/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 =
  f6#(x1, x2, x3) -> f4#(x1, x2, x3) 
  f5#(I0, I1, I2) -> f3#(I0, I1, I2) 
  f3#(I3, I4, I5) -> f5#(I3, I4, -1 + I5) [0 <= -31 + I5]
  f3#(I6, I7, I8) -> f1#(I6, -1 + I7, I8) [-30 + I8 <= 0]
  f4#(I9, I10, I11) -> f1#(I9, I10, I11) 
  f1#(I12, I13, I14) -> f3#(I12, I13, I14) [0 <= -21 + I13]
R = 
  f6(x1, x2, x3) -> f4(x1, x2, x3) 
  f5(I0, I1, I2) -> f3(I0, I1, I2) 
  f3(I3, I4, I5) -> f5(I3, I4, -1 + I5) [0 <= -31 + I5]
  f3(I6, I7, I8) -> f1(I6, -1 + I7, I8) [-30 + I8 <= 0]
  f4(I9, I10, I11) -> f1(I9, I10, I11) 
  f1(I12, I13, I14) -> f3(I12, I13, I14) [0 <= -21 + I13]
  f1(I15, I16, I17) -> f2(rnd1, I16, I17) [rnd1 = rnd1 /\ -20 + I16 <= 0]

The dependency graph for this problem is:
  0 -> 4
  1 -> 2, 3
  2 -> 1
  3 -> 5
  4 -> 5
  5 -> 2, 3
Where:
  0) f6#(x1, x2, x3) -> f4#(x1, x2, x3) 
  1) f5#(I0, I1, I2) -> f3#(I0, I1, I2) 
  2) f3#(I3, I4, I5) -> f5#(I3, I4, -1 + I5) [0 <= -31 + I5]
  3) f3#(I6, I7, I8) -> f1#(I6, -1 + I7, I8) [-30 + I8 <= 0]
  4) f4#(I9, I10, I11) -> f1#(I9, I10, I11) 
  5) f1#(I12, I13, I14) -> f3#(I12, I13, I14) [0 <= -21 + I13]

We have the following SCCs.
  { 1, 2, 3, 5 }

DP problem for innermost termination.
P =
  f5#(I0, I1, I2) -> f3#(I0, I1, I2) 
  f3#(I3, I4, I5) -> f5#(I3, I4, -1 + I5) [0 <= -31 + I5]
  f3#(I6, I7, I8) -> f1#(I6, -1 + I7, I8) [-30 + I8 <= 0]
  f1#(I12, I13, I14) -> f3#(I12, I13, I14) [0 <= -21 + I13]
R = 
  f6(x1, x2, x3) -> f4(x1, x2, x3) 
  f5(I0, I1, I2) -> f3(I0, I1, I2) 
  f3(I3, I4, I5) -> f5(I3, I4, -1 + I5) [0 <= -31 + I5]
  f3(I6, I7, I8) -> f1(I6, -1 + I7, I8) [-30 + I8 <= 0]
  f4(I9, I10, I11) -> f1(I9, I10, I11) 
  f1(I12, I13, I14) -> f3(I12, I13, I14) [0 <= -21 + I13]
  f1(I15, I16, I17) -> f2(rnd1, I16, I17) [rnd1 = rnd1 /\ -20 + I16 <= 0]

We use the basic value criterion with the projection function NU:
  NU[f1#(z1,z2,z3)] = z3
  NU[f3#(z1,z2,z3)] = z3
  NU[f5#(z1,z2,z3)] = z3

This gives the following inequalities:
    ==>  I2 (>! \union =) I2
  0 <= -31 + I5  ==>  I5 >! -1 + I5
  -30 + I8 <= 0  ==>  I8 (>! \union =) I8
  0 <= -21 + I13  ==>  I14 (>! \union =) I14

We remove all the strictly oriented dependency pairs.

