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


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YES

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

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

DP problem for innermost termination.
P =
  f#(true, x, y, z) -> f_2#(x, y, z) 
  f#(true, I0, I1, I2) -> f_1#(I0, I1, I2) 
  f_1#(I0, I1, I2) -> f#(I0 > I1 + I2, I0, I1 + 1, I2) 
  f_2#(x, y, z) -> f#(x > y + z, x, y, z + 1) 
  f_2#(x, y, z) -> f_1#(x, y, z + 1) [x > y + z]
  f_2#(x, y, z) -> f_2#(x, y, z + 1) [x > y + z]
  f_1#(I0, I1, I2) -> f_1#(I0, I1 + 1, I2) [I0 > I1 + I2]
  f_1#(I0, I1, I2) -> f_2#(I0, I1 + 1, I2) [I0 > I1 + I2]
R = 
  f(true, x, y, z) -> f(x > y + z, x, y, z + 1) 
  f(true, I0, I1, I2) -> f(I0 > I1 + I2, I0, I1 + 1, I2) 

The dependency graph for this problem is:
  0 -> 5, 4, 3
  1 -> 7, 6, 2
  2 -> 
  3 -> 
  4 -> 7, 6, 2
  5 -> 5, 3, 4
  6 -> 7, 6, 2
  7 -> 3, 4, 5
Where:
  0) f#(true, x, y, z) -> f_2#(x, y, z) 
  1) f#(true, I0, I1, I2) -> f_1#(I0, I1, I2) 
  2) f_1#(I0, I1, I2) -> f#(I0 > I1 + I2, I0, I1 + 1, I2) 
  3) f_2#(x, y, z) -> f#(x > y + z, x, y, z + 1) 
  4) f_2#(x, y, z) -> f_1#(x, y, z + 1) [x > y + z]
  5) f_2#(x, y, z) -> f_2#(x, y, z + 1) [x > y + z]
  6) f_1#(I0, I1, I2) -> f_1#(I0, I1 + 1, I2) [I0 > I1 + I2]
  7) f_1#(I0, I1, I2) -> f_2#(I0, I1 + 1, I2) [I0 > I1 + I2]

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

DP problem for innermost termination.
P =
  f_2#(x, y, z) -> f_1#(x, y, z + 1) [x > y + z]
  f_2#(x, y, z) -> f_2#(x, y, z + 1) [x > y + z]
  f_1#(I0, I1, I2) -> f_1#(I0, I1 + 1, I2) [I0 > I1 + I2]
  f_1#(I0, I1, I2) -> f_2#(I0, I1 + 1, I2) [I0 > I1 + I2]
R = 
  f(true, x, y, z) -> f(x > y + z, x, y, z + 1) 
  f(true, I0, I1, I2) -> f(I0 > I1 + I2, I0, I1 + 1, I2) 

We use the reverse value criterion with the projection function NU:
  NU[f_1#(z1,z2,z3)] = z1 + -1 * (z2 + 1 + z3)
  NU[f_2#(z1,z2,z3)] = z1 + -1 * (z2 + (z3 + 1))

This gives the following inequalities:
  x > y + z  ==>  x + -1 * (y + (z + 1)) > x + -1 * (y + 1 + (z + 1))  with  x + -1 * (y + (z + 1)) >= 0
  x > y + z  ==>  x + -1 * (y + (z + 1)) > x + -1 * (y + (z + 1 + 1))  with  x + -1 * (y + (z + 1)) >= 0
  I0 > I1 + I2  ==>  I0 + -1 * (I1 + 1 + I2) > I0 + -1 * (I1 + 1 + 1 + I2)  with  I0 + -1 * (I1 + 1 + I2) >= 0
  I0 > I1 + I2  ==>  I0 + -1 * (I1 + 1 + I2) > I0 + -1 * (I1 + 1 + (I2 + 1))  with  I0 + -1 * (I1 + 1 + I2) >= 0

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