/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) -> f4#(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) -> f4#(I24, I27, I26, 0) 
  f4#(I28, I29, I30, I31) -> f3#(I28, I31, I30, I31) [I31 <= I28]
  f4#(I32, I33, I34, I35) -> f1#(I32, I35, I34, I35) [1 + I32 <= I35]
  f3#(I36, I37, I38, I39) -> f4#(I36, I39, I38, 1 + I39) 
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) -> f4(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) -> f2(I16, I17, I18, I19) 
  f6(I20, I21, I22, I23) -> f5(I20, I23, rnd3, rnd4) [rnd4 = rnd3 /\ rnd3 = rnd3]
  f5(I24, I25, I26, I27) -> f4(I24, I27, I26, 0) 
  f4(I28, I29, I30, I31) -> f3(I28, I31, I30, I31) [I31 <= I28]
  f4(I32, I33, I34, I35) -> f1(I32, I35, I34, I35) [1 + I32 <= I35]
  f3(I36, I37, I38, I39) -> f4(I36, I39, I38, 1 + I39) 
  f1(I40, I41, I42, I43) -> f2(I40, I43, I44, I45) [I45 = I44 /\ I44 = I44]

The dependency graph for this problem is:
  0 -> 1, 2, 3, 4, 5
  1 -> 6
  2 -> 7, 8
  3 -> 9
  4 -> 
  5 -> 6
  6 -> 7, 8
  7 -> 9
  8 -> 
  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) -> f4#(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) -> f4#(I24, I27, I26, 0) 
  7) f4#(I28, I29, I30, I31) -> f3#(I28, I31, I30, I31) [I31 <= I28]
  8) f4#(I32, I33, I34, I35) -> f1#(I32, I35, I34, I35) [1 + I32 <= I35]
  9) f3#(I36, I37, I38, I39) -> f4#(I36, I39, I38, 1 + I39) 

We have the following SCCs.
  { 7, 9 }

DP problem for innermost termination.
P =
  f4#(I28, I29, I30, I31) -> f3#(I28, I31, I30, I31) [I31 <= I28]
  f3#(I36, I37, I38, I39) -> f4#(I36, I39, I38, 1 + I39) 
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) -> f4(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) -> f2(I16, I17, I18, I19) 
  f6(I20, I21, I22, I23) -> f5(I20, I23, rnd3, rnd4) [rnd4 = rnd3 /\ rnd3 = rnd3]
  f5(I24, I25, I26, I27) -> f4(I24, I27, I26, 0) 
  f4(I28, I29, I30, I31) -> f3(I28, I31, I30, I31) [I31 <= I28]
  f4(I32, I33, I34, I35) -> f1(I32, I35, I34, I35) [1 + I32 <= I35]
  f3(I36, I37, I38, I39) -> f4(I36, I39, I38, 1 + I39) 
  f1(I40, I41, I42, I43) -> f2(I40, I43, I44, I45) [I45 = I44 /\ I44 = I44]

We use the reverse value criterion with the projection function NU:
  NU[f3#(z1,z2,z3,z4)] = z1 + -1 * (1 + z4)
  NU[f4#(z1,z2,z3,z4)] = z1 + -1 * z4

This gives the following inequalities:
  I31 <= I28  ==>  I28 + -1 * I31 > I28 + -1 * (1 + I31)  with  I28 + -1 * I31 >= 0
    ==>  I36 + -1 * (1 + I39) >= I36 + -1 * (1 + I39)

We remove all the strictly oriented dependency pairs.

DP problem for innermost termination.
P =
  f3#(I36, I37, I38, I39) -> f4#(I36, I39, I38, 1 + I39) 
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) -> f4(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) -> f2(I16, I17, I18, I19) 
  f6(I20, I21, I22, I23) -> f5(I20, I23, rnd3, rnd4) [rnd4 = rnd3 /\ rnd3 = rnd3]
  f5(I24, I25, I26, I27) -> f4(I24, I27, I26, 0) 
  f4(I28, I29, I30, I31) -> f3(I28, I31, I30, I31) [I31 <= I28]
  f4(I32, I33, I34, I35) -> f1(I32, I35, I34, I35) [1 + I32 <= I35]
  f3(I36, I37, I38, I39) -> f4(I36, I39, I38, 1 + I39) 
  f1(I40, I41, I42, I43) -> f2(I40, I43, I44, I45) [I45 = I44 /\ I44 = I44]

The dependency graph for this problem is:
  9 -> 
Where:
  9) f3#(I36, I37, I38, I39) -> f4#(I36, I39, I38, 1 + I39) 

We have the following SCCs.