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

The dependency graph for this problem is:
  0 -> 1
  1 -> 3
  2 -> 3
  3 -> 2
Where:
  0) f4#(x1, x2) -> f3#(x1, x2) 
  1) f3#(I0, I1) -> f1#(I0, I1) 
  2) f2#(I2, I3) -> f1#(I2, I3) 
  3) f1#(I4, I5) -> f2#(I4, 1 + I5) [I5 <= 0 /\ 1 <= I4]

We have the following SCCs.
  { 2, 3 }

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

We use the reverse value criterion with the projection function NU:
  NU[f1#(z1,z2)] = 0 + -1 * z2
  NU[f2#(z1,z2)] = 0 + -1 * z2

This gives the following inequalities:
    ==>  0 + -1 * I3 >= 0 + -1 * I3
  I5 <= 0 /\ 1 <= I4  ==>  0 + -1 * I5 > 0 + -1 * (1 + I5)  with  0 + -1 * I5 >= 0

We remove all the strictly oriented dependency pairs.

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

The dependency graph for this problem is:
  2 -> 
Where:
  2) f2#(I2, I3) -> f1#(I2, I3) 

We have the following SCCs.