/export/starexec/sandbox2/solver/bin/starexec_run_Itrs /export/starexec/sandbox2/benchmark/theBenchmark.itrs /export/starexec/sandbox2/output/output_files


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YES

DP problem for innermost termination.
P =
  conv#(x) -> conv#(x / 2) [x > 0]
  conv#(x) -> lastbit#(x) [x > 0]
  lastbit#(I0) -> lastbit#(I0 - 2) [I0 > 1]
R = 
  conv(x) -> cons(conv(x / 2), lastbit(x)) [x > 0]
  conv(0) -> cons(nil, 0) 
  lastbit(I0) -> lastbit(I0 - 2) [I0 > 1]
  lastbit(1) -> 1 
  lastbit(0) -> 0 

The dependency graph for this problem is:
  0 -> 0, 1
  1 -> 2
  2 -> 2
Where:
  0) conv#(x) -> conv#(x / 2) [x > 0]
  1) conv#(x) -> lastbit#(x) [x > 0]
  2) lastbit#(I0) -> lastbit#(I0 - 2) [I0 > 1]

We have the following SCCs.
  { 0 }
  { 2 }

DP problem for innermost termination.
P =
  lastbit#(I0) -> lastbit#(I0 - 2) [I0 > 1]
R = 
  conv(x) -> cons(conv(x / 2), lastbit(x)) [x > 0]
  conv(0) -> cons(nil, 0) 
  lastbit(I0) -> lastbit(I0 - 2) [I0 > 1]
  lastbit(1) -> 1 
  lastbit(0) -> 0 

We use the reverse value criterion with the projection function NU:
  NU[lastbit#(z1)] = z1

This gives the following inequalities:
  I0 > 1  ==>  I0 > I0 - 2  with  I0 >= 0

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

DP problem for innermost termination.
P =
  conv#(x) -> conv#(x / 2) [x > 0]
R = 
  conv(x) -> cons(conv(x / 2), lastbit(x)) [x > 0]
  conv(0) -> cons(nil, 0) 
  lastbit(I0) -> lastbit(I0 - 2) [I0 > 1]
  lastbit(1) -> 1 
  lastbit(0) -> 0 

We use the reverse value criterion with the projection function NU:
  NU[conv#(z1)] = z1

This gives the following inequalities:
  x > 0  ==>  x > x / 2  with  x >= 0

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