/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) -> f5#(I0, I1, I2) 
  f6#(I3, I4, I5) -> f2#(I3, I4, I5) 
  f6#(I6, I7, I8) -> f3#(I6, I7, I8) 
  f6#(I9, I10, I11) -> f1#(I9, I10, I11) 
  f6#(I15, I16, I17) -> f5#(I17, rnd2, rnd3) [rnd3 = rnd2 /\ rnd2 = rnd2]
  f5#(I18, I19, I20) -> f2#(I20, I19, 0) 
  f2#(I21, I22, I23) -> f3#(I23, I22, I23) [11 <= I23]
  f2#(I24, I25, I26) -> f1#(I26, I25, I26) [I26 <= 10]
  f1#(I32, I33, I34) -> f2#(I34, I33, 1 + I34) 
R = 
  f7(x1, x2, x3) -> f6(x1, x2, x3) 
  f6(I0, I1, I2) -> f5(I0, I1, I2) 
  f6(I3, I4, I5) -> f2(I3, I4, I5) 
  f6(I6, I7, I8) -> f3(I6, I7, I8) 
  f6(I9, I10, I11) -> f1(I9, I10, I11) 
  f6(I12, I13, I14) -> f4(I12, I13, I14) 
  f6(I15, I16, I17) -> f5(I17, rnd2, rnd3) [rnd3 = rnd2 /\ rnd2 = rnd2]
  f5(I18, I19, I20) -> f2(I20, I19, 0) 
  f2(I21, I22, I23) -> f3(I23, I22, I23) [11 <= I23]
  f2(I24, I25, I26) -> f1(I26, I25, I26) [I26 <= 10]
  f3(I27, I28, I29) -> f4(I29, I30, I31) [I31 = I30 /\ I30 = I30]
  f1(I32, I33, I34) -> f2(I34, I33, 1 + I34) 

The dependency graph for this problem is:
  0 -> 1, 2, 3, 4, 5
  1 -> 6
  2 -> 7, 8
  3 -> 
  4 -> 9
  5 -> 6
  6 -> 8
  7 -> 
  8 -> 9
  9 -> 7, 8
Where:
  0) f7#(x1, x2, x3) -> f6#(x1, x2, x3) 
  1) f6#(I0, I1, I2) -> f5#(I0, I1, I2) 
  2) f6#(I3, I4, I5) -> f2#(I3, I4, I5) 
  3) f6#(I6, I7, I8) -> f3#(I6, I7, I8) 
  4) f6#(I9, I10, I11) -> f1#(I9, I10, I11) 
  5) f6#(I15, I16, I17) -> f5#(I17, rnd2, rnd3) [rnd3 = rnd2 /\ rnd2 = rnd2]
  6) f5#(I18, I19, I20) -> f2#(I20, I19, 0) 
  7) f2#(I21, I22, I23) -> f3#(I23, I22, I23) [11 <= I23]
  8) f2#(I24, I25, I26) -> f1#(I26, I25, I26) [I26 <= 10]
  9) f1#(I32, I33, I34) -> f2#(I34, I33, 1 + I34) 

We have the following SCCs.
  { 8, 9 }

DP problem for innermost termination.
P =
  f2#(I24, I25, I26) -> f1#(I26, I25, I26) [I26 <= 10]
  f1#(I32, I33, I34) -> f2#(I34, I33, 1 + I34) 
R = 
  f7(x1, x2, x3) -> f6(x1, x2, x3) 
  f6(I0, I1, I2) -> f5(I0, I1, I2) 
  f6(I3, I4, I5) -> f2(I3, I4, I5) 
  f6(I6, I7, I8) -> f3(I6, I7, I8) 
  f6(I9, I10, I11) -> f1(I9, I10, I11) 
  f6(I12, I13, I14) -> f4(I12, I13, I14) 
  f6(I15, I16, I17) -> f5(I17, rnd2, rnd3) [rnd3 = rnd2 /\ rnd2 = rnd2]
  f5(I18, I19, I20) -> f2(I20, I19, 0) 
  f2(I21, I22, I23) -> f3(I23, I22, I23) [11 <= I23]
  f2(I24, I25, I26) -> f1(I26, I25, I26) [I26 <= 10]
  f3(I27, I28, I29) -> f4(I29, I30, I31) [I31 = I30 /\ I30 = I30]
  f1(I32, I33, I34) -> f2(I34, I33, 1 + I34) 

We use the reverse value criterion with the projection function NU:
  NU[f1#(z1,z2,z3)] = 10 + -1 * (1 + z3)
  NU[f2#(z1,z2,z3)] = 10 + -1 * z3

This gives the following inequalities:
  I26 <= 10  ==>  10 + -1 * I26 > 10 + -1 * (1 + I26)  with  10 + -1 * I26 >= 0
    ==>  10 + -1 * (1 + I34) >= 10 + -1 * (1 + I34)

We remove all the strictly oriented dependency pairs.

DP problem for innermost termination.
P =
  f1#(I32, I33, I34) -> f2#(I34, I33, 1 + I34) 
R = 
  f7(x1, x2, x3) -> f6(x1, x2, x3) 
  f6(I0, I1, I2) -> f5(I0, I1, I2) 
  f6(I3, I4, I5) -> f2(I3, I4, I5) 
  f6(I6, I7, I8) -> f3(I6, I7, I8) 
  f6(I9, I10, I11) -> f1(I9, I10, I11) 
  f6(I12, I13, I14) -> f4(I12, I13, I14) 
  f6(I15, I16, I17) -> f5(I17, rnd2, rnd3) [rnd3 = rnd2 /\ rnd2 = rnd2]
  f5(I18, I19, I20) -> f2(I20, I19, 0) 
  f2(I21, I22, I23) -> f3(I23, I22, I23) [11 <= I23]
  f2(I24, I25, I26) -> f1(I26, I25, I26) [I26 <= 10]
  f3(I27, I28, I29) -> f4(I29, I30, I31) [I31 = I30 /\ I30 = I30]
  f1(I32, I33, I34) -> f2(I34, I33, 1 + I34) 

The dependency graph for this problem is:
  9 -> 
Where:
  9) f1#(I32, I33, I34) -> f2#(I34, I33, 1 + I34) 

We have the following SCCs.