/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 =
  init#(x1, x2, x3) -> f3#(rnd1, rnd2, rnd3) 
  f5#(I0, I1, I2) -> f5#(I3, 0, I4) [1 = I1 /\ 0 <= I4 - 1 /\ 0 <= I3 - 1 /\ 0 <= I2 - 1 /\ 0 <= I0 - 1 /\ I4 <= I2 /\ I4 <= I0 /\ I3 <= I2 /\ I3 <= I0]
  f5#(I5, I6, I7) -> f5#(I8, 1, I9) [0 = I6 /\ -1 <= I9 - 1 /\ -1 <= I8 - 1 /\ 0 <= I7 - 1 /\ 0 <= I5 - 1 /\ I9 + 1 <= I7 /\ I9 + 1 <= I5 /\ I8 + 1 <= I7 /\ I8 + 1 <= I5]
  f2#(I10, I11, I12) -> f5#(I13, 1, I14) [I13 <= I11 /\ 0 <= y1 - 1 /\ I14 <= I11 /\ 0 <= I10 - 1 /\ -1 <= I11 - 1 /\ -1 <= I13 - 1 /\ -1 <= I14 - 1]
  f4#(I15, I16, I17) -> f4#(I15 - 1, I18, I19) [0 <= I15 - 1]
  f3#(I20, I21, I22) -> f4#(I23, I24, I25) [0 <= I20 - 1 /\ -1 <= I23 - 1 /\ -1 <= I21 - 1]
  f3#(I26, I27, I28) -> f2#(I29, I30, I31) [-1 <= I30 - 1 /\ 0 <= I29 - 1 /\ 0 <= I26 - 1 /\ I29 <= I26]
  f1#(I32, I33, I34) -> f2#(I35, I36, I37) [-1 <= I36 - 1 /\ 0 <= I35 - 1 /\ -1 <= I33 - 1 /\ 0 <= I32 - 1 /\ I36 <= I33 /\ I35 - 1 <= I33 /\ I35 <= I32]
R = 
  init(x1, x2, x3) -> f3(rnd1, rnd2, rnd3) 
  f5(I0, I1, I2) -> f5(I3, 0, I4) [1 = I1 /\ 0 <= I4 - 1 /\ 0 <= I3 - 1 /\ 0 <= I2 - 1 /\ 0 <= I0 - 1 /\ I4 <= I2 /\ I4 <= I0 /\ I3 <= I2 /\ I3 <= I0]
  f5(I5, I6, I7) -> f5(I8, 1, I9) [0 = I6 /\ -1 <= I9 - 1 /\ -1 <= I8 - 1 /\ 0 <= I7 - 1 /\ 0 <= I5 - 1 /\ I9 + 1 <= I7 /\ I9 + 1 <= I5 /\ I8 + 1 <= I7 /\ I8 + 1 <= I5]
  f2(I10, I11, I12) -> f5(I13, 1, I14) [I13 <= I11 /\ 0 <= y1 - 1 /\ I14 <= I11 /\ 0 <= I10 - 1 /\ -1 <= I11 - 1 /\ -1 <= I13 - 1 /\ -1 <= I14 - 1]
  f4(I15, I16, I17) -> f4(I15 - 1, I18, I19) [0 <= I15 - 1]
  f3(I20, I21, I22) -> f4(I23, I24, I25) [0 <= I20 - 1 /\ -1 <= I23 - 1 /\ -1 <= I21 - 1]
  f3(I26, I27, I28) -> f2(I29, I30, I31) [-1 <= I30 - 1 /\ 0 <= I29 - 1 /\ 0 <= I26 - 1 /\ I29 <= I26]
  f1(I32, I33, I34) -> f2(I35, I36, I37) [-1 <= I36 - 1 /\ 0 <= I35 - 1 /\ -1 <= I33 - 1 /\ 0 <= I32 - 1 /\ I36 <= I33 /\ I35 - 1 <= I33 /\ I35 <= I32]

