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


--------------------------------------------------------------------------------


YES

DP problem for innermost termination.
P =
  f8#(x1, x2, x3, x4) -> f7#(x1, x2, x3, x4) 
  f7#(I0, I1, I2, I3) -> f4#(I0, I1, 0, I0) 
  f2#(I4, I5, I6, I7) -> f6#(I4, I5, I6, I7) [2 * I7 <= I6]
  f2#(I8, I9, I10, I11) -> f6#(I8, I9, I10, I11) [1 + I10 <= 2 * I11]
  f3#(I16, I17, I18, I19) -> f1#(I16, I17, I18, I19) 
  f4#(I24, I25, I26, I27) -> f3#(I24, 0, I26, I27) [2 <= I27]
  f1#(I28, I29, I30, I31) -> f3#(I28, 1 + I29, 2 + I30, I31) [1 + I29 <= I31]
  f1#(I32, I33, I34, I35) -> f2#(I32, I33, I34, I35) [I35 <= I33]
R = 
  f8(x1, x2, x3, x4) -> f7(x1, x2, x3, x4) 
  f7(I0, I1, I2, I3) -> f4(I0, I1, 0, I0) 
  f2(I4, I5, I6, I7) -> f6(I4, I5, I6, I7) [2 * I7 <= I6]
  f2(I8, I9, I10, I11) -> f6(I8, I9, I10, I11) [1 + I10 <= 2 * I11]
  f6(I12, I13, I14, I15) -> f5(I12, I13, I14, I15) 
  f3(I16, I17, I18, I19) -> f1(I16, I17, I18, I19) 
  f4(I20, I21, I22, I23) -> f5(I20, I21, I22, I23) [I23 <= 1]
  f4(I24, I25, I26, I27) -> f3(I24, 0, I26, I27) [2 <= I27]
  f1(I28, I29, I30, I31) -> f3(I28, 1 + I29, 2 + I30, I31) [1 + I29 <= I31]
  f1(I32, I33, I34, I35) -> f2(I32, I33, I34, I35) [I35 <= I33]

The dependency graph for this problem is:
  0 -> 1
  1 -> 5
  2 -> 
  3 -> 
  4 -> 6, 7
  5 -> 4
  6 -> 4
  7 -> 2, 3
Where:
  0) f8#(x1, x2, x3, x4) -> f7#(x1, x2, x3, x4) 
  1) f7#(I0, I1, I2, I3) -> f4#(I0, I1, 0, I0) 
  2) f2#(I4, I5, I6, I7) -> f6#(I4, I5, I6, I7) [2 * I7 <= I6]
  3) f2#(I8, I9, I10, I11) -> f6#(I8, I9, I10, I11) [1 + I10 <= 2 * I11]
  4) f3#(I16, I17, I18, I19) -> f1#(I16, I17, I18, I19) 
  5) f4#(I24, I25, I26, I27) -> f3#(I24, 0, I26, I27) [2 <= I27]
  6) f1#(I28, I29, I30, I31) -> f3#(I28, 1 + I29, 2 + I30, I31) [1 + I29 <= I31]
  7) f1#(I32, I33, I34, I35) -> f2#(I32, I33, I34, I35) [I35 <= I33]

We have the following SCCs.
  { 4, 6 }

DP problem for innermost termination.
P =
  f3#(I16, I17, I18, I19) -> f1#(I16, I17, I18, I19) 
  f1#(I28, I29, I30, I31) -> f3#(I28, 1 + I29, 2 + I30, I31) [1 + I29 <= I31]
R = 
  f8(x1, x2, x3, x4) -> f7(x1, x2, x3, x4) 
  f7(I0, I1, I2, I3) -> f4(I0, I1, 0, I0) 
  f2(I4, I5, I6, I7) -> f6(I4, I5, I6, I7) [2 * I7 <= I6]
  f2(I8, I9, I10, I11) -> f6(I8, I9, I10, I11) [1 + I10 <= 2 * I11]
  f6(I12, I13, I14, I15) -> f5(I12, I13, I14, I15) 
  f3(I16, I17, I18, I19) -> f1(I16, I17, I18, I19) 
  f4(I20, I21, I22, I23) -> f5(I20, I21, I22, I23) [I23 <= 1]
  f4(I24, I25, I26, I27) -> f3(I24, 0, I26, I27) [2 <= I27]
  f1(I28, I29, I30, I31) -> f3(I28, 1 + I29, 2 + I30, I31) [1 + I29 <= I31]
  f1(I32, I33, I34, I35) -> f2(I32, I33, I34, I35) [I35 <= I33]

We use the reverse value criterion with the projection function NU:
  NU[f1#(z1,z2,z3,z4)] = z4 + -1 * (1 + z2)
  NU[f3#(z1,z2,z3,z4)] = z4 + -1 * (1 + z2)

This gives the following inequalities:
    ==>  I19 + -1 * (1 + I17) >= I19 + -1 * (1 + I17)
  1 + I29 <= I31  ==>  I31 + -1 * (1 + I29) > I31 + -1 * (1 + (1 + I29))  with  I31 + -1 * (1 + I29) >= 0

We remove all the strictly oriented dependency pairs.

DP problem for innermost termination.
P =
  f3#(I16, I17, I18, I19) -> f1#(I16, I17, I18, I19) 
R = 
  f8(x1, x2, x3, x4) -> f7(x1, x2, x3, x4) 
  f7(I0, I1, I2, I3) -> f4(I0, I1, 0, I0) 
  f2(I4, I5, I6, I7) -> f6(I4, I5, I6, I7) [2 * I7 <= I6]
  f2(I8, I9, I10, I11) -> f6(I8, I9, I10, I11) [1 + I10 <= 2 * I11]
  f6(I12, I13, I14, I15) -> f5(I12, I13, I14, I15) 
  f3(I16, I17, I18, I19) -> f1(I16, I17, I18, I19) 
  f4(I20, I21, I22, I23) -> f5(I20, I21, I22, I23) [I23 <= 1]
  f4(I24, I25, I26, I27) -> f3(I24, 0, I26, I27) [2 <= I27]
  f1(I28, I29, I30, I31) -> f3(I28, 1 + I29, 2 + I30, I31) [1 + I29 <= I31]
  f1(I32, I33, I34, I35) -> f2(I32, I33, I34, I35) [I35 <= I33]

The dependency graph for this problem is:
  4 -> 
Where:
  4) f3#(I16, I17, I18, I19) -> f1#(I16, I17, I18, I19) 

We have the following SCCs.