= 2]

The dependency graph for this problem is:
  0 -> 3
  1 -> 2
  2 -> 1
  3 -> 2
Where:
  0) f5#(x1, x2, x3, x4) -> f1#(x1, x2, x3, x4) 
  1) f4#(I0, I1, I2, I3) -> f2#(I0, I1, I2, I3) 
  2) f2#(I4, I5, I6, I7) -> f4#(1 + I4, I5, I6, I7) [1 + I5 <= 10 /\ 1 + I5 <= 1 + I4 /\ 1 + I4 <= 1 + I5 /\ 1 + I4 <= 10]
  3) f1#(I12, I13, I14, I15) -> f2#(rnd1, I13, I14, I15) [y1 = 0 /\ 0 <= y1 /\ y1 <= 0 /\ 0 <= y1 /\ y1 <= 0 /\ 1 + y1 <= 10 /\ y2 = 1 + y1 /\ 1 <= y2 /\ y2 <= 1 /\ 1 <= y2 /\ y2 <= 1 /\ 1 + y2 <= 10 /\ rnd1 = 1 + y2 /\ 2 <= rnd1 /\ rnd1 <= 2]

We have the following SCCs.
  { 1, 2 }

DP problem for innermost termination.
P =
  f4#(I0, I1, I2, I3) -> f2#(I0, I1, I2, I3) 
  f2#(I4, I5, I6, I7) -> f4#(1 + I4, I5, I6, I7) [1 + I5 <= 10 /\ 1 + I5 <= 1 + I4 /\ 1 + I4 <= 1 + I5 /\ 1 + I4 <= 10]
R = 
  f5(x1, x2, x3, x4) -> f1(x1, x2, x3, x4) 
  f4(I0, I1, I2, I3) -> f2(I0, I1, I2, I3) 
  f2(I4, I5, I6, I7) -> f4(1 + I4, I5, I6, I7) [1 + I5 <= 10 /\ 1 + I5 <= 1 + I4 /\ 1 + I4 <= 1 + I5 /\ 1 + I4 <= 10]
  f2(I8, I9, I10, I11) -> f3(I8, I9, I11, I11) [10 <= I8]
  f1(I12, I13, I14, I15) -> f2(rnd1, I13, I14, I15) [y1 = 0 /\ 0 <= y1 /\ y1 <= 0 /\ 0 <= y1 /\ y1 <= 0 /\ 1 + y1 <= 10 /\ y2 = 1 + y1 /\ 1 <= y2 /\ y2 <= 1 /\ 1 <= y2 /\ y2 <= 1 /\ 1 + y2 <= 10 /\ rnd1 = 1 + y2 /\ 2 <= rnd1 /\ rnd1 <= 2]

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

This gives the following inequalities:
    ==>  1 + I1 + -1 * (1 + I0) >= 1 + I1 + -1 * (1 + I0)
  1 + I5 <= 10 /\ 1 + I5 <= 1 + I4 /\ 1 + I4 <= 1 + I5 /\ 1 + I4 <= 10  ==>  1 + I5 + -1 * (1 + I4) > 1 + I5 + -1 * (1 + (1 + I4))  with  1 + I5 + -1 * (1 + I4) >= 0

We remove all the strictly oriented dependency pairs.

DP problem for innermost termination.
P =
  f4#(I0, I1, I2, I3) -> f2#(I0, I1, I2, I3) 
R = 
  f5(x1, x2, x3, x4) -> f1(x1, x2, x3, x4) 
  f4(I0, I1, I2, I3) -> f2(I0, I1, I2, I3) 
  f2(I4, I5, I6, I7) -> f4(1 + I4, I5, I6, I7) [1 + I5 <= 10 /\ 1 + I5 <= 1 + I4 /\ 1 + I4 <= 1 + I5 /\ 1 + I4 <= 10]
  f2(I8, I9, I10, I11) -> f3(I8, I9, I11, I11) [10 <= I8]
  f1(I12, I13, I14, I15) -> f2(rnd1, I13, I14, I15) [y1 = 0 /\ 0 <= y1 /\ y1 <= 0 /\ 0 <= y1 /\ y1 <= 0 /\ 1 + y1 <= 10 /\ y2 = 1 + y1 /\ 1 <= y2 /\ y2 <= 1 /\ 1 <= y2 /\ y2 <= 1 /\ 1 + y2 <= 10 /\ rnd1 = 1 + y2 /\ 2 <= rnd1 /\ rnd1 <= 2]

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

We have the following SCCs.