/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) -> f6#(x1, x2, x3) 
  f6#(I0, I1, I2) -> f2#(I0, I1, I2) [I0 <= 0]
  f2#(I3, I4, I5) -> f5#(I3, I4, -1 + I5) [1 <= I5]
  f5#(I6, I7, I8) -> f1#(I6, I7, I8) [1 <= I6]
  f5#(I9, I10, I11) -> f4#(I9, I10, I11) [I9 <= 0]
  f4#(I12, I13, I14) -> f2#(1, I14, I14) 
  f4#(I15, I16, I17) -> f2#(I15, I16, I17) [I15 <= 0]
  f1#(I21, I22, I23) -> f2#(I21, I22, I23) [1 + I23 <= I22]
R = 
  f7(x1, x2, x3) -> f6(x1, x2, x3) 
  f6(I0, I1, I2) -> f2(I0, I1, I2) [I0 <= 0]
  f2(I3, I4, I5) -> f5(I3, I4, -1 + I5) [1 <= I5]
  f5(I6, I7, I8) -> f1(I6, I7, I8) [1 <= I6]
  f5(I9, I10, I11) -> f4(I9, I10, I11) [I9 <= 0]
  f4(I12, I13, I14) -> f2(1, I14, I14) 
  f4(I15, I16, I17) -> f2(I15, I16, I17) [I15 <= 0]
  f1(I18, I19, I20) -> f3(I18, I19, I20) [I19 <= I20]
  f1(I21, I22, I23) -> f2(I21, I22, I23) [1 + I23 <= I22]

The dependency graph for this problem is:
  0 -> 1
  1 -> 2
  2 -> 3, 4
  3 -> 7
  4 -> 5, 6
  5 -> 2
  6 -> 2
  7 -> 2
Where:
  0) f7#(x1, x2, x3) -> f6#(x1, x2, x3) 
  1) f6#(I0, I1, I2) -> f2#(I0, I1, I2) [I0 <= 0]
  2) f2#(I3, I4, I5) -> f5#(I3, I4, -1 + I5) [1 <= I5]
  3) f5#(I6, I7, I8) -> f1#(I6, I7, I8) [1 <= I6]
  4) f5#(I9, I10, I11) -> f4#(I9, I10, I11) [I9 <= 0]
  5) f4#(I12, I13, I14) -> f2#(1, I14, I14) 
  6) f4#(I15, I16, I17) -> f2#(I15, I16, I17) [I15 <= 0]
  7) f1#(I21, I22, I23) -> f2#(I21, I22, I23) [1 + I23 <= I22]

We have the following SCCs.
  { 2, 3, 4, 5, 6, 7 }

DP problem for innermost termination.
P =
  f2#(I3, I4, I5) -> f5#(I3, I4, -1 + I5) [1 <= I5]
  f5#(I6, I7, I8) -> f1#(I6, I7, I8) [1 <= I6]
  f5#(I9, I10, I11) -> f4#(I9, I10, I11) [I9 <= 0]
  f4#(I12, I13, I14) -> f2#(1, I14, I14) 
  f4#(I15, I16, I17) -> f2#(I15, I16, I17) [I15 <= 0]
  f1#(I21, I22, I23) -> f2#(I21, I22, I23) [1 + I23 <= I22]
R = 
  f7(x1, x2, x3) -> f6(x1, x2, x3) 
  f6(I0, I1, I2) -> f2(I0, I1, I2) [I0 <= 0]
  f2(I3, I4, I5) -> f5(I3, I4, -1 + I5) [1 <= I5]
  f5(I6, I7, I8) -> f1(I6, I7, I8) [1 <= I6]
  f5(I9, I10, I11) -> f4(I9, I10, I11) [I9 <= 0]
  f4(I12, I13, I14) -> f2(1, I14, I14) 
  f4(I15, I16, I17) -> f2(I15, I16, I17) [I15 <= 0]
  f1(I18, I19, I20) -> f3(I18, I19, I20) [I19 <= I20]
  f1(I21, I22, I23) -> f2(I21, I22, I23) [1 + I23 <= I22]

We use the basic value criterion with the projection function NU:
  NU[f4#(z1,z2,z3)] = z3
  NU[f1#(z1,z2,z3)] = z3
  NU[f5#(z1,z2,z3)] = z3
  NU[f2#(z1,z2,z3)] = z3

This gives the following inequalities:
  1 <= I5  ==>  I5 >! -1 + I5
  1 <= I6  ==>  I8 (>! \union =) I8
  I9 <= 0  ==>  I11 (>! \union =) I11
    ==>  I14 (>! \union =) I14
  I15 <= 0  ==>  I17 (>! \union =) I17
  1 + I23 <= I22  ==>  I23 (>! \union =) I23

We remove all the strictly oriented dependency pairs.

DP problem for innermost termination.
P =
  f5#(I6, I7, I8) -> f1#(I6, I7, I8) [1 <= I6]
  f5#(I9, I10, I11) -> f4#(I9, I10, I11) [I9 <= 0]
  f4#(I12, I13, I14) -> f2#(1, I14, I14) 
  f4#(I15, I16, I17) -> f2#(I15, I16, I17) [I15 <= 0]
  f1#(I21, I22, I23) -> f2#(I21, I22, I23) [1 + I23 <= I22]
R = 
  f7(x1, x2, x3) -> f6(x1, x2, x3) 
  f6(I0, I1, I2) -> f2(I0, I1, I2) [I0 <= 0]
  f2(I3, I4, I5) -> f5(I3, I4, -1 + I5) [1 <= I5]
  f5(I6, I7, I8) -> f1(I6, I7, I8) [1 <= I6]
  f5(I9, I10, I11) -> f4(I9, I10, I11) [I9 <= 0]
  f4(I12, I13, I14) -> f2(1, I14, I14) 
  f4(I15, I16, I17) -> f2(I15, I16, I17) [I15 <= 0]
  f1(I18, I19, I20) -> f3(I18, I19, I20) [I19 <= I20]
  f1(I21, I22, I23) -> f2(I21, I22, I23) [1 + I23 <= I22]

The dependency graph for this problem is:
  3 -> 7
  4 -> 5, 6
  5 -> 
  6 -> 
  7 -> 
Where:
  3) f5#(I6, I7, I8) -> f1#(I6, I7, I8) [1 <= I6]
  4) f5#(I9, I10, I11) -> f4#(I9, I10, I11) [I9 <= 0]
  5) f4#(I12, I13, I14) -> f2#(1, I14, I14) 
  6) f4#(I15, I16, I17) -> f2#(I15, I16, I17) [I15 <= 0]
  7) f1#(I21, I22, I23) -> f2#(I21, I22, I23) [1 + I23 <= I22]

We have the following SCCs.