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


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YES

DP problem for innermost termination.
P =
  f4#(x1) -> f3#(x1) 
  f3#(I0) -> f1#(I0) 
  f2#(I1) -> f1#(I1) 
  f1#(I2) -> f2#(-1000 + I2) [201 <= -1000 + I2]
R = 
  f4(x1) -> f3(x1) 
  f3(I0) -> f1(I0) 
  f2(I1) -> f1(I1) 
  f1(I2) -> f2(-1000 + I2) [201 <= -1000 + I2]

The dependency graph for this problem is:
  0 -> 1
  1 -> 3
  2 -> 3
  3 -> 2
Where:
  0) f4#(x1) -> f3#(x1) 
  1) f3#(I0) -> f1#(I0) 
  2) f2#(I1) -> f1#(I1) 
  3) f1#(I2) -> f2#(-1000 + I2) [201 <= -1000 + I2]

We have the following SCCs.
  { 2, 3 }

DP problem for innermost termination.
P =
  f2#(I1) -> f1#(I1) 
  f1#(I2) -> f2#(-1000 + I2) [201 <= -1000 + I2]
R = 
  f4(x1) -> f3(x1) 
  f3(I0) -> f1(I0) 
  f2(I1) -> f1(I1) 
  f1(I2) -> f2(-1000 + I2) [201 <= -1000 + I2]

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

This gives the following inequalities:
    ==>  I1 (>! \union =) I1
  201 <= -1000 + I2  ==>  I2 >! -1000 + I2

We remove all the strictly oriented dependency pairs.

DP problem for innermost termination.
P =
  f2#(I1) -> f1#(I1) 
R = 
  f4(x1) -> f3(x1) 
  f3(I0) -> f1(I0) 
  f2(I1) -> f1(I1) 
  f1(I2) -> f2(-1000 + I2) [201 <= -1000 + I2]

The dependency graph for this problem is:
  2 -> 
Where:
  2) f2#(I1) -> f1#(I1) 

We have the following SCCs.