/export/starexec/sandbox2/solver/bin/starexec_run_Transition /export/starexec/sandbox2/benchmark/theBenchmark.smt2 /export/starexec/sandbox2/output/output_files


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YES

DP problem for innermost termination.
P =
  f6#(x1, x2, x3, x4, x5, x6) -> f4#(x1, x2, x3, x4, x5, x6) 
  f5#(I0, I1, I2, I3, I4, I5) -> f1#(I0, I1, I2, I3, 1 + I4, I5) [0 <= -1 + I5 /\ 0 <= I2 - I4]
  f4#(I12, I13, I14, I15, I16, I17) -> f5#(I12, I13, I14, I15, 0, I17) 
  f3#(I18, I19, I20, I21, I22, I23) -> f1#(I18, I19, I20, I21, I22, I23) 
  f1#(I24, I25, I26, I27, I28, I29) -> f3#(I24, I25, I26, I27, 1 + I28, I29) [0 <= -1 + I29 /\ 0 <= I26 - I28]
R = 
  f6(x1, x2, x3, x4, x5, x6) -> f4(x1, x2, x3, x4, x5, x6) 
  f5(I0, I1, I2, I3, I4, I5) -> f1(I0, I1, I2, I3, 1 + I4, I5) [0 <= -1 + I5 /\ 0 <= I2 - I4]
  f5(I6, I7, I8, I9, I10, I11) -> f2(rnd1, rnd2, I8, 0, I10, I11) [rnd1 = rnd2 /\ rnd2 = 0 /\ I11 <= 0 /\ 0 <= I8 - I10]
  f4(I12, I13, I14, I15, I16, I17) -> f5(I12, I13, I14, I15, 0, I17) 
  f3(I18, I19, I20, I21, I22, I23) -> f1(I18, I19, I20, I21, I22, I23) 
  f1(I24, I25, I26, I27, I28, I29) -> f3(I24, I25, I26, I27, 1 + I28, I29) [0 <= -1 + I29 /\ 0 <= I26 - I28]
  f1(I30, I31, I32, I33, I34, I35) -> f2(I36, I37, I32, 0, I34, I35) [I36 = I37 /\ I37 = 0 /\ 1 + I32 - I34 <= 0]

The dependency graph for this problem is:
  0 -> 2
  1 -> 4
  2 -> 1
  3 -> 4
  4 -> 3
Where:
  0) f6#(x1, x2, x3, x4, x5, x6) -> f4#(x1, x2, x3, x4, x5, x6) 
  1) f5#(I0, I1, I2, I3, I4, I5) -> f1#(I0, I1, I2, I3, 1 + I4, I5) [0 <= -1 + I5 /\ 0 <= I2 - I4]
  2) f4#(I12, I13, I14, I15, I16, I17) -> f5#(I12, I13, I14, I15, 0, I17) 
  3) f3#(I18, I19, I20, I21, I22, I23) -> f1#(I18, I19, I20, I21, I22, I23) 
  4) f1#(I24, I25, I26, I27, I28, I29) -> f3#(I24, I25, I26, I27, 1 + I28, I29) [0 <= -1 + I29 /\ 0 <= I26 - I28]

We have the following SCCs.
  { 3, 4 }

DP problem for innermost termination.
P =
  f3#(I18, I19, I20, I21, I22, I23) -> f1#(I18, I19, I20, I21, I22, I23) 
  f1#(I24, I25, I26, I27, I28, I29) -> f3#(I24, I25, I26, I27, 1 + I28, I29) [0 <= -1 + I29 /\ 0 <= I26 - I28]
R = 
  f6(x1, x2, x3, x4, x5, x6) -> f4(x1, x2, x3, x4, x5, x6) 
  f5(I0, I1, I2, I3, I4, I5) -> f1(I0, I1, I2, I3, 1 + I4, I5) [0 <= -1 + I5 /\ 0 <= I2 - I4]
  f5(I6, I7, I8, I9, I10, I11) -> f2(rnd1, rnd2, I8, 0, I10, I11) [rnd1 = rnd2 /\ rnd2 = 0 /\ I11 <= 0 /\ 0 <= I8 - I10]
  f4(I12, I13, I14, I15, I16, I17) -> f5(I12, I13, I14, I15, 0, I17) 
  f3(I18, I19, I20, I21, I22, I23) -> f1(I18, I19, I20, I21, I22, I23) 
  f1(I24, I25, I26, I27, I28, I29) -> f3(I24, I25, I26, I27, 1 + I28, I29) [0 <= -1 + I29 /\ 0 <= I26 - I28]
  f1(I30, I31, I32, I33, I34, I35) -> f2(I36, I37, I32, 0, I34, I35) [I36 = I37 /\ I37 = 0 /\ 1 + I32 - I34 <= 0]

We use the reverse value criterion with the projection function NU:
  NU[f1#(z1,z2,z3,z4,z5,z6)] = z3 - z5 + -1 * 0
  NU[f3#(z1,z2,z3,z4,z5,z6)] = z3 - z5 + -1 * 0

This gives the following inequalities:
    ==>  I20 - I22 + -1 * 0 >= I20 - I22 + -1 * 0
  0 <= -1 + I29 /\ 0 <= I26 - I28  ==>  I26 - I28 + -1 * 0 > I26 - (1 + I28) + -1 * 0  with  I26 - I28 + -1 * 0 >= 0

We remove all the strictly oriented dependency pairs.

DP problem for innermost termination.
P =
  f3#(I18, I19, I20, I21, I22, I23) -> f1#(I18, I19, I20, I21, I22, I23) 
R = 
  f6(x1, x2, x3, x4, x5, x6) -> f4(x1, x2, x3, x4, x5, x6) 
  f5(I0, I1, I2, I3, I4, I5) -> f1(I0, I1, I2, I3, 1 + I4, I5) [0 <= -1 + I5 /\ 0 <= I2 - I4]
  f5(I6, I7, I8, I9, I10, I11) -> f2(rnd1, rnd2, I8, 0, I10, I11) [rnd1 = rnd2 /\ rnd2 = 0 /\ I11 <= 0 /\ 0 <= I8 - I10]
  f4(I12, I13, I14, I15, I16, I17) -> f5(I12, I13, I14, I15, 0, I17) 
  f3(I18, I19, I20, I21, I22, I23) -> f1(I18, I19, I20, I21, I22, I23) 
  f1(I24, I25, I26, I27, I28, I29) -> f3(I24, I25, I26, I27, 1 + I28, I29) [0 <= -1 + I29 /\ 0 <= I26 - I28]
  f1(I30, I31, I32, I33, I34, I35) -> f2(I36, I37, I32, 0, I34, I35) [I36 = I37 /\ I37 = 0 /\ 1 + I32 - I34 <= 0]

The dependency graph for this problem is:
  3 -> 
Where:
  3) f3#(I18, I19, I20, I21, I22, I23) -> f1#(I18, I19, I20, I21, I22, I23) 

We have the following SCCs.