/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 =
  if#(true, I0, I1) -> pow#(I0, I1 - 1) 
  pow#(I2, I3) -> if#(I3 > 0, I2, I3) 
R = 
  if(false, x, y) -> 1 
  if(true, I0, I1) -> I0 * pow(I0, I1 - 1) 
  pow(I2, I3) -> if(I3 > 0, I2, I3) 

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

DP problem for innermost termination.
P =
  if#(true, I0, I1) -> pow#(I0, I1 - 1) 
  pow#(I2, I3) -> if#(I3 > 0, I2, I3) 
  pow#(I2, I3) -> pow#(I2, I3 - 1) [I3 > 0]
R = 
  if(false, x, y) -> 1 
  if(true, I0, I1) -> I0 * pow(I0, I1 - 1) 
  pow(I2, I3) -> if(I3 > 0, I2, I3) 

The dependency graph for this problem is:
  0 -> 2, 1
  1 -> 
  2 -> 2, 1
Where:
  0) if#(true, I0, I1) -> pow#(I0, I1 - 1) 
  1) pow#(I2, I3) -> if#(I3 > 0, I2, I3) 
  2) pow#(I2, I3) -> pow#(I2, I3 - 1) [I3 > 0]

We have the following SCCs.
  { 2 }

DP problem for innermost termination.
P =
  pow#(I2, I3) -> pow#(I2, I3 - 1) [I3 > 0]
R = 
  if(false, x, y) -> 1 
  if(true, I0, I1) -> I0 * pow(I0, I1 - 1) 
  pow(I2, I3) -> if(I3 > 0, I2, I3) 

We use the reverse value criterion with the projection function NU:
  NU[pow#(z1,z2)] = z2 + -1 * 0

This gives the following inequalities:
  I3 > 0  ==>  I3 + -1 * 0 > I3 - 1 + -1 * 0  with  I3 + -1 * 0 >= 0

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