The dependency graph for this problem is:
  0 -> 5, 6
  1 -> 2
  2 -> 1
  3 -> 1
  4 -> 4
  5 -> 4
  6 -> 3
  7 -> 3
Where:
  0) init#(x1, x2, x3) -> f3#(rnd1, rnd2, rnd3) 
  1) f5#(I0, I1, I2) -> f5#(I3, 0, I4) [1 = I1 /\ 0 <= I4 - 1 /\ 0 <= I3 - 1 /\ 0 <= I2 - 1 /\ 0 <= I0 - 1 /\ I4 <= I2 /\ I4 <= I0 /\ I3 <= I2 /\ I3 <= I0]
  2) f5#(I5, I6, I7) -> f5#(I8, 1, I9) [0 = I6 /\ -1 <= I9 - 1 /\ -1 <= I8 - 1 /\ 0 <= I7 - 1 /\ 0 <= I5 - 1 /\ I9 + 1 <= I7 /\ I9 + 1 <= I5 /\ I8 + 1 <= I7 /\ I8 + 1 <= I5]
  3) f2#(I10, I11, I12) -> f5#(I13, 1, I14) [I13 <= I11 /\ 0 <= y1 - 1 /\ I14 <= I11 /\ 0 <= I10 - 1 /\ -1 <= I11 - 1 /\ -1 <= I13 - 1 /\ -1 <= I14 - 1]
  4) f4#(I15, I16, I17) -> f4#(I15 - 1, I18, I19) [0 <= I15 - 1]
  5) f3#(I20, I21, I22) -> f4#(I23, I24, I25) [0 <= I20 - 1 /\ -1 <= I23 - 1 /\ -1 <= I21 - 1]
  6) f3#(I26, I27, I28) -> f2#(I29, I30, I31) [-1 <= I30 - 1 /\ 0 <= I29 - 1 /\ 0 <= I26 - 1 /\ I29 <= I26]
  7) f1#(I32, I33, I34) -> f2#(I35, I36, I37) [-1 <= I36 - 1 /\ 0 <= I35 - 1 /\ -1 <= I33 - 1 /\ 0 <= I32 - 1 /\ I36 <= I33 /\ I35 - 1 <= I33 /\ I35 <= I32]

We have the following SCCs.
  { 4 }
  { 1, 2 }

DP problem for innermost termination.
P =
  f5#(I0, I1, I2) -> f5#(I3, 0, I4) [1 = I1 /\ 0 <= I4 - 1 /\ 0 <= I3 - 1 /\ 0 <= I2 - 1 /\ 0 <= I0 - 1 /\ I4 <= I2 /\ I4 <= I0 /\ I3 <= I2 /\ I3 <= I0]
  f5#(I5, I6, I7) -> f5#(I8, 1, I9) [0 = I6 /\ -1 <= I9 - 1 /\ -1 <= I8 - 1 /\ 0 <= I7 - 1 /\ 0 <= I5 - 1 /\ I9 + 1 <= I7 /\ I9 + 1 <= I5 /\ I8 + 1 <= I7 /\ I8 + 1 <= I5]
R = 
  init(x1, x2, x3) -> f3(rnd1, rnd2, rnd3) 
  f5(I0, I1, I2) -> f5(I3, 0, I4) [1 = I1 /\ 0 <= I4 - 1 /\ 0 <= I3 - 1 /\ 0 <= I2 - 1 /\ 0 <= I0 - 1 /\ I4 <= I2 /\ I4 <= I0 /\ I3 <= I2 /\ I3 <= I0]
  f5(I5, I6, I7) -> f5(I8, 1, I9) [0 = I6 /\ -1 <= I9 - 1 /\ -1 <= I8 - 1 /\ 0 <= I7 - 1 /\ 0 <= I5 - 1 /\ I9 + 1 <= I7 /\ I9 + 1 <= I5 /\ I8 + 1 <= I7 /\ I8 + 1 <= I5]
  f2(I10, I11, I12) -> f5(I13, 1, I14) [I13 <= I11 /\ 0 <= y1 - 1 /\ I14 <= I11 /\ 0 <= I10 - 1 /\ -1 <= I11 - 1 /\ -1 <= I13 - 1 /\ -1 <= I14 - 1]
  f4(I15, I16, I17) -> f4(I15 - 1, I18, I19) [0 <= I15 - 1]
  f3(I20, I21, I22) -> f4(I23, I24, I25) [0 <= I20 - 1 /\ -1 <= I23 - 1 /\ -1 <= I21 - 1]
  f3(I26, I27, I28) -> f2(I29, I30, I31) [-1 <= I30 - 1 /\ 0 <= I29 - 1 /\ 0 <= I26 - 1 /\ I29 <= I26]
  f1(I32, I33, I34) -> f2(I35, I36, I37) [-1 <= I36 - 1 /\ 0 <= I35 - 1 /\ -1 <= I33 - 1 /\ 0 <= I32 - 1 /\ I36 <= I33 /\ I35 - 1 <= I33 /\ I35 <= I32]

We use the basic value criterion with the projection function NU:
  NU[f5#(z1,z2,z3)] = z3

This gives the following inequalities:
  1 = I1 /\ 0 <= I4 - 1 /\ 0 <= I3 - 1 /\ 0 <= I2 - 1 /\ 0 <= I0 - 1 /\ I4 <= I2 /\ I4 <= I0 /\ I3 <= I2 /\ I3 <= I0  ==>  I2 (>! \union =) I4
  0 = I6 /\ -1 <= I9 - 1 /\ -1 <= I8 - 1 /\ 0 <= I7 - 1 /\ 0 <= I5 - 1 /\ I9 + 1 <= I7 /\ I9 + 1 <= I5 /\ I8 + 1 <= I7 /\ I8 + 1 <= I5  ==>  I7 >! I9

We remove all the strictly oriented dependency pairs.

