/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 =
  init#(x1, x2) -> f1#(rnd1, rnd2) 
  f3#(I0, I1) -> f2#(I1, I0) [I0 <= I1 - 1]
  f3#(I2, I3) -> f3#(I2 - I3, I3) [0 <= I3 - 1 /\ 0 <= I2 - 1 /\ I3 <= I2]
  f2#(I4, I5) -> f3#(I4, I5) [0 <= I4 - 1 /\ 0 <= I5 - 1]
  f1#(I6, I7) -> f2#(I8, I9) [0 <= I6 - 1 /\ -1 <= I8 - 1 /\ -1 <= I7 - 1 /\ -1 <= I9 - 1]
R = 
  init(x1, x2) -> f1(rnd1, rnd2) 
  f3(I0, I1) -> f2(I1, I0) [I0 <= I1 - 1]
  f3(I2, I3) -> f3(I2 - I3, I3) [0 <= I3 - 1 /\ 0 <= I2 - 1 /\ I3 <= I2]
  f2(I4, I5) -> f3(I4, I5) [0 <= I4 - 1 /\ 0 <= I5 - 1]
  f1(I6, I7) -> f2(I8, I9) [0 <= I6 - 1 /\ -1 <= I8 - 1 /\ -1 <= I7 - 1 /\ -1 <= I9 - 1]

The dependency graph for this problem is:
  0 -> 4
  1 -> 3
  2 -> 1, 2
  3 -> 1, 2
  4 -> 3
Where:
  0) init#(x1, x2) -> f1#(rnd1, rnd2) 
  1) f3#(I0, I1) -> f2#(I1, I0) [I0 <= I1 - 1]
  2) f3#(I2, I3) -> f3#(I2 - I3, I3) [0 <= I3 - 1 /\ 0 <= I2 - 1 /\ I3 <= I2]
  3) f2#(I4, I5) -> f3#(I4, I5) [0 <= I4 - 1 /\ 0 <= I5 - 1]
  4) f1#(I6, I7) -> f2#(I8, I9) [0 <= I6 - 1 /\ -1 <= I8 - 1 /\ -1 <= I7 - 1 /\ -1 <= I9 - 1]

We have the following SCCs.
  { 1, 2, 3 }

DP problem for innermost termination.
P =
  f3#(I0, I1) -> f2#(I1, I0) [I0 <= I1 - 1]
  f3#(I2, I3) -> f3#(I2 - I3, I3) [0 <= I3 - 1 /\ 0 <= I2 - 1 /\ I3 <= I2]
  f2#(I4, I5) -> f3#(I4, I5) [0 <= I4 - 1 /\ 0 <= I5 - 1]
R = 
  init(x1, x2) -> f1(rnd1, rnd2) 
  f3(I0, I1) -> f2(I1, I0) [I0 <= I1 - 1]
  f3(I2, I3) -> f3(I2 - I3, I3) [0 <= I3 - 1 /\ 0 <= I2 - 1 /\ I3 <= I2]
  f2(I4, I5) -> f3(I4, I5) [0 <= I4 - 1 /\ 0 <= I5 - 1]
  f1(I6, I7) -> f2(I8, I9) [0 <= I6 - 1 /\ -1 <= I8 - 1 /\ -1 <= I7 - 1 /\ -1 <= I9 - 1]

We use the reverse value criterion with the projection function NU:
  NU[f2#(z1,z2)] = z2
  NU[f3#(z1,z2)] = z2 - 1 + -1 * 0

This gives the following inequalities:
  I0 <= I1 - 1  ==>  I1 - 1 + -1 * 0 >= I0
  0 <= I3 - 1 /\ 0 <= I2 - 1 /\ I3 <= I2  ==>  I3 - 1 + -1 * 0 >= I3 - 1 + -1 * 0
  0 <= I4 - 1 /\ 0 <= I5 - 1  ==>  I5 > I5 - 1 + -1 * 0  with  I5 >= 0

We remove all the strictly oriented dependency pairs.

DP problem for innermost termination.
P =
  f3#(I0, I1) -> f2#(I1, I0) [I0 <= I1 - 1]
  f3#(I2, I3) -> f3#(I2 - I3, I3) [0 <= I3 - 1 /\ 0 <= I2 - 1 /\ I3 <= I2]
R = 
  init(x1, x2) -> f1(rnd1, rnd2) 
  f3(I0, I1) -> f2(I1, I0) [I0 <= I1 - 1]
  f3(I2, I3) -> f3(I2 - I3, I3) [0 <= I3 - 1 /\ 0 <= I2 - 1 /\ I3 <= I2]
  f2(I4, I5) -> f3(I4, I5) [0 <= I4 - 1 /\ 0 <= I5 - 1]
  f1(I6, I7) -> f2(I8, I9) [0 <= I6 - 1 /\ -1 <= I8 - 1 /\ -1 <= I7 - 1 /\ -1 <= I9 - 1]

The dependency graph for this problem is:
  1 -> 
  2 -> 1, 2
Where:
  1) f3#(I0, I1) -> f2#(I1, I0) [I0 <= I1 - 1]
  2) f3#(I2, I3) -> f3#(I2 - I3, I3) [0 <= I3 - 1 /\ 0 <= I2 - 1 /\ I3 <= I2]

We have the following SCCs.
  { 2 }

DP problem for innermost termination.
P =
  f3#(I2, I3) -> f3#(I2 - I3, I3) [0 <= I3 - 1 /\ 0 <= I2 - 1 /\ I3 <= I2]
R = 
  init(x1, x2) -> f1(rnd1, rnd2) 
  f3(I0, I1) -> f2(I1, I0) [I0 <= I1 - 1]
  f3(I2, I3) -> f3(I2 - I3, I3) [0 <= I3 - 1 /\ 0 <= I2 - 1 /\ I3 <= I2]
  f2(I4, I5) -> f3(I4, I5) [0 <= I4 - 1 /\ 0 <= I5 - 1]
  f1(I6, I7) -> f2(I8, I9) [0 <= I6 - 1 /\ -1 <= I8 - 1 /\ -1 <= I7 - 1 /\ -1 <= I9 - 1]

We use the basic value criterion with the projection function NU:
  NU[f3#(z1,z2)] = z1

This gives the following inequalities:
  0 <= I3 - 1 /\ 0 <= I2 - 1 /\ I3 <= I2  ==>  I2 >! I2 - I3

All dependency pairs are strictly oriented, so the entire dependency pair problem may be removed.