/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 =
  cond2#(false, x) -> f#(3 * x + 1) 
  cond2#(true, I0) -> f#(I0 / 2) 
  cond1#(true, I2) -> cond2#(I2 % 2 = 0, I2) 
  f#(I3) -> cond1#(I3 > 1, I3) 
R = 
  cond2(false, x) -> f(3 * x + 1) 
  cond2(true, I0) -> f(I0 / 2) 
  cond1(false, I1) -> I1 
  cond1(true, I2) -> cond2(I2 % 2 = 0, I2) 
  f(I3) -> cond1(I3 > 1, I3) 

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

DP problem for innermost termination.
P =
  cond2#(false, x) -> f#(3 * x + 1) 
  cond2#(true, I0) -> f#(I0 / 2) 
  cond1#(true, I2) -> cond2#(I2 % 2 = 0, I2) 
  f#(I3) -> cond1#(I3 > 1, I3) 
  f#(I3) -> cond2#(I3 % 2 = 0, I3) [I3 > 1]
  f#(I3) -> f#(I3 / 2) [I3 > 1, I3 % 2 = 0]
  f#(I3) -> f#(3 * I3 + 1) [I3 > 1, not(I3 % 2 = 0)]
  cond1#(true, I2) -> f#(3 * I2 + 1) [not(I2 % 2 = 0)]
  cond1#(true, I2) -> f#(I2 / 2) [I2 % 2 = 0]
R = 
  cond2(false, x) -> f(3 * x + 1) 
  cond2(true, I0) -> f(I0 / 2) 
  cond1(false, I1) -> I1 
  cond1(true, I2) -> cond2(I2 % 2 = 0, I2) 
  f(I3) -> cond1(I3 > 1, I3) 

The dependency graph for this problem is:
  0 -> 6, 5, 4, 3
  1 -> 6, 5, 4, 3
  2 -> 
  3 -> 
  4 -> 
  5 -> 6, 5, 3, 4
  6 -> 3, 4, 5
  7 -> 3, 4, 5
  8 -> 3, 4, 5, 6
Where:
  0) cond2#(false, x) -> f#(3 * x + 1) 
  1) cond2#(true, I0) -> f#(I0 / 2) 
  2) cond1#(true, I2) -> cond2#(I2 % 2 = 0, I2) 
  3) f#(I3) -> cond1#(I3 > 1, I3) 
  4) f#(I3) -> cond2#(I3 % 2 = 0, I3) [I3 > 1]
  5) f#(I3) -> f#(I3 / 2) [I3 > 1, I3 % 2 = 0]
  6) f#(I3) -> f#(3 * I3 + 1) [I3 > 1, not(I3 % 2 = 0)]
  7) cond1#(true, I2) -> f#(3 * I2 + 1) [not(I2 % 2 = 0)]
  8) cond1#(true, I2) -> f#(I2 / 2) [I2 % 2 = 0]

We have the following SCCs.
  { 5, 6 }

DP problem for innermost termination.
P =
  f#(I3) -> f#(I3 / 2) [I3 > 1, I3 % 2 = 0]
  f#(I3) -> f#(3 * I3 + 1) [I3 > 1, not(I3 % 2 = 0)]
R = 
  cond2(false, x) -> f(3 * x + 1) 
  cond2(true, I0) -> f(I0 / 2) 
  cond1(false, I1) -> I1 
  cond1(true, I2) -> cond2(I2 % 2 = 0, I2) 
  f(I3) -> cond1(I3 > 1, I3)