/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 =
  if2#(false, x, y) -> f#(x, y) 
  if1#(true, I4, I5) -> h#(I4, I5) 
  h#(I6, I7) -> if2#(I6 > I7, I6, I7) 
  f#(I8, I9) -> if1#(I8 > I9, I8, I9) 
R = 
  if2(false, x, y) -> f(x, y) 
  if2(true, I0, I1) -> 0 
  if1(false, I2, I3) -> 0 
  if1(true, I4, I5) -> h(I4, I5) 
  h(I6, I7) -> if2(I6 > I7, I6, I7) 
  f(I8, I9) -> if1(I8 > I9, I8, I9) 

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

DP problem for innermost termination.
P =
  if2#(false, x, y) -> f#(x, y) 
  if1#(true, I4, I5) -> h#(I4, I5) 
  h#(I6, I7) -> if2#(I6 > I7, I6, I7) 
  f#(I8, I9) -> if1#(I8 > I9, I8, I9) 
  f#(I8, I9) -> h#(I8, I9) [I8 > I9]
  h#(I6, I7) -> f#(I6, I7) [not(I6 > I7)]
R = 
  if2(false, x, y) -> f(x, y) 
  if2(true, I0, I1) -> 0 
  if1(false, I2, I3) -> 0 
  if1(true, I4, I5) -> h(I4, I5) 
  h(I6, I7) -> if2(I6 > I7, I6, I7) 
  f(I8, I9) -> if1(I8 > I9, I8, I9) 

The dependency graph for this problem is:
  0 -> 4, 3
  1 -> 5, 2
  2 -> 
  3 -> 
  4 -> 2
  5 -> 3
Where:
  0) if2#(false, x, y) -> f#(x, y) 
  1) if1#(true, I4, I5) -> h#(I4, I5) 
  2) h#(I6, I7) -> if2#(I6 > I7, I6, I7) 
  3) f#(I8, I9) -> if1#(I8 > I9, I8, I9) 
  4) f#(I8, I9) -> h#(I8, I9) [I8 > I9]
  5) h#(I6, I7) -> f#(I6, I7) [not(I6 > I7)]

We have the following SCCs.