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


--------------------------------------------------------------------------------


YES

DP problem for innermost termination.
P =
  f6#(x1, x2, x3, x4, x5, x6) -> f5#(x1, x2, x3, x4, x5, x6) 
  f5#(I6, I7, I8, I9, I10, I11) -> f1#(I6, I7, I8, I9, I10, I11) [0 <= -2 + I9 /\ 0 <= -2 + I8]
  f3#(I19, I20, I21, I22, I23, I24) -> f1#(I19, I20, I21, I22, I23, I24) 
  f2#(I25, I26, I27, I28, I29, I30) -> f3#(I25, I26, I27, I28, I29, I30) [I26 = I26]
  f1#(I31, I32, I33, I34, I35, I36) -> f2#(I31, I32, I33, I34, 1 + I35, 1 + I36) [0 <= -2 - I35]
R = 
  f6(x1, x2, x3, x4, x5, x6) -> f5(x1, x2, x3, x4, x5, x6) 
  f5(I0, I1, I2, I3, I4, I5) -> f4(rnd1, I1, I2, I3, I4, I5) [rnd1 = rnd1 /\ -1 + I2 <= 0]
  f5(I6, I7, I8, I9, I10, I11) -> f1(I6, I7, I8, I9, I10, I11) [0 <= -2 + I9 /\ 0 <= -2 + I8]
  f1(I12, I13, I14, I15, I16, I17) -> f4(I18, I13, I14, I15, I16, I17) [I18 = I18 /\ -1 - I16 <= 0]
  f3(I19, I20, I21, I22, I23, I24) -> f1(I19, I20, I21, I22, I23, I24) 
  f2(I25, I26, I27, I28, I29, I30) -> f3(I25, I26, I27, I28, I29, I30) [I26 = I26]
  f1(I31, I32, I33, I34, I35, I36) -> f2(I31, I32, I33, I34, 1 + I35, 1 + I36) [0 <= -2 - I35]

The dependency graph for this problem is:
  0 -> 1
  1 -> 4
  2 -> 4
  3 -> 2
  4 -> 3
Where:
  0) f6#(x1, x2, x3, x4, x5, x6) -> f5#(x1, x2, x3, x4, x5, x6) 
  1) f5#(I6, I7, I8, I9, I10, I11) -> f1#(I6, I7, I8, I9, I10, I11) [0 <= -2 + I9 /\ 0 <= -2 + I8]
  2) f3#(I19, I20, I21, I22, I23, I24) -> f1#(I19, I20, I21, I22, I23, I24) 
  3) f2#(I25, I26, I27, I28, I29, I30) -> f3#(I25, I26, I27, I28, I29, I30) [I26 = I26]
  4) f1#(I31, I32, I33, I34, I35, I36) -> f2#(I31, I32, I33, I34, 1 + I35, 1 + I36) [0 <= -2 - I35]

We have the following SCCs.
  { 2, 3, 4 }

DP problem for innermost termination.
P =
  f3#(I19, I20, I21, I22, I23, I24) -> f1#(I19, I20, I21, I22, I23, I24) 
  f2#(I25, I26, I27, I28, I29, I30) -> f3#(I25, I26, I27, I28, I29, I30) [I26 = I26]
  f1#(I31, I32, I33, I34, I35, I36) -> f2#(I31, I32, I33, I34, 1 + I35, 1 + I36) [0 <= -2 - I35]
R = 
  f6(x1, x2, x3, x4, x5, x6) -> f5(x1, x2, x3, x4, x5, x6) 
  f5(I0, I1, I2, I3, I4, I5) -> f4(rnd1, I1, I2, I3, I4, I5) [rnd1 = rnd1 /\ -1 + I2 <= 0]
  f5(I6, I7, I8, I9, I10, I11) -> f1(I6, I7, I8, I9, I10, I11) [0 <= -2 + I9 /\ 0 <= -2 + I8]
  f1(I12, I13, I14, I15, I16, I17) -> f4(I18, I13, I14, I15, I16, I17) [I18 = I18 /\ -1 - I16 <= 0]
  f3(I19, I20, I21, I22, I23, I24) -> f1(I19, I20, I21, I22, I23, I24) 
  f2(I25, I26, I27, I28, I29, I30) -> f3(I25, I26, I27, I28, I29, I30) [I26 = I26]
  f1(I31, I32, I33, I34, I35, I36) -> f2(I31, I32, I33, I34, 1 + I35, 1 + I36) [0 <= -2 - I35]

We use the extended value criterion with the projection function NU:
  NU[f2#(x0,x1,x2,x3,x4,x5)] = -x4 - 2
  NU[f1#(x0,x1,x2,x3,x4,x5)] = -x4 - 2
  NU[f3#(x0,x1,x2,x3,x4,x5)] = -x4 - 2

This gives the following inequalities:
    ==>  -I23 - 2 >= -I23 - 2
  I26 = I26  ==>  -I29 - 2 >= -I29 - 2
  0 <= -2 - I35  ==>  -I35 - 2 > -(1 + I35) - 2  with  -I35 - 2 >= 0

We remove all the strictly oriented dependency pairs.

DP problem for innermost termination.
P =
  f3#(I19, I20, I21, I22, I23, I24) -> f1#(I19, I20, I21, I22, I23, I24) 
  f2#(I25, I26, I27, I28, I29, I30) -> f3#(I25, I26, I27, I28, I29, I30) [I26 = I26]
R = 
  f6(x1, x2, x3, x4, x5, x6) -> f5(x1, x2, x3, x4, x5, x6) 
  f5(I0, I1, I2, I3, I4, I5) -> f4(rnd1, I1, I2, I3, I4, I5) [rnd1 = rnd1 /\ -1 + I2 <= 0]
  f5(I6, I7, I8, I9, I10, I11) -> f1(I6, I7, I8, I9, I10, I11) [0 <= -2 + I9 /\ 0 <= -2 + I8]
  f1(I12, I13, I14, I15, I16, I17) -> f4(I18, I13, I14, I15, I16, I17) [I18 = I18 /\ -1 - I16 <= 0]
  f3(I19, I20, I21, I22, I23, I24) -> f1(I19, I20, I21, I22, I23, I24) 
  f2(I25, I26, I27, I28, I29, I30) -> f3(I25, I26, I27, I28, I29, I30) [I26 = I26]
  f1(I31, I32, I33, I34, I35, I36) -> f2(I31, I32, I33, I34, 1 + I35, 1 + I36) [0 <= -2 - I35]

The dependency graph for this problem is:
  2 -> 
  3 -> 2
Where:
  2) f3#(I19, I20, I21, I22, I23, I24) -> f1#(I19, I20, I21, I22, I23, I24) 
  3) f2#(I25, I26, I27, I28, I29, I30) -> f3#(I25, I26, I27, I28, I29, I30) [I26 = I26]

We have the following SCCs.