/export/starexec/sandbox2/solver/bin/starexec_run_Transition /export/starexec/sandbox2/benchmark/theBenchmark.smt2 /export/starexec/sandbox2/output/output_files


--------------------------------------------------------------------------------


YES

DP problem for innermost termination.
P =
  f3#(x1) -> f1#(x1) 
  f2#(I0) -> f1#(I0) 
  f1#(I1) -> f2#(-1 + I1) [1 <= I1]
R = 
  f3(x1) -> f1(x1) 
  f2(I0) -> f1(I0) 
  f1(I1) -> f2(-1 + I1) [1 <= I1]

The dependency graph for this problem is:
  0 -> 2
  1 -> 2
  2 -> 1
Where:
  0) f3#(x1) -> f1#(x1) 
  1) f2#(I0) -> f1#(I0) 
  2) f1#(I1) -> f2#(-1 + I1) [1 <= I1]

We have the following SCCs.
  { 1, 2 }

DP problem for innermost termination.
P =
  f2#(I0) -> f1#(I0) 
  f1#(I1) -> f2#(-1 + I1) [1 <= I1]
R = 
  f3(x1) -> f1(x1) 
  f2(I0) -> f1(I0) 
  f1(I1) -> f2(-1 + I1) [1 <= I1]

We use the basic value criterion with the projection function NU:
  NU[f1#(z1)] = z1
  NU[f2#(z1)] = z1

This gives the following inequalities:
    ==>  I0 (>! \union =) I0
  1 <= I1  ==>  I1 >! -1 + I1

We remove all the strictly oriented dependency pairs.

DP problem for innermost termination.
P =
  f2#(I0) -> f1#(I0) 
R = 
  f3(x1) -> f1(x1) 
  f2(I0) -> f1(I0) 
  f1(I1) -> f2(-1 + I1) [1 <= I1]

The dependency graph for this problem is:
  1 -> 
Where:
  1) f2#(I0) -> f1#(I0) 

We have the following SCCs.