/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 =
  f7#(x1, x2, x3, x4) -> f6#(x1, x2, x3, x4) 
  f6#(I0, I1, I2, I3) -> f5#(I0, I1, I2, I3) 
  f6#(I4, I5, I6, I7) -> f2#(I4, I5, I6, I7) 
  f6#(I8, I9, I10, I11) -> f3#(I8, I9, I10, I11) 
  f6#(I12, I13, I14, I15) -> f1#(I12, I13, I14, I15) 
  f6#(I20, I21, I22, I23) -> f5#(I20, I23, rnd3, rnd4) [rnd4 = rnd3 /\ rnd3 = rnd3]
  f5#(I24, I25, I26, I27) -> f2#(I24, I27, I26, 0) 
  f2#(I28, I29, I30, I31) -> f3#(I28, I31, I30, I31) [1 + I28 <= I31]
  f2#(I32, I33, I34, I35) -> f1#(I32, I35, I34, I35) [I35 <= I32]
  f1#(I42, I43, I44, I45) -> f2#(I42, I45, I44, 1 + I45) 
R = 
  f7(x1, x2, x3, x4) -> f6(x1, x2, x3, x4) 
  f6(I0, I1, I2, I3) -> f5(I0, I1, I2, I3) 
  f6(I4, I5, I6, I7) -> f2(I4, I5, I6, I7) 
  f6(I8, I9, I10, I11) -> f3(I8, I9, I10, I11) 
  f6(I12, I13, I14, I15) -> f1(I12, I13, I14, I15) 
  f6(I16, I17, I18, I19) -> f4(I16, I17, I18, I19) 
  f6(I20, I21, I22, I23) -> f5(I20, I23, rnd3, rnd4) [rnd4 = rnd3 /\ rnd3 = rnd3]
  f5(I24, I25, I26, I27) -> f2(I24, I27, I26, 0) 
  f2(I28, I29, I30, I31) -> f3(I28, I31, I30, I31) [1 + I28 <= I31]
  f2(I32, I33, I34, I35) -> f1(I32, I35, I34, I35) [I35 <= I32]
  f3(I36, I37, I38, I39) -> f4(I36, I39, I40, I41) [I41 = I40 /\ I40 = I40]
  f1(I42, I43, I44, I45) -> f2(I42, I45, I44, 1 + I45) 

The dependency graph for this problem is:
  0 -> 1, 2, 3, 4, 5
  1 -> 6
  2 -> 7, 8
  3 -> 
  4 -> 9
  5 -> 6
  6 -> 7, 8
  7 -> 
  8 -> 9
  9 -> 7, 8
Where:
  0) f7#(x1, x2, x3, x4) -> f6#(x1, x2, x3, x4) 
  1) f6#(I0, I1, I2, I3) -> f5#(I0, I1, I2, I3) 
  2) f6#(I4, I5, I6, I7) -> f2#(I4, I5, I6, I7) 
  3) f6#(I8, I9, I10, I11) -> f3#(I8, I9, I10, I11) 
  4) f6#(I12, I13, I14, I15) -> f1#(I12, I13, I14, I15) 
  5) f6#(I20, I21, I22, I23) -> f5#(I20, I23, rnd3, rnd4) [rnd4 = rnd3 /\ rnd3 = rnd3]
  6) f5#(I24, I25, I26, I27) -> f2#(I24, I27, I26, 0) 
  7) f2#(I28, I29, I30, I31) -> f3#(I28, I31, I30, I31) [1 + I28 <= I31]
  8) f2#(I32, I33, I34, I35) -> f1#(I32, I35, I34, I35) [I35 <= I32]
  9) f1#(I42, I43, I44, I45) -> f2#(I42, I45, I44, 1 + I45) 

We have the following SCCs.
  { 8, 9 }

DP problem for innermost termination.
P =
  f2#(I32, I33, I34, I35) -> f1#(I32, I35, I34, I35) [I35 <= I32]
  f1#(I42, I43, I44, I45) -> f2#(I42, I45, I44, 1 + I45) 
R = 
  f7(x1, x2, x3, x4) -> f6(x1, x2, x3, x4) 
  f6(I0, I1, I2, I3) -> f5(I0, I1, I2, I3) 
  f6(I4, I5, I6, I7) -> f2(I4, I5, I6, I7) 
  f6(I8, I9, I10, I11) -> f3(I8, I9, I10, I11) 
  f6(I12, I13, I14, I15) -> f1(I12, I13, I14, I15) 
  f6(I16, I17, I18, I19) -> f4(I16, I17, I18, I19) 
  f6(I20, I21, I22, I23) -> f5(I20, I23, rnd3, rnd4) [rnd4 = rnd3 /\ rnd3 = rnd3]
  f5(I24, I25, I26, I27) -> f2(I24, I27, I26, 0) 
  f2(I28, I29, I30, I31) -> f3(I28, I31, I30, I31) [1 + I28 <= I31]
  f2(I32, I33, I34, I35) -> f1(I32, I35, I34, I35) [I35 <= I32]
  f3(I36, I37, I38, I39) -> f4(I36, I39, I40, I41) [I41 = I40 /\ I40 = I40]
  f1(I42, I43, I44, I45) -> f2(I42, I45, I44, 1 + I45) 

We use the reverse value criterion with the projection function NU:
  NU[f1#(z1,z2,z3,z4)] = z1 + -1 * (1 + z4)
  NU[f2#(z1,z2,z3,z4)] = z1 + -1 * z4

This gives the following inequalities:
  I35 <= I32  ==>  I32 + -1 * I35 > I32 + -1 * (1 + I35)  with  I32 + -1 * I35 >= 0
    ==>  I42 + -1 * (1 + I45) >= I42 + -1 * (1 + I45)

We remove all the strictly oriented dependency pairs.

DP problem for innermost termination.
P =
  f1#(I42, I43, I44, I45) -> f2#(I42, I45, I44, 1 + I45) 
R = 
  f7(x1, x2, x3, x4) -> f6(x1, x2, x3, x4) 
  f6(I0, I1, I2, I3) -> f5(I0, I1, I2, I3) 
  f6(I4, I5, I6, I7) -> f2(I4, I5, I6, I7) 
  f6(I8, I9, I10, I11) -> f3(I8, I9, I10, I11) 
  f6(I12, I13, I14, I15) -> f1(I12, I13, I14, I15) 
  f6(I16, I17, I18, I19) -> f4(I16, I17, I18, I19) 
  f6(I20, I21, I22, I23) -> f5(I20, I23, rnd3, rnd4) [rnd4 = rnd3 /\ rnd3 = rnd3]
  f5(I24, I25, I26, I27) -> f2(I24, I27, I26, 0) 
  f2(I28, I29, I30, I31) -> f3(I28, I31, I30, I31) [1 + I28 <= I31]
  f2(I32, I33, I34, I35) -> f1(I32, I35, I34, I35) [I35 <= I32]
  f3(I36, I37, I38, I39) -> f4(I36, I39, I40, I41) [I41 = I40 /\ I40 = I40]
  f1(I42, I43, I44, I45) -> f2(I42, I45, I44, 1 + I45) 

The dependency graph for this problem is:
  9 -> 
Where:
  9) f1#(I42, I43, I44, I45) -> f2#(I42, I45, I44, 1 + I45) 

We have the following SCCs.