/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) -> f7#(x1, x2) 
  f7#(I0, I1) -> f2#(I0, 0) 
  f2#(I2, I3) -> f5#(I2, I3) 
  f5#(I4, I5) -> f4#(I4, I5) [1 + I5 <= I4]
  f4#(I8, I9) -> f3#(I8, I9) 
  f4#(I10, I11) -> f1#(I10, I11) 
  f4#(I12, I13) -> f3#(I12, I13) 
  f3#(I14, I15) -> f1#(I14, I15) 
  f1#(I16, I17) -> f2#(I16, 1 + I17) 
R = 
  f8(x1, x2) -> f7(x1, x2) 
  f7(I0, I1) -> f2(I0, 0) 
  f2(I2, I3) -> f5(I2, I3) 
  f5(I4, I5) -> f4(I4, I5) [1 + I5 <= I4]
  f5(I6, I7) -> f6(I6, I7) [I6 <= I7]
  f4(I8, I9) -> f3(I8, I9) 
  f4(I10, I11) -> f1(I10, I11) 
  f4(I12, I13) -> f3(I12, I13) 
  f3(I14, I15) -> f1(I14, I15) 
  f1(I16, I17) -> f2(I16, 1 + I17) 

The dependency graph for this problem is:
  0 -> 1
  1 -> 2
  2 -> 3
  3 -> 4, 5, 6
  4 -> 7
  5 -> 8
  6 -> 7
  7 -> 8
  8 -> 2
Where:
  0) f8#(x1, x2) -> f7#(x1, x2) 
  1) f7#(I0, I1) -> f2#(I0, 0) 
  2) f2#(I2, I3) -> f5#(I2, I3) 
  3) f5#(I4, I5) -> f4#(I4, I5) [1 + I5 <= I4]
  4) f4#(I8, I9) -> f3#(I8, I9) 
  5) f4#(I10, I11) -> f1#(I10, I11) 
  6) f4#(I12, I13) -> f3#(I12, I13) 
  7) f3#(I14, I15) -> f1#(I14, I15) 
  8) f1#(I16, I17) -> f2#(I16, 1 + I17) 

We have the following SCCs.
  { 2, 3, 4, 5, 6, 7, 8 }

DP problem for innermost termination.
P =
  f2#(I2, I3) -> f5#(I2, I3) 
  f5#(I4, I5) -> f4#(I4, I5) [1 + I5 <= I4]
  f4#(I8, I9) -> f3#(I8, I9) 
  f4#(I10, I11) -> f1#(I10, I11) 
  f4#(I12, I13) -> f3#(I12, I13) 
  f3#(I14, I15) -> f1#(I14, I15) 
  f1#(I16, I17) -> f2#(I16, 1 + I17) 
R = 
  f8(x1, x2) -> f7(x1, x2) 
  f7(I0, I1) -> f2(I0, 0) 
  f2(I2, I3) -> f5(I2, I3) 
  f5(I4, I5) -> f4(I4, I5) [1 + I5 <= I4]
  f5(I6, I7) -> f6(I6, I7) [I6 <= I7]
  f4(I8, I9) -> f3(I8, I9) 
  f4(I10, I11) -> f1(I10, I11) 
  f4(I12, I13) -> f3(I12, I13) 
  f3(I14, I15) -> f1(I14, I15) 
  f1(I16, I17) -> f2(I16, 1 + I17) 

We use the extended value criterion with the projection function NU:
  NU[f1#(x0,x1)] = x0 - x1 - 2
  NU[f3#(x0,x1)] = x0 - x1 - 2
  NU[f4#(x0,x1)] = x0 - x1 - 2
  NU[f5#(x0,x1)] = x0 - x1 - 1
  NU[f2#(x0,x1)] = x0 - x1 - 1

This gives the following inequalities:
    ==>  I2 - I3 - 1 >= I2 - I3 - 1
  1 + I5 <= I4  ==>  I4 - I5 - 1 > I4 - I5 - 2  with  I4 - I5 - 1 >= 0
    ==>  I8 - I9 - 2 >= I8 - I9 - 2
    ==>  I10 - I11 - 2 >= I10 - I11 - 2
    ==>  I12 - I13 - 2 >= I12 - I13 - 2
    ==>  I14 - I15 - 2 >= I14 - I15 - 2
    ==>  I16 - I17 - 2 >= I16 - (1 + I17) - 1

We remove all the strictly oriented dependency pairs.

DP problem for innermost termination.
P =
  f2#(I2, I3) -> f5#(I2, I3) 
  f4#(I8, I9) -> f3#(I8, I9) 
  f4#(I10, I11) -> f1#(I10, I11) 
  f4#(I12, I13) -> f3#(I12, I13) 
  f3#(I14, I15) -> f1#(I14, I15) 
  f1#(I16, I17) -> f2#(I16, 1 + I17) 
R = 
  f8(x1, x2) -> f7(x1, x2) 
  f7(I0, I1) -> f2(I0, 0) 
  f2(I2, I3) -> f5(I2, I3) 
  f5(I4, I5) -> f4(I4, I5) [1 + I5 <= I4]
  f5(I6, I7) -> f6(I6, I7) [I6 <= I7]
  f4(I8, I9) -> f3(I8, I9) 
  f4(I10, I11) -> f1(I10, I11) 
  f4(I12, I13) -> f3(I12, I13) 
  f3(I14, I15) -> f1(I14, I15) 
  f1(I16, I17) -> f2(I16, 1 + I17) 

The dependency graph for this problem is:
  2 -> 
  4 -> 7
  5 -> 8
  6 -> 7
  7 -> 8
  8 -> 2
Where:
  2) f2#(I2, I3) -> f5#(I2, I3) 
  4) f4#(I8, I9) -> f3#(I8, I9) 
  5) f4#(I10, I11) -> f1#(I10, I11) 
  6) f4#(I12, I13) -> f3#(I12, I13) 
  7) f3#(I14, I15) -> f1#(I14, I15) 
  8) f1#(I16, I17) -> f2#(I16, 1 + I17) 

We have the following SCCs.