/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 =
  f6#(x1, x2, x3, x4, x5) -> f5#(x1, x2, x3, x4, x5) 
  f5#(I0, I1, I2, I3, I4) -> f4#(I0, I1, 0, I3, I4) 
  f3#(I5, I6, I7, I8, I9) -> f1#(I5, I6, I7, I8, I9) 
  f4#(I15, I16, I17, I18, I19) -> f3#(I15, I16, I17, I18, rnd5) [rnd5 = rnd5 /\ I18 <= I15]
  f1#(I20, I21, I22, I23, I24) -> f3#(I20, I21, 1 + I22, I23, 1 + I24) [1 + I22 <= I21]
R = 
  f6(x1, x2, x3, x4, x5) -> f5(x1, x2, x3, x4, x5) 
  f5(I0, I1, I2, I3, I4) -> f4(I0, I1, 0, I3, I4) 
  f3(I5, I6, I7, I8, I9) -> f1(I5, I6, I7, I8, I9) 
  f4(I10, I11, I12, I13, I14) -> f2(I10, I11, I12, I13, I14) [1 + I10 <= I13]
  f4(I15, I16, I17, I18, I19) -> f3(I15, I16, I17, I18, rnd5) [rnd5 = rnd5 /\ I18 <= I15]
  f1(I20, I21, I22, I23, I24) -> f3(I20, I21, 1 + I22, I23, 1 + I24) [1 + I22 <= I21]
  f1(I25, I26, I27, I28, I29) -> f2(I25, I26, I27, I28, I29) [I26 <= I27]

The dependency graph for this problem is:
  0 -> 1
  1 -> 3
  2 -> 4
  3 -> 2
  4 -> 2
Where:
  0) f6#(x1, x2, x3, x4, x5) -> f5#(x1, x2, x3, x4, x5) 
  1) f5#(I0, I1, I2, I3, I4) -> f4#(I0, I1, 0, I3, I4) 
  2) f3#(I5, I6, I7, I8, I9) -> f1#(I5, I6, I7, I8, I9) 
  3) f4#(I15, I16, I17, I18, I19) -> f3#(I15, I16, I17, I18, rnd5) [rnd5 = rnd5 /\ I18 <= I15]
  4) f1#(I20, I21, I22, I23, I24) -> f3#(I20, I21, 1 + I22, I23, 1 + I24) [1 + I22 <= I21]

We have the following SCCs.
  { 2, 4 }

DP problem for innermost termination.
P =
  f3#(I5, I6, I7, I8, I9) -> f1#(I5, I6, I7, I8, I9) 
  f1#(I20, I21, I22, I23, I24) -> f3#(I20, I21, 1 + I22, I23, 1 + I24) [1 + I22 <= I21]
R = 
  f6(x1, x2, x3, x4, x5) -> f5(x1, x2, x3, x4, x5) 
  f5(I0, I1, I2, I3, I4) -> f4(I0, I1, 0, I3, I4) 
  f3(I5, I6, I7, I8, I9) -> f1(I5, I6, I7, I8, I9) 
  f4(I10, I11, I12, I13, I14) -> f2(I10, I11, I12, I13, I14) [1 + I10 <= I13]
  f4(I15, I16, I17, I18, I19) -> f3(I15, I16, I17, I18, rnd5) [rnd5 = rnd5 /\ I18 <= I15]
  f1(I20, I21, I22, I23, I24) -> f3(I20, I21, 1 + I22, I23, 1 + I24) [1 + I22 <= I21]
  f1(I25, I26, I27, I28, I29) -> f2(I25, I26, I27, I28, I29) [I26 <= I27]

We use the reverse value criterion with the projection function NU:
  NU[f1#(z1,z2,z3,z4,z5)] = z2 + -1 * (1 + z3)
  NU[f3#(z1,z2,z3,z4,z5)] = z2 + -1 * (1 + z3)

This gives the following inequalities:
    ==>  I6 + -1 * (1 + I7) >= I6 + -1 * (1 + I7)
  1 + I22 <= I21  ==>  I21 + -1 * (1 + I22) > I21 + -1 * (1 + (1 + I22))  with  I21 + -1 * (1 + I22) >= 0

We remove all the strictly oriented dependency pairs.

DP problem for innermost termination.
P =
  f3#(I5, I6, I7, I8, I9) -> f1#(I5, I6, I7, I8, I9) 
R = 
  f6(x1, x2, x3, x4, x5) -> f5(x1, x2, x3, x4, x5) 
  f5(I0, I1, I2, I3, I4) -> f4(I0, I1, 0, I3, I4) 
  f3(I5, I6, I7, I8, I9) -> f1(I5, I6, I7, I8, I9) 
  f4(I10, I11, I12, I13, I14) -> f2(I10, I11, I12, I13, I14) [1 + I10 <= I13]
  f4(I15, I16, I17, I18, I19) -> f3(I15, I16, I17, I18, rnd5) [rnd5 = rnd5 /\ I18 <= I15]
  f1(I20, I21, I22, I23, I24) -> f3(I20, I21, 1 + I22, I23, 1 + I24) [1 + I22 <= I21]
  f1(I25, I26, I27, I28, I29) -> f2(I25, I26, I27, I28, I29) [I26 <= I27]

The dependency graph for this problem is:
  2 -> 
Where:
  2) f3#(I5, I6, I7, I8, I9) -> f1#(I5, I6, I7, I8, I9) 

We have the following SCCs.