/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 =
  condAcc#(false, x, y) -> sqrtAcc#(x, y + 1) 
  sqrtAcc#(I2, I3) -> condAcc#(I3 * I3 >= I2 || I3 < 0, I2, I3) 
  sqrt#(I4) -> sqrtAcc#(I4, 0) 
R = 
  condAcc(false, x, y) -> sqrtAcc(x, y + 1) 
  condAcc(true, I0, I1) -> I1 
  sqrtAcc(I2, I3) -> condAcc(I3 * I3 >= I2 || I3 < 0, I2, I3) 
  sqrt(I4) -> sqrtAcc(I4, 0) 

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

DP problem for innermost termination.
P =
  condAcc#(false, x, y) -> sqrtAcc#(x, y + 1) 
  sqrtAcc#(I2, I3) -> condAcc#(I3 * I3 >= I2 || I3 < 0, I2, I3) 
  sqrt#(I4) -> sqrtAcc#(I4, 0) 
  sqrtAcc#(I2, I3) -> sqrtAcc#(I2, I3 + 1) [not(I3 * I3 >= I2 || I3 < 0)]
R = 
  condAcc(false, x, y) -> sqrtAcc(x, y + 1) 
  condAcc(true, I0, I1) -> I1 
  sqrtAcc(I2, I3) -> condAcc(I3 * I3 >= I2 || I3 < 0, I2, I3) 
  sqrt(I4) -> sqrtAcc(I4, 0) 

The dependency graph for this problem is:
  0 -> 3, 1
  1 -> 
  2 -> 3, 1
  3 -> 3, 1
Where:
  0) condAcc#(false, x, y) -> sqrtAcc#(x, y + 1) 
  1) sqrtAcc#(I2, I3) -> condAcc#(I3 * I3 >= I2 || I3 < 0, I2, I3) 
  2) sqrt#(I4) -> sqrtAcc#(I4, 0) 
  3) sqrtAcc#(I2, I3) -> sqrtAcc#(I2, I3 + 1) [not(I3 * I3 >= I2 || I3 < 0)]

We have the following SCCs.
  { 3 }

DP problem for innermost termination.
P =
  sqrtAcc#(I2, I3) -> sqrtAcc#(I2, I3 + 1) [not(I3 * I3 >= I2 || I3 < 0)]
R = 
  condAcc(false, x, y) -> sqrtAcc(x, y + 1) 
  condAcc(true, I0, I1) -> I1 
  sqrtAcc(I2, I3) -> condAcc(I3 * I3 >= I2 || I3 < 0, I2, I3) 
  sqrt(I4) -> sqrtAcc(I4, 0)