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

The dependency graph for this problem is:
  0 -> 1, 2
  1 -> 4
  2 -> 3
  3 -> 4
  4 -> 3
Where:
  0) f4#(x1, x2) -> f3#(x1, x2) 
  1) f3#(I0, I1) -> f1#(I0, I1) 
  2) f3#(I2, I3) -> f2#(I2, -1 + I3) 
  3) f2#(I4, I5) -> f1#(-1 + I4, I5) 
  4) f1#(I6, I7) -> f2#(I6, I7) [1 <= I7 /\ 1 <= I6]

We have the following SCCs.
  { 3, 4 }

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

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

This gives the following inequalities:
    ==>  -1 + I4 + -1 * 1 >= -1 + I4 + -1 * 1
  1 <= I7 /\ 1 <= I6  ==>  I6 + -1 * 1 > -1 + I6 + -1 * 1  with  I6 + -1 * 1 >= 0

We remove all the strictly oriented dependency pairs.

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

The dependency graph for this problem is:
  3 -> 
Where:
  3) f2#(I4, I5) -> f1#(-1 + I4, I5) 

We have the following SCCs.