/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, x3) -> f1#(rnd1, rnd2, rnd3) 
  f3#(I0, I1, I2) -> f2#(I0 - 1, I1, I2) [I2 <= I0 - 1 /\ I1 <= I2]
  f3#(I3, I4, I5) -> f2#(I3, I4, I5) [I3 <= I5 /\ I4 <= I5]
  f3#(I6, I7, I8) -> f2#(I6, I7 - 1, I8) [I8 <= I7 - 1]
  f2#(I9, I10, I11) -> f3#(I9, I10, I11) [I11 <= I9 - 1 /\ I10 <= I11]
  f2#(I12, I13, I14) -> f3#(I12, I13, I14) [I14 <= I13 - 1]
  f1#(I15, I16, I17) -> f2#(I18, I19, I20) [0 <= I15 - 1 /\ -1 <= I18 - 1 /\ -1 <= I20 - 1 /\ -1 <= I16 - 1 /\ -1 <= I19 - 1]
R = 
  init(x1, x2, x3) -> f1(rnd1, rnd2, rnd3) 
  f3(I0, I1, I2) -> f2(I0 - 1, I1, I2) [I2 <= I0 - 1 /\ I1 <= I2]
  f3(I3, I4, I5) -> f2(I3, I4, I5) [I3 <= I5 /\ I4 <= I5]
  f3(I6, I7, I8) -> f2(I6, I7 - 1, I8) [I8 <= I7 - 1]
  f2(I9, I10, I11) -> f3(I9, I10, I11) [I11 <= I9 - 1 /\ I10 <= I11]
  f2(I12, I13, I14) -> f3(I12, I13, I14) [I14 <= I13 - 1]
  f1(I15, I16, I17) -> f2(I18, I19, I20) [0 <= I15 - 1 /\ -1 <= I18 - 1 /\ -1 <= I20 - 1 /\ -1 <= I16 - 1 /\ -1 <= I19 - 1]

The dependency graph for this problem is:
  0 -> 6
  1 -> 4
  2 -> 
  3 -> 4, 5
  4 -> 1
  5 -> 3
  6 -> 4, 5
Where:
  0) init#(x1, x2, x3) -> f1#(rnd1, rnd2, rnd3) 
  1) f3#(I0, I1, I2) -> f2#(I0 - 1, I1, I2) [I2 <= I0 - 1 /\ I1 <= I2]
  2) f3#(I3, I4, I5) -> f2#(I3, I4, I5) [I3 <= I5 /\ I4 <= I5]
  3) f3#(I6, I7, I8) -> f2#(I6, I7 - 1, I8) [I8 <= I7 - 1]
  4) f2#(I9, I10, I11) -> f3#(I9, I10, I11) [I11 <= I9 - 1 /\ I10 <= I11]
  5) f2#(I12, I13, I14) -> f3#(I12, I13, I14) [I14 <= I13 - 1]
  6) f1#(I15, I16, I17) -> f2#(I18, I19, I20) [0 <= I15 - 1 /\ -1 <= I18 - 1 /\ -1 <= I20 - 1 /\ -1 <= I16 - 1 /\ -1 <= I19 - 1]

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

DP problem for innermost termination.
P =
  f3#(I0, I1, I2) -> f2#(I0 - 1, I1, I2) [I2 <= I0 - 1 /\ I1 <= I2]
  f2#(I9, I10, I11) -> f3#(I9, I10, I11) [I11 <= I9 - 1 /\ I10 <= I11]
R = 
  init(x1, x2, x3) -> f1(rnd1, rnd2, rnd3) 
  f3(I0, I1, I2) -> f2(I0 - 1, I1, I2) [I2 <= I0 - 1 /\ I1 <= I2]
  f3(I3, I4, I5) -> f2(I3, I4, I5) [I3 <= I5 /\ I4 <= I5]
  f3(I6, I7, I8) -> f2(I6, I7 - 1, I8) [I8 <= I7 - 1]
  f2(I9, I10, I11) -> f3(I9, I10, I11) [I11 <= I9 - 1 /\ I10 <= I11]
  f2(I12, I13, I14) -> f3(I12, I13, I14) [I14 <= I13 - 1]
  f1(I15, I16, I17) -> f2(I18, I19, I20) [0 <= I15 - 1 /\ -1 <= I18 - 1 /\ -1 <= I20 - 1 /\ -1 <= I16 - 1 /\ -1 <= I19 - 1]

We use the reverse value criterion with the projection function NU:
  NU[f2#(z1,z2,z3)] = z1 - 1 + -1 * z3
  NU[f3#(z1,z2,z3)] = z1 - 1 - 1 + -1 * z3

This gives the following inequalities:
  I2 <= I0 - 1 /\ I1 <= I2  ==>  I0 - 1 - 1 + -1 * I2 >= I0 - 1 - 1 + -1 * I2
  I11 <= I9 - 1 /\ I10 <= I11  ==>  I9 - 1 + -1 * I11 > I9 - 1 - 1 + -1 * I11  with  I9 - 1 + -1 * I11 >= 0

We remove all the strictly oriented dependency pairs.

