/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 =
  lif#(true, I0, I1) -> log#(I0 / I1, I1) 
  log#(I2, I3) -> lif#(I2 >= I3 && I3 > 1, I2, I3) 
R = 
  lif(false, x, y) -> 0 
  lif(true, I0, I1) -> 1 + log(I0 / I1, I1) 
  log(I2, I3) -> lif(I2 >= I3 && I3 > 1, I2, I3) 

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

DP problem for innermost termination.
P =
  lif#(true, I0, I1) -> log#(I0 / I1, I1) 
  log#(I2, I3) -> lif#(I2 >= I3 && I3 > 1, I2, I3) 
  log#(I2, I3) -> log#(I2 / I3, I3) [I2 >= I3 && I3 > 1]
R = 
  lif(false, x, y) -> 0 
  lif(true, I0, I1) -> 1 + log(I0 / I1, I1) 
  log(I2, I3) -> lif(I2 >= I3 && I3 > 1, I2, I3) 

The dependency graph for this problem is:
  0 -> 2, 1
  1 -> 
  2 -> 2, 1
Where:
  0) lif#(true, I0, I1) -> log#(I0 / I1, I1) 
  1) log#(I2, I3) -> lif#(I2 >= I3 && I3 > 1, I2, I3) 
  2) log#(I2, I3) -> log#(I2 / I3, I3) [I2 >= I3 && I3 > 1]

We have the following SCCs.
  { 2 }

DP problem for innermost termination.
P =
  log#(I2, I3) -> log#(I2 / I3, I3) [I2 >= I3 && I3 > 1]
R = 
  lif(false, x, y) -> 0 
  lif(true, I0, I1) -> 1 + log(I0 / I1, I1) 
  log(I2, I3) -> lif(I2 >= I3 && I3 > 1, I2, I3) 

We use the reverse value criterion with the projection function NU:
  NU[log#(z1,z2)] = z1

This gives the following inequalities:
  I2 >= I3 && I3 > 1  ==>  I2 > I2 / I3  with  I2 >= 0

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