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


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YES

DP problem for innermost termination.
P =
  f5#(x1, x2, x3) -> f1#(x1, x2, x3) 
  f2#(I0, I1, I2) -> f3#(I0, I1, I2) [1 + I0 <= I1]
  f4#(I3, I4, I5) -> f3#(I3, I4, I5) 
  f3#(I6, I7, I8) -> f4#(I6, 1 + I7, I8) [1 + I7 <= I8]
  f3#(I9, I10, I11) -> f2#(1 + I9, I10, I11) [I11 <= I10]
  f1#(I12, I13, I14) -> f2#(I12, I13, I14) 
R = 
  f5(x1, x2, x3) -> f1(x1, x2, x3) 
  f2(I0, I1, I2) -> f3(I0, I1, I2) [1 + I0 <= I1]
  f4(I3, I4, I5) -> f3(I3, I4, I5) 
  f3(I6, I7, I8) -> f4(I6, 1 + I7, I8) [1 + I7 <= I8]
  f3(I9, I10, I11) -> f2(1 + I9, I10, I11) [I11 <= I10]
  f1(I12, I13, I14) -> f2(I12, I13, I14) 

The dependency graph for this problem is:
  0 -> 5
  1 -> 3, 4
  2 -> 3, 4
  3 -> 2
  4 -> 1
  5 -> 1
Where:
  0) f5#(x1, x2, x3) -> f1#(x1, x2, x3) 
  1) f2#(I0, I1, I2) -> f3#(I0, I1, I2) [1 + I0 <= I1]
  2) f4#(I3, I4, I5) -> f3#(I3, I4, I5) 
  3) f3#(I6, I7, I8) -> f4#(I6, 1 + I7, I8) [1 + I7 <= I8]
  4) f3#(I9, I10, I11) -> f2#(1 + I9, I10, I11) [I11 <= I10]
  5) f1#(I12, I13, I14) -> f2#(I12, I13, I14) 

We have the following SCCs.
  { 1, 2, 3, 4 }

DP problem for innermost termination.
P =
  f2#(I0, I1, I2) -> f3#(I0, I1, I2) [1 + I0 <= I1]
  f4#(I3, I4, I5) -> f3#(I3, I4, I5) 
  f3#(I6, I7, I8) -> f4#(I6, 1 + I7, I8) [1 + I7 <= I8]
  f3#(I9, I10, I11) -> f2#(1 + I9, I10, I11) [I11 <= I10]
R = 
  f5(x1, x2, x3) -> f1(x1, x2, x3) 
  f2(I0, I1, I2) -> f3(I0, I1, I2) [1 + I0 <= I1]
  f4(I3, I4, I5) -> f3(I3, I4, I5) 
  f3(I6, I7, I8) -> f4(I6, 1 + I7, I8) [1 + I7 <= I8]
  f3(I9, I10, I11) -> f2(1 + I9, I10, I11) [I11 <= I10]
  f1(I12, I13, I14) -> f2(I12, I13, I14) 

We use the reverse value criterion with the projection function NU:
  NU[f4#(z1,z2,z3)] = z3 + -1 * (1 + z2)
  NU[f3#(z1,z2,z3)] = z3 + -1 * (1 + z2)
  NU[f2#(z1,z2,z3)] = z3 + -1 * (1 + z2)

This gives the following inequalities:
  1 + I0 <= I1  ==>  I2 + -1 * (1 + I1) >= I2 + -1 * (1 + I1)
    ==>  I5 + -1 * (1 + I4) >= I5 + -1 * (1 + I4)
  1 + I7 <= I8  ==>  I8 + -1 * (1 + I7) > I8 + -1 * (1 + (1 + I7))  with  I8 + -1 * (1 + I7) >= 0
  I11 <= I10  ==>  I11 + -1 * (1 + I10) >= I11 + -1 * (1 + I10)

We remove all the strictly oriented dependency pairs.

DP problem for innermost termination.
P =
  f2#(I0, I1, I2) -> f3#(I0, I1, I2) [1 + I0 <= I1]
  f4#(I3, I4, I5) -> f3#(I3, I4, I5) 
  f3#(I9, I10, I11) -> f2#(1 + I9, I10, I11) [I11 <= I10]
R = 
  f5(x1, x2, x3) -> f1(x1, x2, x3) 
  f2(I0, I1, I2) -> f3(I0, I1, I2) [1 + I0 <= I1]
  f4(I3, I4, I5) -> f3(I3, I4, I5) 
  f3(I6, I7, I8) -> f4(I6, 1 + I7, I8) [1 + I7 <= I8]
  f3(I9, I10, I11) -> f2(1 + I9, I10, I11) [I11 <= I10]
  f1(I12, I13, I14) -> f2(I12, I13, I14) 

The dependency graph for this problem is:
  1 -> 4
  2 -> 4
  4 -> 1
Where:
  1) f2#(I0, I1, I2) -> f3#(I0, I1, I2) [1 + I0 <= I1]
  2) f4#(I3, I4, I5) -> f3#(I3, I4, I5) 
  4) f3#(I9, I10, I11) -> f2#(1 + I9, I10, I11) [I11 <= I10]

We have the following SCCs.
  { 1, 4 }

DP problem for innermost termination.
P =
  f2#(I0, I1, I2) -> f3#(I0, I1, I2) [1 + I0 <= I1]
  f3#(I9, I10, I11) -> f2#(1 + I9, I10, I11) [I11 <= I10]
R = 
  f5(x1, x2, x3) -> f1(x1, x2, x3) 
  f2(I0, I1, I2) -> f3(I0, I1, I2) [1 + I0 <= I1]
  f4(I3, I4, I5) -> f3(I3, I4, I5) 
  f3(I6, I7, I8) -> f4(I6, 1 + I7, I8) [1 + I7 <= I8]
  f3(I9, I10, I11) -> f2(1 + I9, I10, I11) [I11 <= I10]
  f1(I12, I13, I14) -> f2(I12, I13, I14) 

We use the reverse value criterion with the projection function NU:
  NU[f3#(z1,z2,z3)] = z2 + -1 * (1 + (1 + z1))
  NU[f2#(z1,z2,z3)] = z2 + -1 * (1 + z1)

This gives the following inequalities:
  1 + I0 <= I1  ==>  I1 + -1 * (1 + I0) > I1 + -1 * (1 + (1 + I0))  with  I1 + -1 * (1 + I0) >= 0
  I11 <= I10  ==>  I10 + -1 * (1 + (1 + I9)) >= I10 + -1 * (1 + (1 + I9))

We remove all the strictly oriented dependency pairs.

DP problem for innermost termination.
P =
  f3#(I9, I10, I11) -> f2#(1 + I9, I10, I11) [I11 <= I10]
R = 
  f5(x1, x2, x3) -> f1(x1, x2, x3) 
  f2(I0, I1, I2) -> f3(I0, I1, I2) [1 + I0 <= I1]
  f4(I3, I4, I5) -> f3(I3, I4, I5) 
  f3(I6, I7, I8) -> f4(I6, 1 + I7, I8) [1 + I7 <= I8]
  f3(I9, I10, I11) -> f2(1 + I9, I10, I11) [I11 <= I10]
  f1(I12, I13, I14) -> f2(I12, I13, I14) 

The dependency graph for this problem is:
  4 -> 
Where:
  4) f3#(I9, I10, I11) -> f2#(1 + I9, I10, I11) [I11 <= I10]

We have the following SCCs.