/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) -> 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, I6, I7, -1 + I8, I9) [1 <= I9 /\ -1 + I9 <= -1 + I8 /\ -1 + I8 <= -1 + I9 /\ 1 <= I8]
  f1#(I15, I16, I17, I18, I19) -> f2#(rnd1, I16, I17, rnd4, I19) [y1 = y1 /\ rnd4 = y1 /\ rnd1 = rnd1]
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, I6, I7, -1 + I8, I9) [1 <= I9 /\ -1 + I9 <= -1 + I8 /\ -1 + I8 <= -1 + I9 /\ 1 <= I8]
  f2(I10, I11, I12, I13, I14) -> f3(I10, I12, I12, I13, I14) [I13 <= 0]
  f1(I15, I16, I17, I18, I19) -> f2(rnd1, I16, I17, rnd4, I19) [y1 = y1 /\ rnd4 = y1 /\ rnd1 = rnd1]

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, I6, I7, -1 + I8, I9) [1 <= I9 /\ -1 + I9 <= -1 + I8 /\ -1 + I8 <= -1 + I9 /\ 1 <= I8]
  3) f1#(I15, I16, I17, I18, I19) -> f2#(rnd1, I16, I17, rnd4, I19) [y1 = y1 /\ rnd4 = y1 /\ rnd1 = rnd1]

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, I6, I7, -1 + I8, I9) [1 <= I9 /\ -1 + I9 <= -1 + I8 /\ -1 + I8 <= -1 + I9 /\ 1 <= I8]
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, I6, I7, -1 + I8, I9) [1 <= I9 /\ -1 + I9 <= -1 + I8 /\ -1 + I8 <= -1 + I9 /\ 1 <= I8]
  f2(I10, I11, I12, I13, I14) -> f3(I10, I12, I12, I13, I14) [I13 <= 0]
  f1(I15, I16, I17, I18, I19) -> f2(rnd1, I16, I17, rnd4, I19) [y1 = y1 /\ rnd4 = y1 /\ rnd1 = rnd1]

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

This gives the following inequalities:
    ==>  I3 (>! \union =) I3
  1 <= I9 /\ -1 + I9 <= -1 + I8 /\ -1 + I8 <= -1 + I9 /\ 1 <= I8  ==>  I8 >! -1 + I8

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, I6, I7, -1 + I8, I9) [1 <= I9 /\ -1 + I9 <= -1 + I8 /\ -1 + I8 <= -1 + I9 /\ 1 <= I8]
  f2(I10, I11, I12, I13, I14) -> f3(I10, I12, I12, I13, I14) [I13 <= 0]
  f1(I15, I16, I17, I18, I19) -> f2(rnd1, I16, I17, rnd4, I19) [y1 = y1 /\ rnd4 = y1 /\ rnd1 = rnd1]

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.