/export/starexec/sandbox/solver/bin/starexec_run_Itrs /export/starexec/sandbox/benchmark/theBenchmark.itrs /export/starexec/sandbox/output/output_files


--------------------------------------------------------------------------------


YES

DP problem for innermost termination.
P =
  g#(true, x, y, z) -> f#(x > z, x, y, z + 1) 
  g#(true, I0, I1, I2) -> f#(I0 > I2, I0, I1 + 1, I2) 
  f#(true, I3, I4, I5) -> g#(I3 > I4, I3, I4, I5) 
R = 
  g(true, x, y, z) -> f(x > z, x, y, z + 1) 
  g(true, I0, I1, I2) -> f(I0 > I2, I0, I1 + 1, I2) 
  f(true, I3, I4, I5) -> g(I3 > I4, I3, I4, I5) 

This problem is converted using chaining, where edges between chained DPs are removed.

DP problem for innermost termination.
P =
  g#(true, x, y, z) -> f#(x > z, x, y, z + 1) 
  g#(true, I0, I1, I2) -> f#(I0 > I2, I0, I1 + 1, I2) 
  f#(true, I3, I4, I5) -> f_1#(I3, I4, I5) 
  f_1#(I3, I4, I5) -> g#(I3 > I4, I3, I4, I5) 
  f_1#(I3, I4, I5) -> f#(I3 > I5, I3, I4 + 1, I5) [I3 > I4]
  f_1#(I3, I4, I5) -> f_1#(I3, I4 + 1, I5) [I3 > I4, I3 > I5]
  f_1#(I3, I4, I5) -> f#(I3 > I5, I3, I4, I5 + 1) [I3 > I4]
  f_1#(I3, I4, I5) -> f_1#(I3, I4, I5 + 1) [I3 > I4, I3 > I5]
  g#(true, I0, I1, I2) -> f_1#(I0, I1 + 1, I2) [I0 > I2]
  g#(true, x, y, z) -> f_1#(x, y, z + 1) [x > z]
R = 
  g(true, x, y, z) -> f(x > z, x, y, z + 1) 
  g(true, I0, I1, I2) -> f(I0 > I2, I0, I1 + 1, I2) 
  f(true, I3, I4, I5) -> g(I3 > I4, I3, I4, I5) 

The dependency graph for this problem is:
  0 -> 
  1 -> 
  2 -> 7, 6, 5, 4, 3
  3 -> 
  4 -> 
  5 -> 7, 6, 5, 3, 4
  6 -> 
  7 -> 7, 3, 4, 5, 6
  8 -> 3, 4, 5, 6, 7
  9 -> 3, 4, 5, 6, 7
Where:
  0) g#(true, x, y, z) -> f#(x > z, x, y, z + 1) 
  1) g#(true, I0, I1, I2) -> f#(I0 > I2, I0, I1 + 1, I2) 
  2) f#(true, I3, I4, I5) -> f_1#(I3, I4, I5) 
  3) f_1#(I3, I4, I5) -> g#(I3 > I4, I3, I4, I5) 
  4) f_1#(I3, I4, I5) -> f#(I3 > I5, I3, I4 + 1, I5) [I3 > I4]
  5) f_1#(I3, I4, I5) -> f_1#(I3, I4 + 1, I5) [I3 > I4, I3 > I5]
  6) f_1#(I3, I4, I5) -> f#(I3 > I5, I3, I4, I5 + 1) [I3 > I4]
  7) f_1#(I3, I4, I5) -> f_1#(I3, I4, I5 + 1) [I3 > I4, I3 > I5]
  8) g#(true, I0, I1, I2) -> f_1#(I0, I1 + 1, I2) [I0 > I2]
  9) g#(true, x, y, z) -> f_1#(x, y, z + 1) [x > z]

We have the following SCCs.
  { 5, 7 }

DP problem for innermost termination.
P =
  f_1#(I3, I4, I5) -> f_1#(I3, I4 + 1, I5) [I3 > I4, I3 > I5]
  f_1#(I3, I4, I5) -> f_1#(I3, I4, I5 + 1) [I3 > I4, I3 > I5]
R = 
  g(true, x, y, z) -> f(x > z, x, y, z + 1) 
  g(true, I0, I1, I2) -> f(I0 > I2, I0, I1 + 1, I2) 
  f(true, I3, I4, I5) -> g(I3 > I4, I3, I4, I5) 

We use the reverse value criterion with the projection function NU:
  NU[f_1#(z1,z2,z3)] = z1 + -1 * z2

This gives the following inequalities:
  I3 > I4/\I3 > I5  ==>  I3 + -1 * I4 > I3 + -1 * (I4 + 1)  with  I3 + -1 * I4 >= 0
  I3 > I4/\I3 > I5  ==>  I3 + -1 * I4 >= I3 + -1 * I4

We remove all the strictly oriented dependency pairs.

DP problem for innermost termination.
P =
  f_1#(I3, I4, I5) -> f_1#(I3, I4, I5 + 1) [I3 > I4, I3 > I5]
R = 
  g(true, x, y, z) -> f(x > z, x, y, z + 1) 
  g(true, I0, I1, I2) -> f(I0 > I2, I0, I1 + 1, I2) 
  f(true, I3, I4, I5) -> g(I3 > I4, I3, I4, I5) 

We use the reverse value criterion with the projection function NU:
  NU[f_1#(z1,z2,z3)] = z1 + -1 * z3

This gives the following inequalities:
  I3 > I4/\I3 > I5  ==>  I3 + -1 * I5 > I3 + -1 * (I5 + 1)  with  I3 + -1 * I5 >= 0

All dependency pairs are strictly oriented, so the entire dependency pair problem may be removed.