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


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


YES

DP problem for innermost termination.
P =
  f9#(x1, x2) -> f8#(x1, x2) 
  f8#(I0, I1) -> f3#(0, I1) 
  f4#(I2, I3) -> f2#(I2, 1 + I3) [1 + I3 <= 2]
  f4#(I4, I5) -> f7#(I4, I5) [2 <= I5]
  f7#(I6, I7) -> f5#(I6, I7) 
  f7#(I8, I9) -> f5#(I8, I9) 
  f2#(I12, I13) -> f4#(I12, I13) 
  f3#(I14, I15) -> f1#(I14, I15) 
  f1#(I16, I17) -> f3#(1 + I16, I17) [1 + I16 <= 2]
  f1#(I18, I19) -> f2#(I18, 0) [2 <= I18]
R = 
  f9(x1, x2) -> f8(x1, x2) 
  f8(I0, I1) -> f3(0, I1) 
  f4(I2, I3) -> f2(I2, 1 + I3) [1 + I3 <= 2]
  f4(I4, I5) -> f7(I4, I5) [2 <= I5]
  f7(I6, I7) -> f5(I6, I7) 
  f7(I8, I9) -> f5(I8, I9) 
  f5(I10, I11) -> f6(I10, I11) 
  f2(I12, I13) -> f4(I12, I13) 
  f3(I14, I15) -> f1(I14, I15) 
  f1(I16, I17) -> f3(1 + I16, I17) [1 + I16 <= 2]
  f1(I18, I19) -> f2(I18, 0) [2 <= I18]

The dependency graph for this problem is:
  0 -> 1
  1 -> 7
  2 -> 6
  3 -> 4, 5
  4 -> 
  5 -> 
  6 -> 2, 3
  7 -> 8, 9
  8 -> 7
  9 -> 6
Where:
  0) f9#(x1, x2) -> f8#(x1, x2) 
  1) f8#(I0, I1) -> f3#(0, I1) 
  2) f4#(I2, I3) -> f2#(I2, 1 + I3) [1 + I3 <= 2]
  3) f4#(I4, I5) -> f7#(I4, I5) [2 <= I5]
  4) f7#(I6, I7) -> f5#(I6, I7) 
  5) f7#(I8, I9) -> f5#(I8, I9) 
  6) f2#(I12, I13) -> f4#(I12, I13) 
  7) f3#(I14, I15) -> f1#(I14, I15) 
  8) f1#(I16, I17) -> f3#(1 + I16, I17) [1 + I16 <= 2]
  9) f1#(I18, I19) -> f2#(I18, 0) [2 <= I18]

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

DP problem for innermost termination.
P =
  f4#(I2, I3) -> f2#(I2, 1 + I3) [1 + I3 <= 2]
  f2#(I12, I13) -> f4#(I12, I13) 
R = 
  f9(x1, x2) -> f8(x1, x2) 
  f8(I0, I1) -> f3(0, I1) 
  f4(I2, I3) -> f2(I2, 1 + I3) [1 + I3 <= 2]
  f4(I4, I5) -> f7(I4, I5) [2 <= I5]
  f7(I6, I7) -> f5(I6, I7) 
  f7(I8, I9) -> f5(I8, I9) 
  f5(I10, I11) -> f6(I10, I11) 
  f2(I12, I13) -> f4(I12, I13) 
  f3(I14, I15) -> f1(I14, I15) 
  f1(I16, I17) -> f3(1 + I16, I17) [1 + I16 <= 2]
  f1(I18, I19) -> f2(I18, 0) [2 <= I18]

We use the reverse value criterion with the projection function NU:
  NU[f2#(z1,z2)] = 2 + -1 * (1 + z2)
  NU[f4#(z1,z2)] = 2 + -1 * (1 + z2)

This gives the following inequalities:
  1 + I3 <= 2  ==>  2 + -1 * (1 + I3) > 2 + -1 * (1 + (1 + I3))  with  2 + -1 * (1 + I3) >= 0
    ==>  2 + -1 * (1 + I13) >= 2 + -1 * (1 + I13)

We remove all the strictly oriented dependency pairs.

DP problem for innermost termination.
P =
  f2#(I12, I13) -> f4#(I12, I13) 
R = 
  f9(x1, x2) -> f8(x1, x2) 
  f8(I0, I1) -> f3(0, I1) 
  f4(I2, I3) -> f2(I2, 1 + I3) [1 + I3 <= 2]
  f4(I4, I5) -> f7(I4, I5) [2 <= I5]
  f7(I6, I7) -> f5(I6, I7) 
  f7(I8, I9) -> f5(I8, I9) 
  f5(I10, I11) -> f6(I10, I11) 
  f2(I12, I13) -> f4(I12, I13) 
  f3(I14, I15) -> f1(I14, I15) 
  f1(I16, I17) -> f3(1 + I16, I17) [1 + I16 <= 2]
  f1(I18, I19) -> f2(I18, 0) [2 <= I18]

The dependency graph for this problem is:
  6 -> 
Where:
  6) f2#(I12, I13) -> f4#(I12, I13) 

We have the following SCCs.
  

DP problem for innermost termination.
P =
  f3#(I14, I15) -> f1#(I14, I15) 
  f1#(I16, I17) -> f3#(1 + I16, I17) [1 + I16 <= 2]
R = 
  f9(x1, x2) -> f8(x1, x2) 
  f8(I0, I1) -> f3(0, I1) 
  f4(I2, I3) -> f2(I2, 1 + I3) [1 + I3 <= 2]
  f4(I4, I5) -> f7(I4, I5) [2 <= I5]
  f7(I6, I7) -> f5(I6, I7) 
  f7(I8, I9) -> f5(I8, I9) 
  f5(I10, I11) -> f6(I10, I11) 
  f2(I12, I13) -> f4(I12, I13) 
  f3(I14, I15) -> f1(I14, I15) 
  f1(I16, I17) -> f3(1 + I16, I17) [1 + I16 <= 2]
  f1(I18, I19) -> f2(I18, 0) [2 <= I18]

We use the reverse value criterion with the projection function NU:
  NU[f1#(z1,z2)] = 2 + -1 * (1 + z1)
  NU[f3#(z1,z2)] = 2 + -1 * (1 + z1)

This gives the following inequalities:
    ==>  2 + -1 * (1 + I14) >= 2 + -1 * (1 + I14)
  1 + I16 <= 2  ==>  2 + -1 * (1 + I16) > 2 + -1 * (1 + (1 + I16))  with  2 + -1 * (1 + I16) >= 0

We remove all the strictly oriented dependency pairs.

DP problem for innermost termination.
P =
  f3#(I14, I15) -> f1#(I14, I15) 
R = 
  f9(x1, x2) -> f8(x1, x2) 
  f8(I0, I1) -> f3(0, I1) 
  f4(I2, I3) -> f2(I2, 1 + I3) [1 + I3 <= 2]
  f4(I4, I5) -> f7(I4, I5) [2 <= I5]
  f7(I6, I7) -> f5(I6, I7) 
  f7(I8, I9) -> f5(I8, I9) 
  f5(I10, I11) -> f6(I10, I11) 
  f2(I12, I13) -> f4(I12, I13) 
  f3(I14, I15) -> f1(I14, I15) 
  f1(I16, I17) -> f3(1 + I16, I17) [1 + I16 <= 2]
  f1(I18, I19) -> f2(I18, 0) [2 <= I18]

The dependency graph for this problem is:
  7 -> 
Where:
  7) f3#(I14, I15) -> f1#(I14, I15) 

We have the following SCCs.