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


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MAYBE

DP problem for innermost termination.
P =
  cond#(false, x, y) -> log#(x, y * y) 
  logNat#(true, I2, I3) -> cond#(I2 <= I3, I2, I3) 
  log#(I4, I5) -> logNat#(I4 >= 0 && I5 >= 2, I4, I5) 
R = 
  cond(false, x, y) -> 2 * log(x, y * y) 
  cond(true, I0, I1) -> 1 
  logNat(true, I2, I3) -> cond(I2 <= I3, I2, I3) 
  log(I4, I5) -> logNat(I4 >= 0 && I5 >= 2, I4, I5) 

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

DP problem for innermost termination.
P =
  cond#(false, x, y) -> log#(x, y * y) 
  logNat#(true, I2, I3) -> cond#(I2 <= I3, I2, I3) 
  log#(I4, I5) -> logNat#(I4 >= 0 && I5 >= 2, I4, I5) 
  log#(I4, I5) -> cond#(I4 <= I5, I4, I5) [I4 >= 0 && I5 >= 2]
  log#(I4, I5) -> log#(I4, I5 * I5) [I4 >= 0 && I5 >= 2, not(I4 <= I5)]
  logNat#(true, I2, I3) -> log#(I2, I3 * I3) [not(I2 <= I3)]
R = 
  cond(false, x, y) -> 2 * log(x, y * y) 
  cond(true, I0, I1) -> 1 
  logNat(true, I2, I3) -> cond(I2 <= I3, I2, I3) 
  log(I4, I5) -> logNat(I4 >= 0 && I5 >= 2, I4, I5) 

The dependency graph for this problem is:
  0 -> 4, 3, 2
  1 -> 
  2 -> 
  3 -> 
  4 -> 4, 2, 3
  5 -> 2, 3, 4
Where:
  0) cond#(false, x, y) -> log#(x, y * y) 
  1) logNat#(true, I2, I3) -> cond#(I2 <= I3, I2, I3) 
  2) log#(I4, I5) -> logNat#(I4 >= 0 && I5 >= 2, I4, I5) 
  3) log#(I4, I5) -> cond#(I4 <= I5, I4, I5) [I4 >= 0 && I5 >= 2]
  4) log#(I4, I5) -> log#(I4, I5 * I5) [I4 >= 0 && I5 >= 2, not(I4 <= I5)]
  5) logNat#(true, I2, I3) -> log#(I2, I3 * I3) [not(I2 <= I3)]

We have the following SCCs.
  { 4 }

DP problem for innermost termination.
P =
  log#(I4, I5) -> log#(I4, I5 * I5) [I4 >= 0 && I5 >= 2, not(I4 <= I5)]
R = 
  cond(false, x, y) -> 2 * log(x, y * y) 
  cond(true, I0, I1) -> 1 
  logNat(true, I2, I3) -> cond(I2 <= I3, I2, I3) 
  log(I4, I5) -> logNat(I4 >= 0 && I5 >= 2, I4, I5)