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


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YES

DP problem for innermost termination.
P =
  f4#(x1) -> f1#(x1) 
  f3#(I0) -> f2#(I0) 
  f2#(I1) -> f3#(rnd1) [0 <= I1 /\ y1 = I1 /\ rnd1 = -1 + y1]
  f1#(I2) -> f2#(I2) 
R = 
  f4(x1) -> f1(x1) 
  f3(I0) -> f2(I0) 
  f2(I1) -> f3(rnd1) [0 <= I1 /\ y1 = I1 /\ rnd1 = -1 + y1]
  f1(I2) -> f2(I2) 

The dependency graph for this problem is:
  0 -> 3
  1 -> 2
  2 -> 1
  3 -> 2
Where:
  0) f4#(x1) -> f1#(x1) 
  1) f3#(I0) -> f2#(I0) 
  2) f2#(I1) -> f3#(rnd1) [0 <= I1 /\ y1 = I1 /\ rnd1 = -1 + y1]
  3) f1#(I2) -> f2#(I2) 

We have the following SCCs.
  { 1, 2 }

DP problem for innermost termination.
P =
  f3#(I0) -> f2#(I0) 
  f2#(I1) -> f3#(rnd1) [0 <= I1 /\ y1 = I1 /\ rnd1 = -1 + y1]
R = 
  f4(x1) -> f1(x1) 
  f3(I0) -> f2(I0) 
  f2(I1) -> f3(rnd1) [0 <= I1 /\ y1 = I1 /\ rnd1 = -1 + y1]
  f1(I2) -> f2(I2) 

We use the basic value criterion with the projection function NU:
  NU[f2#(z1)] = z1
  NU[f3#(z1)] = z1

This gives the following inequalities:
    ==>  I0 (>! \union =) I0
  0 <= I1 /\ y1 = I1 /\ rnd1 = -1 + y1  ==>  I1 >! rnd1

We remove all the strictly oriented dependency pairs.

DP problem for innermost termination.
P =
  f3#(I0) -> f2#(I0) 
R = 
  f4(x1) -> f1(x1) 
  f3(I0) -> f2(I0) 
  f2(I1) -> f3(rnd1) [0 <= I1 /\ y1 = I1 /\ rnd1 = -1 + y1]
  f1(I2) -> f2(I2) 

The dependency graph for this problem is:
  1 -> 
Where:
  1) f3#(I0) -> f2#(I0) 

We have the following SCCs.