/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) -> f6#(x1, x2, x3) 
  f6#(I0, I1, I2) -> f3#(I0, I1, I2) 
  f6#(I3, I4, I5) -> f5#(I3, I4, I5) 
  f6#(I6, I7, I8) -> f4#(I6, I7, I8) 
  f6#(I9, I10, I11) -> f1#(I9, I10, I11) 
  f3#(I15, I16, I17) -> f5#(I17, I16, I17) 
  f5#(I18, I19, I20) -> f4#(I20, I19, I20) [1 + I20 <= 0]
  f5#(I21, I22, I23) -> f1#(I23, I22, I23) [0 <= I23]
  f1#(I27, I28, I29) -> f3#(I29, I28, -1 + I29) 
R = 
  f7(x1, x2, x3) -> f6(x1, x2, x3) 
  f6(I0, I1, I2) -> f3(I0, I1, I2) 
  f6(I3, I4, I5) -> f5(I3, I4, I5) 
  f6(I6, I7, I8) -> f4(I6, I7, I8) 
  f6(I9, I10, I11) -> f1(I9, I10, I11) 
  f6(I12, I13, I14) -> f2(I12, I13, I14) 
  f3(I15, I16, I17) -> f5(I17, I16, I17) 
  f5(I18, I19, I20) -> f4(I20, I19, I20) [1 + I20 <= 0]
  f5(I21, I22, I23) -> f1(I23, I22, I23) [0 <= I23]
  f4(I24, I25, I26) -> f2(I26, rnd2, rnd3) [rnd3 = rnd2 /\ rnd2 = rnd2]
  f1(I27, I28, I29) -> f3(I29, I28, -1 + I29) 
  f1(I30, I31, I32) -> f2(I32, I33, I34) [I34 = I33 /\ I33 = I33]

The dependency graph for this problem is:
  0 -> 1, 2, 3, 4
  1 -> 5
  2 -> 6, 7
  3 -> 
  4 -> 8
  5 -> 6, 7
  6 -> 
  7 -> 8
  8 -> 5
Where:
  0) f7#(x1, x2, x3) -> f6#(x1, x2, x3) 
  1) f6#(I0, I1, I2) -> f3#(I0, I1, I2) 
  2) f6#(I3, I4, I5) -> f5#(I3, I4, I5) 
  3) f6#(I6, I7, I8) -> f4#(I6, I7, I8) 
  4) f6#(I9, I10, I11) -> f1#(I9, I10, I11) 
  5) f3#(I15, I16, I17) -> f5#(I17, I16, I17) 
  6) f5#(I18, I19, I20) -> f4#(I20, I19, I20) [1 + I20 <= 0]
  7) f5#(I21, I22, I23) -> f1#(I23, I22, I23) [0 <= I23]
  8) f1#(I27, I28, I29) -> f3#(I29, I28, -1 + I29) 

We have the following SCCs.
  { 5, 7, 8 }

DP problem for innermost termination.
P =
  f3#(I15, I16, I17) -> f5#(I17, I16, I17) 
  f5#(I21, I22, I23) -> f1#(I23, I22, I23) [0 <= I23]
  f1#(I27, I28, I29) -> f3#(I29, I28, -1 + I29) 
R = 
  f7(x1, x2, x3) -> f6(x1, x2, x3) 
  f6(I0, I1, I2) -> f3(I0, I1, I2) 
  f6(I3, I4, I5) -> f5(I3, I4, I5) 
  f6(I6, I7, I8) -> f4(I6, I7, I8) 
  f6(I9, I10, I11) -> f1(I9, I10, I11) 
  f6(I12, I13, I14) -> f2(I12, I13, I14) 
  f3(I15, I16, I17) -> f5(I17, I16, I17) 
  f5(I18, I19, I20) -> f4(I20, I19, I20) [1 + I20 <= 0]
  f5(I21, I22, I23) -> f1(I23, I22, I23) [0 <= I23]
  f4(I24, I25, I26) -> f2(I26, rnd2, rnd3) [rnd3 = rnd2 /\ rnd2 = rnd2]
  f1(I27, I28, I29) -> f3(I29, I28, -1 + I29) 
  f1(I30, I31, I32) -> f2(I32, I33, I34) [I34 = I33 /\ I33 = I33]

We use the extended value criterion with the projection function NU:
  NU[f1#(x0,x1,x2)] = x2
  NU[f5#(x0,x1,x2)] = x2 + 1
  NU[f3#(x0,x1,x2)] = x2 + 1

This gives the following inequalities:
    ==>  I17 + 1 >= I17 + 1
  0 <= I23  ==>  I23 + 1 > I23  with  I23 + 1 >= 0
    ==>  I29 >= (-1 + I29) + 1

We remove all the strictly oriented dependency pairs.

DP problem for innermost termination.
P =
  f3#(I15, I16, I17) -> f5#(I17, I16, I17) 
  f1#(I27, I28, I29) -> f3#(I29, I28, -1 + I29) 
R = 
  f7(x1, x2, x3) -> f6(x1, x2, x3) 
  f6(I0, I1, I2) -> f3(I0, I1, I2) 
  f6(I3, I4, I5) -> f5(I3, I4, I5) 
  f6(I6, I7, I8) -> f4(I6, I7, I8) 
  f6(I9, I10, I11) -> f1(I9, I10, I11) 
  f6(I12, I13, I14) -> f2(I12, I13, I14) 
  f3(I15, I16, I17) -> f5(I17, I16, I17) 
  f5(I18, I19, I20) -> f4(I20, I19, I20) [1 + I20 <= 0]
  f5(I21, I22, I23) -> f1(I23, I22, I23) [0 <= I23]
  f4(I24, I25, I26) -> f2(I26, rnd2, rnd3) [rnd3 = rnd2 /\ rnd2 = rnd2]
  f1(I27, I28, I29) -> f3(I29, I28, -1 + I29) 
  f1(I30, I31, I32) -> f2(I32, I33, I34) [I34 = I33 /\ I33 = I33]

The dependency graph for this problem is:
  5 -> 
  8 -> 5
Where:
  5) f3#(I15, I16, I17) -> f5#(I17, I16, I17) 
  8) f1#(I27, I28, I29) -> f3#(I29, I28, -1 + I29) 

We have the following SCCs.