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


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MAYBE

DP problem for innermost termination.
P =
  if#(true, I0, I1) -> exp#(I0, I1 - 1) 
  exp#(I2, I3) -> if#(I3 > 0, I2, I3) 
  cu#(true, I4) -> cu#(I4 < exp(10, 2), I4 + 1) 
  cu#(true, I4) -> exp#(10, 2) 
R = 
  if(false, x, y) -> 1 
  if(true, I0, I1) -> I0 * exp(I0, I1 - 1) 
  exp(I2, I3) -> if(I3 > 0, I2, I3) 
  cu(true, I4) -> cu(I4 < exp(10, 2), I4 + 1) 

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

DP problem for innermost termination.
P =
  if#(true, I0, I1) -> exp#(I0, I1 - 1) 
  exp#(I2, I3) -> if#(I3 > 0, I2, I3) 
  cu#(true, I4) -> cu_1#(I4) 
  cu#(true, I4) -> exp#(10, 2) 
  cu_1#(I4) -> cu#(I4 < exp(10, 2), I4 + 1) 
  exp#(I2, I3) -> exp#(I2, I3 - 1) [I3 > 0]
R = 
  if(false, x, y) -> 1 
  if(true, I0, I1) -> I0 * exp(I0, I1 - 1) 
  exp(I2, I3) -> if(I3 > 0, I2, I3) 
  cu(true, I4) -> cu(I4 < exp(10, 2), I4 + 1) 

The dependency graph for this problem is:
  0 -> 5, 1
  1 -> 
  2 -> 4
  3 -> 5, 1
  4 -> 3, 2
  5 -> 5, 1
Where:
  0) if#(true, I0, I1) -> exp#(I0, I1 - 1) 
  1) exp#(I2, I3) -> if#(I3 > 0, I2, I3) 
  2) cu#(true, I4) -> cu_1#(I4) 
  3) cu#(true, I4) -> exp#(10, 2) 
  4) cu_1#(I4) -> cu#(I4 < exp(10, 2), I4 + 1) 
  5) exp#(I2, I3) -> exp#(I2, I3 - 1) [I3 > 0]

We have the following SCCs.
  { 2, 4 }
  { 5 }

DP problem for innermost termination.
P =
  exp#(I2, I3) -> exp#(I2, I3 - 1) [I3 > 0]
R = 
  if(false, x, y) -> 1 
  if(true, I0, I1) -> I0 * exp(I0, I1 - 1) 
  exp(I2, I3) -> if(I3 > 0, I2, I3) 
  cu(true, I4) -> cu(I4 < exp(10, 2), I4 + 1) 

We use the reverse value criterion with the projection function NU:
  NU[exp#(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.

DP problem for innermost termination.
P =
  cu#(true, I4) -> cu_1#(I4) 
  cu_1#(I4) -> cu#(I4 < exp(10, 2), I4 + 1) 
R = 
  if(false, x, y) -> 1 
  if(true, I0, I1) -> I0 * exp(I0, I1 - 1) 
  exp(I2, I3) -> if(I3 > 0, I2, I3) 
  cu(true, I4) -> cu(I4 < exp(10, 2), I4 + 1)