/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 =
  f5#(x1, x2) -> f1#(x1, x2) 
  f4#(I0, I1) -> f2#(I0, I1) 
  f2#(I2, I3) -> f4#(I2, -1 + I3) [0 <= -1 + -1 + I3]
  f1#(I6, I7) -> f2#(I6, I7) 
R = 
  f5(x1, x2) -> f1(x1, x2) 
  f4(I0, I1) -> f2(I0, I1) 
  f2(I2, I3) -> f4(I2, -1 + I3) [0 <= -1 + -1 + I3]
  f2(I4, I5) -> f3(rnd1, -1 + I5) [rnd1 = rnd1 /\ -1 + I5 <= 0]
  f1(I6, I7) -> f2(I6, I7) 

The dependency graph for this problem is:
  0 -> 3
  1 -> 2
  2 -> 1
  3 -> 2
Where:
  0) f5#(x1, x2) -> f1#(x1, x2) 
  1) f4#(I0, I1) -> f2#(I0, I1) 
  2) f2#(I2, I3) -> f4#(I2, -1 + I3) [0 <= -1 + -1 + I3]
  3) f1#(I6, I7) -> f2#(I6, I7) 

We have the following SCCs.
  { 1, 2 }

DP problem for innermost termination.
P =
  f4#(I0, I1) -> f2#(I0, I1) 
  f2#(I2, I3) -> f4#(I2, -1 + I3) [0 <= -1 + -1 + I3]
R = 
  f5(x1, x2) -> f1(x1, x2) 
  f4(I0, I1) -> f2(I0, I1) 
  f2(I2, I3) -> f4(I2, -1 + I3) [0 <= -1 + -1 + I3]
  f2(I4, I5) -> f3(rnd1, -1 + I5) [rnd1 = rnd1 /\ -1 + I5 <= 0]
  f1(I6, I7) -> f2(I6, I7) 

We use the basic value criterion with the projection function NU:
  NU[f2#(z1,z2)] = z2
  NU[f4#(z1,z2)] = z2

This gives the following inequalities:
    ==>  I1 (>! \union =) I1
  0 <= -1 + -1 + I3  ==>  I3 >! -1 + I3

We remove all the strictly oriented dependency pairs.

DP problem for innermost termination.
P =
  f4#(I0, I1) -> f2#(I0, I1) 
R = 
  f5(x1, x2) -> f1(x1, x2) 
  f4(I0, I1) -> f2(I0, I1) 
  f2(I2, I3) -> f4(I2, -1 + I3) [0 <= -1 + -1 + I3]
  f2(I4, I5) -> f3(rnd1, -1 + I5) [rnd1 = rnd1 /\ -1 + I5 <= 0]
  f1(I6, I7) -> f2(I6, I7) 

The dependency graph for this problem is:
  1 -> 
Where:
  1) f4#(I0, I1) -> f2#(I0, I1) 

We have the following SCCs.