DP problem for innermost termination.
P =
  f3#(I0, I1, I2) -> f2#(I0 - 1, I1, I2) [I2 <= I0 - 1 /\ I1 <= I2]
R = 
  init(x1, x2, x3) -> f1(rnd1, rnd2, rnd3) 
  f3(I0, I1, I2) -> f2(I0 - 1, I1, I2) [I2 <= I0 - 1 /\ I1 <= I2]
  f3(I3, I4, I5) -> f2(I3, I4, I5) [I3 <= I5 /\ I4 <= I5]
  f3(I6, I7, I8) -> f2(I6, I7 - 1, I8) [I8 <= I7 - 1]
  f2(I9, I10, I11) -> f3(I9, I10, I11) [I11 <= I9 - 1 /\ I10 <= I11]
  f2(I12, I13, I14) -> f3(I12, I13, I14) [I14 <= I13 - 1]
  f1(I15, I16, I17) -> f2(I18, I19, I20) [0 <= I15 - 1 /\ -1 <= I18 - 1 /\ -1 <= I20 - 1 /\ -1 <= I16 - 1 /\ -1 <= I19 - 1]

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

We have the following SCCs.
  

DP problem for innermost termination.
P =
  f3#(I6, I7, I8) -> f2#(I6, I7 - 1, I8) [I8 <= I7 - 1]
  f2#(I12, I13, I14) -> f3#(I12, I13, I14) [I14 <= I13 - 1]
R = 
  init(x1, x2, x3) -> f1(rnd1, rnd2, rnd3) 
  f3(I0, I1, I2) -> f2(I0 - 1, I1, I2) [I2 <= I0 - 1 /\ I1 <= I2]
  f3(I3, I4, I5) -> f2(I3, I4, I5) [I3 <= I5 /\ I4 <= I5]
  f3(I6, I7, I8) -> f2(I6, I7 - 1, I8) [I8 <= I7 - 1]
  f2(I9, I10, I11) -> f3(I9, I10, I11) [I11 <= I9 - 1 /\ I10 <= I11]
  f2(I12, I13, I14) -> f3(I12, I13, I14) [I14 <= I13 - 1]
  f1(I15, I16, I17) -> f2(I18, I19, I20) [0 <= I15 - 1 /\ -1 <= I18 - 1 /\ -1 <= I20 - 1 /\ -1 <= I16 - 1 /\ -1 <= I19 - 1]

We use the reverse value criterion with the projection function NU:
  NU[f2#(z1,z2,z3)] = z2 - 1 + -1 * z3
  NU[f3#(z1,z2,z3)] = z2 - 1 - 1 + -1 * z3

This gives the following inequalities:
  I8 <= I7 - 1  ==>  I7 - 1 - 1 + -1 * I8 >= I7 - 1 - 1 + -1 * I8
  I14 <= I13 - 1  ==>  I13 - 1 + -1 * I14 > I13 - 1 - 1 + -1 * I14  with  I13 - 1 + -1 * I14 >= 0

We remove all the strictly oriented dependency pairs.

DP problem for innermost termination.
P =
  f3#(I6, I7, I8) -> f2#(I6, I7 - 1, I8) [I8 <= I7 - 1]
R = 
  init(x1, x2, x3) -> f1(rnd1, rnd2, rnd3) 
  f3(I0, I1, I2) -> f2(I0 - 1, I1, I2) [I2 <= I0 - 1 /\ I1 <= I2]
  f3(I3, I4, I5) -> f2(I3, I4, I5) [I3 <= I5 /\ I4 <= I5]
  f3(I6, I7, I8) -> f2(I6, I7 - 1, I8) [I8 <= I7 - 1]
  f2(I9, I10, I11) -> f3(I9, I10, I11) [I11 <= I9 - 1 /\ I10 <= I11]
  f2(I12, I13, I14) -> f3(I12, I13, I14) [I14 <= I13 - 1]
  f1(I15, I16, I17) -> f2(I18, I19, I20) [0 <= I15 - 1 /\ -1 <= I18 - 1 /\ -1 <= I20 - 1 /\ -1 <= I16 - 1 /\ -1 <= I19 - 1]

The dependency graph for this problem is:
  3 -> 
Where:
  3) f3#(I6, I7, I8) -> f2#(I6, I7 - 1, I8) [I8 <= I7 - 1]

We have the following SCCs.