/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 =
  f8#(x1, x2, x3) -> f7#(x1, x2, x3) 
  f7#(I0, I1, I2) -> f4#(I0, 0, 10) 
  f2#(I3, I4, I5) -> f6#(I3, I4, I5) [2 * I5 <= I4]
  f2#(I6, I7, I8) -> f6#(I6, I7, I8) [1 + I7 <= 2 * I8]
  f3#(I12, I13, I14) -> f1#(I12, I13, I14) 
  f4#(I18, I19, I20) -> f3#(0, I19, I20) [2 <= I20]
  f1#(I21, I22, I23) -> f3#(1 + I21, 2 + I22, I23) [1 + I21 <= I23]
  f1#(I24, I25, I26) -> f2#(I24, I25, I26) [I26 <= I24]
R = 
  f8(x1, x2, x3) -> f7(x1, x2, x3) 
  f7(I0, I1, I2) -> f4(I0, 0, 10) 
  f2(I3, I4, I5) -> f6(I3, I4, I5) [2 * I5 <= I4]
  f2(I6, I7, I8) -> f6(I6, I7, I8) [1 + I7 <= 2 * I8]
  f6(I9, I10, I11) -> f5(I9, I10, I11) 
  f3(I12, I13, I14) -> f1(I12, I13, I14) 
  f4(I15, I16, I17) -> f5(I15, I16, I17) [I17 <= 1]
  f4(I18, I19, I20) -> f3(0, I19, I20) [2 <= I20]
  f1(I21, I22, I23) -> f3(1 + I21, 2 + I22, I23) [1 + I21 <= I23]
  f1(I24, I25, I26) -> f2(I24, I25, I26) [I26 <= I24]

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) -> f7#(x1, x2, x3) 
  1) f7#(I0, I1, I2) -> f4#(I0, 0, 10) 
  2) f2#(I3, I4, I5) -> f6#(I3, I4, I5) [2 * I5 <= I4]
  3) f2#(I6, I7, I8) -> f6#(I6, I7, I8) [1 + I7 <= 2 * I8]
  4) f3#(I12, I13, I14) -> f1#(I12, I13, I14) 
  5) f4#(I18, I19, I20) -> f3#(0, I19, I20) [2 <= I20]
  6) f1#(I21, I22, I23) -> f3#(1 + I21, 2 + I22, I23) [1 + I21 <= I23]
  7) f1#(I24, I25, I26) -> f2#(I24, I25, I26) [I26 <= I24]

We have the following SCCs.
  { 4, 6 }

DP problem for innermost termination.
P =
  f3#(I12, I13, I14) -> f1#(I12, I13, I14) 
  f1#(I21, I22, I23) -> f3#(1 + I21, 2 + I22, I23) [1 + I21 <= I23]
R = 
  f8(x1, x2, x3) -> f7(x1, x2, x3) 
  f7(I0, I1, I2) -> f4(I0, 0, 10) 
  f2(I3, I4, I5) -> f6(I3, I4, I5) [2 * I5 <= I4]
  f2(I6, I7, I8) -> f6(I6, I7, I8) [1 + I7 <= 2 * I8]
  f6(I9, I10, I11) -> f5(I9, I10, I11) 
  f3(I12, I13, I14) -> f1(I12, I13, I14) 
  f4(I15, I16, I17) -> f5(I15, I16, I17) [I17 <= 1]
  f4(I18, I19, I20) -> f3(0, I19, I20) [2 <= I20]
  f1(I21, I22, I23) -> f3(1 + I21, 2 + I22, I23) [1 + I21 <= I23]
  f1(I24, I25, I26) -> f2(I24, I25, I26) [I26 <= I24]

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

This gives the following inequalities:
    ==>  I14 + -1 * (1 + I12) >= I14 + -1 * (1 + I12)
  1 + I21 <= I23  ==>  I23 + -1 * (1 + I21) > I23 + -1 * (1 + (1 + I21))  with  I23 + -1 * (1 + I21) >= 0

We remove all the strictly oriented dependency pairs.

DP problem for innermost termination.
P =
  f3#(I12, I13, I14) -> f1#(I12, I13, I14) 
R = 
  f8(x1, x2, x3) -> f7(x1, x2, x3) 
  f7(I0, I1, I2) -> f4(I0, 0, 10) 
  f2(I3, I4, I5) -> f6(I3, I4, I5) [2 * I5 <= I4]
  f2(I6, I7, I8) -> f6(I6, I7, I8) [1 + I7 <= 2 * I8]
  f6(I9, I10, I11) -> f5(I9, I10, I11) 
  f3(I12, I13, I14) -> f1(I12, I13, I14) 
  f4(I15, I16, I17) -> f5(I15, I16, I17) [I17 <= 1]
  f4(I18, I19, I20) -> f3(0, I19, I20) [2 <= I20]
  f1(I21, I22, I23) -> f3(1 + I21, 2 + I22, I23) [1 + I21 <= I23]
  f1(I24, I25, I26) -> f2(I24, I25, I26) [I26 <= I24]

The dependency graph for this problem is:
  4 -> 
Where:
  4) f3#(I12, I13, I14) -> f1#(I12, I13, I14) 

We have the following SCCs.