DP problem for innermost termination.
P =
  f5#(I0, I1, I2) -> f5#(I3, 0, I4) [1 = I1 /\ 0 <= I4 - 1 /\ 0 <= I3 - 1 /\ 0 <= I2 - 1 /\ 0 <= I0 - 1 /\ I4 <= I2 /\ I4 <= I0 /\ I3 <= I2 /\ I3 <= I0]
R = 
  init(x1, x2, x3) -> f3(rnd1, rnd2, rnd3) 
  f5(I0, I1, I2) -> f5(I3, 0, I4) [1 = I1 /\ 0 <= I4 - 1 /\ 0 <= I3 - 1 /\ 0 <= I2 - 1 /\ 0 <= I0 - 1 /\ I4 <= I2 /\ I4 <= I0 /\ I3 <= I2 /\ I3 <= I0]
  f5(I5, I6, I7) -> f5(I8, 1, I9) [0 = I6 /\ -1 <= I9 - 1 /\ -1 <= I8 - 1 /\ 0 <= I7 - 1 /\ 0 <= I5 - 1 /\ I9 + 1 <= I7 /\ I9 + 1 <= I5 /\ I8 + 1 <= I7 /\ I8 + 1 <= I5]
  f2(I10, I11, I12) -> f5(I13, 1, I14) [I13 <= I11 /\ 0 <= y1 - 1 /\ I14 <= I11 /\ 0 <= I10 - 1 /\ -1 <= I11 - 1 /\ -1 <= I13 - 1 /\ -1 <= I14 - 1]
  f4(I15, I16, I17) -> f4(I15 - 1, I18, I19) [0 <= I15 - 1]
  f3(I20, I21, I22) -> f4(I23, I24, I25) [0 <= I20 - 1 /\ -1 <= I23 - 1 /\ -1 <= I21 - 1]
  f3(I26, I27, I28) -> f2(I29, I30, I31) [-1 <= I30 - 1 /\ 0 <= I29 - 1 /\ 0 <= I26 - 1 /\ I29 <= I26]
  f1(I32, I33, I34) -> f2(I35, I36, I37) [-1 <= I36 - 1 /\ 0 <= I35 - 1 /\ -1 <= I33 - 1 /\ 0 <= I32 - 1 /\ I36 <= I33 /\ I35 - 1 <= I33 /\ I35 <= I32]

The dependency graph for this problem is:
  1 -> 
Where:
  1) f5#(I0, I1, I2) -> f5#(I3, 0, I4) [1 = I1 /\ 0 <= I4 - 1 /\ 0 <= I3 - 1 /\ 0 <= I2 - 1 /\ 0 <= I0 - 1 /\ I4 <= I2 /\ I4 <= I0 /\ I3 <= I2 /\ I3 <= I0]

We have the following SCCs.
  

DP problem for innermost termination.
P =
  f4#(I15, I16, I17) -> f4#(I15 - 1, I18, I19) [0 <= I15 - 1]
R = 
  init(x1, x2, x3) -> f3(rnd1, rnd2, rnd3) 
  f5(I0, I1, I2) -> f5(I3, 0, I4) [1 = I1 /\ 0 <= I4 - 1 /\ 0 <= I3 - 1 /\ 0 <= I2 - 1 /\ 0 <= I0 - 1 /\ I4 <= I2 /\ I4 <= I0 /\ I3 <= I2 /\ I3 <= I0]
  f5(I5, I6, I7) -> f5(I8, 1, I9) [0 = I6 /\ -1 <= I9 - 1 /\ -1 <= I8 - 1 /\ 0 <= I7 - 1 /\ 0 <= I5 - 1 /\ I9 + 1 <= I7 /\ I9 + 1 <= I5 /\ I8 + 1 <= I7 /\ I8 + 1 <= I5]
  f2(I10, I11, I12) -> f5(I13, 1, I14) [I13 <= I11 /\ 0 <= y1 - 1 /\ I14 <= I11 /\ 0 <= I10 - 1 /\ -1 <= I11 - 1 /\ -1 <= I13 - 1 /\ -1 <= I14 - 1]
  f4(I15, I16, I17) -> f4(I15 - 1, I18, I19) [0 <= I15 - 1]
  f3(I20, I21, I22) -> f4(I23, I24, I25) [0 <= I20 - 1 /\ -1 <= I23 - 1 /\ -1 <= I21 - 1]
  f3(I26, I27, I28) -> f2(I29, I30, I31) [-1 <= I30 - 1 /\ 0 <= I29 - 1 /\ 0 <= I26 - 1 /\ I29 <= I26]
  f1(I32, I33, I34) -> f2(I35, I36, I37) [-1 <= I36 - 1 /\ 0 <= I35 - 1 /\ -1 <= I33 - 1 /\ 0 <= I32 - 1 /\ I36 <= I33 /\ I35 - 1 <= I33 /\ I35 <= I32]

We use the basic value criterion with the projection function NU:
  NU[f4#(z1,z2,z3)] = z1

This gives the following inequalities:
  0 <= I15 - 1  ==>  I15 >! I15 - 1

All dependency pairs are strictly oriented, so the entire dependency pair problem may be removed.