DP problem for innermost termination.
P =
  f5#(I0, I1, I2) -> f3#(I0, I1, I2) 
  f3#(I6, I7, I8) -> f1#(I6, -1 + I7, I8) [-30 + I8 <= 0]
  f1#(I12, I13, I14) -> f3#(I12, I13, I14) [0 <= -21 + I13]
R = 
  f6(x1, x2, x3) -> f4(x1, x2, x3) 
  f5(I0, I1, I2) -> f3(I0, I1, I2) 
  f3(I3, I4, I5) -> f5(I3, I4, -1 + I5) [0 <= -31 + I5]
  f3(I6, I7, I8) -> f1(I6, -1 + I7, I8) [-30 + I8 <= 0]
  f4(I9, I10, I11) -> f1(I9, I10, I11) 
  f1(I12, I13, I14) -> f3(I12, I13, I14) [0 <= -21 + I13]
  f1(I15, I16, I17) -> f2(rnd1, I16, I17) [rnd1 = rnd1 /\ -20 + I16 <= 0]

The dependency graph for this problem is:
  1 -> 3
  3 -> 5
  5 -> 3
Where:
  1) f5#(I0, I1, I2) -> f3#(I0, I1, I2) 
  3) f3#(I6, I7, I8) -> f1#(I6, -1 + I7, I8) [-30 + I8 <= 0]
  5) f1#(I12, I13, I14) -> f3#(I12, I13, I14) [0 <= -21 + I13]

We have the following SCCs.
  { 3, 5 }

DP problem for innermost termination.
P =
  f3#(I6, I7, I8) -> f1#(I6, -1 + I7, I8) [-30 + I8 <= 0]
  f1#(I12, I13, I14) -> f3#(I12, I13, I14) [0 <= -21 + I13]
R = 
  f6(x1, x2, x3) -> f4(x1, x2, x3) 
  f5(I0, I1, I2) -> f3(I0, I1, I2) 
  f3(I3, I4, I5) -> f5(I3, I4, -1 + I5) [0 <= -31 + I5]
  f3(I6, I7, I8) -> f1(I6, -1 + I7, I8) [-30 + I8 <= 0]
  f4(I9, I10, I11) -> f1(I9, I10, I11) 
  f1(I12, I13, I14) -> f3(I12, I13, I14) [0 <= -21 + I13]
  f1(I15, I16, I17) -> f2(rnd1, I16, I17) [rnd1 = rnd1 /\ -20 + I16 <= 0]

We use the reverse value criterion with the projection function NU:
  NU[f1#(z1,z2,z3)] = -21 + z2 + -1 * 0
  NU[f3#(z1,z2,z3)] = -21 + (-1 + z2) + -1 * 0

This gives the following inequalities:
  -30 + I8 <= 0  ==>  -21 + (-1 + I7) + -1 * 0 >= -21 + (-1 + I7) + -1 * 0
  0 <= -21 + I13  ==>  -21 + I13 + -1 * 0 > -21 + (-1 + I13) + -1 * 0  with  -21 + I13 + -1 * 0 >= 0

We remove all the strictly oriented dependency pairs.

DP problem for innermost termination.
P =
  f3#(I6, I7, I8) -> f1#(I6, -1 + I7, I8) [-30 + I8 <= 0]
R = 
  f6(x1, x2, x3) -> f4(x1, x2, x3) 
  f5(I0, I1, I2) -> f3(I0, I1, I2) 
  f3(I3, I4, I5) -> f5(I3, I4, -1 + I5) [0 <= -31 + I5]
  f3(I6, I7, I8) -> f1(I6, -1 + I7, I8) [-30 + I8 <= 0]
  f4(I9, I10, I11) -> f1(I9, I10, I11) 
  f1(I12, I13, I14) -> f3(I12, I13, I14) [0 <= -21 + I13]
  f1(I15, I16, I17) -> f2(rnd1, I16, I17) [rnd1 = rnd1 /\ -20 + I16 <= 0]

The dependency graph for this problem is:
  3 -> 
Where:
  3) f3#(I6, I7, I8) -> f1#(I6, -1 + I7, I8) [-30 + I8 <= 0]

We have the following SCCs.