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


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MAYBE

DP problem for innermost termination.
P =
  f5#(x1, x2, x3, x4) -> f4#(x1, x2, x3, x4) 
  f4#(I0, I1, I2, I3) -> f2#(0, 0, rnd3, rnd4) [rnd4 = rnd4 /\ rnd3 = rnd3]
  f2#(I4, I5, I6, I7) -> f1#(0, I5, I6, I7) [I6 <= 0 /\ y1 = 1]
  f3#(I8, I9, I10, I11) -> f1#(I8, I9, I10, I11) 
  f1#(I12, I13, I14, I15) -> f3#(I12, I13, I14, -1 + I15) [1 <= I15]
  f1#(I16, I17, I18, I19) -> f2#(I16, 0, I20, I19) [I19 <= 0 /\ I21 = 0 /\ I20 = I20]
R = 
  f5(x1, x2, x3, x4) -> f4(x1, x2, x3, x4) 
  f4(I0, I1, I2, I3) -> f2(0, 0, rnd3, rnd4) [rnd4 = rnd4 /\ rnd3 = rnd3]
  f2(I4, I5, I6, I7) -> f1(0, I5, I6, I7) [I6 <= 0 /\ y1 = 1]
  f3(I8, I9, I10, I11) -> f1(I8, I9, I10, I11) 
  f1(I12, I13, I14, I15) -> f3(I12, I13, I14, -1 + I15) [1 <= I15]
  f1(I16, I17, I18, I19) -> f2(I16, 0, I20, I19) [I19 <= 0 /\ I21 = 0 /\ I20 = I20]

The dependency graph for this problem is:
  0 -> 1
  1 -> 2
  2 -> 4, 5
  3 -> 4, 5
  4 -> 3
  5 -> 2
Where:
  0) f5#(x1, x2, x3, x4) -> f4#(x1, x2, x3, x4) 
  1) f4#(I0, I1, I2, I3) -> f2#(0, 0, rnd3, rnd4) [rnd4 = rnd4 /\ rnd3 = rnd3]
  2) f2#(I4, I5, I6, I7) -> f1#(0, I5, I6, I7) [I6 <= 0 /\ y1 = 1]
  3) f3#(I8, I9, I10, I11) -> f1#(I8, I9, I10, I11) 
  4) f1#(I12, I13, I14, I15) -> f3#(I12, I13, I14, -1 + I15) [1 <= I15]
  5) f1#(I16, I17, I18, I19) -> f2#(I16, 0, I20, I19) [I19 <= 0 /\ I21 = 0 /\ I20 = I20]

We have the following SCCs.
  { 2, 3, 4, 5 }

DP problem for innermost termination.
P =
  f2#(I4, I5, I6, I7) -> f1#(0, I5, I6, I7) [I6 <= 0 /\ y1 = 1]
  f3#(I8, I9, I10, I11) -> f1#(I8, I9, I10, I11) 
  f1#(I12, I13, I14, I15) -> f3#(I12, I13, I14, -1 + I15) [1 <= I15]
  f1#(I16, I17, I18, I19) -> f2#(I16, 0, I20, I19) [I19 <= 0 /\ I21 = 0 /\ I20 = I20]
R = 
  f5(x1, x2, x3, x4) -> f4(x1, x2, x3, x4) 
  f4(I0, I1, I2, I3) -> f2(0, 0, rnd3, rnd4) [rnd4 = rnd4 /\ rnd3 = rnd3]
  f2(I4, I5, I6, I7) -> f1(0, I5, I6, I7) [I6 <= 0 /\ y1 = 1]
  f3(I8, I9, I10, I11) -> f1(I8, I9, I10, I11) 
  f1(I12, I13, I14, I15) -> f3(I12, I13, I14, -1 + I15) [1 <= I15]
  f1(I16, I17, I18, I19) -> f2(I16, 0, I20, I19) [I19 <= 0 /\ I21 = 0 /\ I20 = I20]

We use the basic value criterion with the projection function NU:
  NU[f3#(z1,z2,z3,z4)] = z4
  NU[f1#(z1,z2,z3,z4)] = z4
  NU[f2#(z1,z2,z3,z4)] = z4

This gives the following inequalities:
  I6 <= 0 /\ y1 = 1  ==>  I7 (>! \union =) I7
    ==>  I11 (>! \union =) I11
  1 <= I15  ==>  I15 >! -1 + I15
  I19 <= 0 /\ I21 = 0 /\ I20 = I20  ==>  I19 (>! \union =) I19

We remove all the strictly oriented dependency pairs.

DP problem for innermost termination.
P =
  f2#(I4, I5, I6, I7) -> f1#(0, I5, I6, I7) [I6 <= 0 /\ y1 = 1]
  f3#(I8, I9, I10, I11) -> f1#(I8, I9, I10, I11) 
  f1#(I16, I17, I18, I19) -> f2#(I16, 0, I20, I19) [I19 <= 0 /\ I21 = 0 /\ I20 = I20]
R = 
  f5(x1, x2, x3, x4) -> f4(x1, x2, x3, x4) 
  f4(I0, I1, I2, I3) -> f2(0, 0, rnd3, rnd4) [rnd4 = rnd4 /\ rnd3 = rnd3]
  f2(I4, I5, I6, I7) -> f1(0, I5, I6, I7) [I6 <= 0 /\ y1 = 1]
  f3(I8, I9, I10, I11) -> f1(I8, I9, I10, I11) 
  f1(I12, I13, I14, I15) -> f3(I12, I13, I14, -1 + I15) [1 <= I15]
  f1(I16, I17, I18, I19) -> f2(I16, 0, I20, I19) [I19 <= 0 /\ I21 = 0 /\ I20 = I20]

The dependency graph for this problem is:
  2 -> 5
  3 -> 5
  5 -> 2
Where:
  2) f2#(I4, I5, I6, I7) -> f1#(0, I5, I6, I7) [I6 <= 0 /\ y1 = 1]
  3) f3#(I8, I9, I10, I11) -> f1#(I8, I9, I10, I11) 
  5) f1#(I16, I17, I18, I19) -> f2#(I16, 0, I20, I19) [I19 <= 0 /\ I21 = 0 /\ I20 = I20]

We have the following SCCs.
  { 2, 5 }

DP problem for innermost termination.
P =
  f2#(I4, I5, I6, I7) -> f1#(0, I5, I6, I7) [I6 <= 0 /\ y1 = 1]
  f1#(I16, I17, I18, I19) -> f2#(I16, 0, I20, I19) [I19 <= 0 /\ I21 = 0 /\ I20 = I20]
R = 
  f5(x1, x2, x3, x4) -> f4(x1, x2, x3, x4) 
  f4(I0, I1, I2, I3) -> f2(0, 0, rnd3, rnd4) [rnd4 = rnd4 /\ rnd3 = rnd3]
  f2(I4, I5, I6, I7) -> f1(0, I5, I6, I7) [I6 <= 0 /\ y1 = 1]
  f3(I8, I9, I10, I11) -> f1(I8, I9, I10, I11) 
  f1(I12, I13, I14, I15) -> f3(I12, I13, I14, -1 + I15) [1 <= I15]
  f1(I16, I17, I18, I19) -> f2(I16, 0, I20, I19) [I19 <= 0 /\ I21 = 0 /\ I20 = I20]