/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, I2) [I1 <= I0 - 1 /\ 1 <= I0 - 1]
  f4#(I3, I4) -> f3#(I3, I3 - 2) [0 <= I3 - 1]
  f2#(I5, I6) -> f3#(I5, I5 - 2) [I5 - 1 <= I5 - 1 /\ 1 <= I5 - 1]
  f2#(I7, I8) -> f3#(2, 0) [2 = I7]
  f2#(I9, I10) -> f2#(I9 - 1, I11) [I9 - 1 <= I9 - 1 /\ 1 <= I9 - 1]
  f1#(I12, I13) -> f2#(I13, I14) [-1 <= I13 - 1 /\ 0 <= I12 - 1]
R = 
  init(x1, x2) -> f1(rnd1, rnd2) 
  f3(I0, I1) -> f2(I1, I2) [I1 <= I0 - 1 /\ 1 <= I0 - 1]
  f4(I3, I4) -> f3(I3, I3 - 2) [0 <= I3 - 1]
  f2(I5, I6) -> f3(I5, I5 - 2) [I5 - 1 <= I5 - 1 /\ 1 <= I5 - 1]
  f2(I7, I8) -> f3(2, 0) [2 = I7]
  f2(I9, I10) -> f2(I9 - 1, I11) [I9 - 1 <= I9 - 1 /\ 1 <= I9 - 1]
  f1(I12, I13) -> f2(I13, I14) [-1 <= I13 - 1 /\ 0 <= I12 - 1]

The dependency graph for this problem is:
  0 -> 6
  1 -> 3, 4, 5
  2 -> 1
  3 -> 1
  4 -> 1
  5 -> 3, 4, 5
  6 -> 3, 4, 5
Where:
  0) init#(x1, x2) -> f1#(rnd1, rnd2) 
  1) f3#(I0, I1) -> f2#(I1, I2) [I1 <= I0 - 1 /\ 1 <= I0 - 1]
  2) f4#(I3, I4) -> f3#(I3, I3 - 2) [0 <= I3 - 1]
  3) f2#(I5, I6) -> f3#(I5, I5 - 2) [I5 - 1 <= I5 - 1 /\ 1 <= I5 - 1]
  4) f2#(I7, I8) -> f3#(2, 0) [2 = I7]
  5) f2#(I9, I10) -> f2#(I9 - 1, I11) [I9 - 1 <= I9 - 1 /\ 1 <= I9 - 1]
  6) f1#(I12, I13) -> f2#(I13, I14) [-1 <= I13 - 1 /\ 0 <= I12 - 1]

We have the following SCCs.
  { 1, 3, 4, 5 }

DP problem for innermost termination.
P =
  f3#(I0, I1) -> f2#(I1, I2) [I1 <= I0 - 1 /\ 1 <= I0 - 1]
  f2#(I5, I6) -> f3#(I5, I5 - 2) [I5 - 1 <= I5 - 1 /\ 1 <= I5 - 1]
  f2#(I7, I8) -> f3#(2, 0) [2 = I7]
  f2#(I9, I10) -> f2#(I9 - 1, I11) [I9 - 1 <= I9 - 1 /\ 1 <= I9 - 1]
R = 
  init(x1, x2) -> f1(rnd1, rnd2) 
  f3(I0, I1) -> f2(I1, I2) [I1 <= I0 - 1 /\ 1 <= I0 - 1]
  f4(I3, I4) -> f3(I3, I3 - 2) [0 <= I3 - 1]
  f2(I5, I6) -> f3(I5, I5 - 2) [I5 - 1 <= I5 - 1 /\ 1 <= I5 - 1]
  f2(I7, I8) -> f3(2, 0) [2 = I7]
  f2(I9, I10) -> f2(I9 - 1, I11) [I9 - 1 <= I9 - 1 /\ 1 <= I9 - 1]
  f1(I12, I13) -> f2(I13, I14) [-1 <= I13 - 1 /\ 0 <= I12 - 1]

We use the basic value criterion with the projection function NU:
  NU[f2#(z1,z2)] = z1
  NU[f3#(z1,z2)] = z2

This gives the following inequalities:
  I1 <= I0 - 1 /\ 1 <= I0 - 1  ==>  I1 (>! \union =) I1
  I5 - 1 <= I5 - 1 /\ 1 <= I5 - 1  ==>  I5 >! I5 - 2
  2 = I7  ==>  I7 >! 0
  I9 - 1 <= I9 - 1 /\ 1 <= I9 - 1  ==>  I9 >! I9 - 1

We remove all the strictly oriented dependency pairs.

DP problem for innermost termination.
P =
  f3#(I0, I1) -> f2#(I1, I2) [I1 <= I0 - 1 /\ 1 <= I0 - 1]
R = 
  init(x1, x2) -> f1(rnd1, rnd2) 
  f3(I0, I1) -> f2(I1, I2) [I1 <= I0 - 1 /\ 1 <= I0 - 1]
  f4(I3, I4) -> f3(I3, I3 - 2) [0 <= I3 - 1]
  f2(I5, I6) -> f3(I5, I5 - 2) [I5 - 1 <= I5 - 1 /\ 1 <= I5 - 1]
  f2(I7, I8) -> f3(2, 0) [2 = I7]
  f2(I9, I10) -> f2(I9 - 1, I11) [I9 - 1 <= I9 - 1 /\ 1 <= I9 - 1]
  f1(I12, I13) -> f2(I13, I14) [-1 <= I13 - 1 /\ 0 <= I12 - 1]

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

We have the following SCCs.