/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 =
  f5#(x1, x2) -> f4#(x1, x2) 
  f4#(I0, I1) -> f1#(I0, I1) 
  f3#(I2, I3) -> f2#(I2, I3) 
  f2#(I4, I5) -> f3#(I4, -1 + I5) [1 <= I5]
  f2#(I6, I7) -> f1#(-1 + I6, 1 + I7) [I7 <= 0]
  f1#(I8, I9) -> f2#(I8, I8) [1 <= I8]
R = 
  f5(x1, x2) -> f4(x1, x2) 
  f4(I0, I1) -> f1(I0, I1) 
  f3(I2, I3) -> f2(I2, I3) 
  f2(I4, I5) -> f3(I4, -1 + I5) [1 <= I5]
  f2(I6, I7) -> f1(-1 + I6, 1 + I7) [I7 <= 0]
  f1(I8, I9) -> f2(I8, I8) [1 <= I8]

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

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

DP problem for innermost termination.
P =
  f3#(I2, I3) -> f2#(I2, I3) 
  f2#(I4, I5) -> f3#(I4, -1 + I5) [1 <= I5]
  f2#(I6, I7) -> f1#(-1 + I6, 1 + I7) [I7 <= 0]
  f1#(I8, I9) -> f2#(I8, I8) [1 <= I8]
R = 
  f5(x1, x2) -> f4(x1, x2) 
  f4(I0, I1) -> f1(I0, I1) 
  f3(I2, I3) -> f2(I2, I3) 
  f2(I4, I5) -> f3(I4, -1 + I5) [1 <= I5]
  f2(I6, I7) -> f1(-1 + I6, 1 + I7) [I7 <= 0]
  f1(I8, I9) -> f2(I8, I8) [1 <= I8]

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

This gives the following inequalities:
    ==>  I2 >= I2
  1 <= I5  ==>  I4 >= I4
  I7 <= 0  ==>  I6 >= (-1 + I6) + 1
  1 <= I8  ==>  I8 + 1 > I8  with  I8 + 1 >= 0

We remove all the strictly oriented dependency pairs.

DP problem for innermost termination.
P =
  f3#(I2, I3) -> f2#(I2, I3) 
  f2#(I4, I5) -> f3#(I4, -1 + I5) [1 <= I5]
  f2#(I6, I7) -> f1#(-1 + I6, 1 + I7) [I7 <= 0]
R = 
  f5(x1, x2) -> f4(x1, x2) 
  f4(I0, I1) -> f1(I0, I1) 
  f3(I2, I3) -> f2(I2, I3) 
  f2(I4, I5) -> f3(I4, -1 + I5) [1 <= I5]
  f2(I6, I7) -> f1(-1 + I6, 1 + I7) [I7 <= 0]
  f1(I8, I9) -> f2(I8, I8) [1 <= I8]

The dependency graph for this problem is:
  2 -> 3, 4
  3 -> 2
  4 -> 
Where:
  2) f3#(I2, I3) -> f2#(I2, I3) 
  3) f2#(I4, I5) -> f3#(I4, -1 + I5) [1 <= I5]
  4) f2#(I6, I7) -> f1#(-1 + I6, 1 + I7) [I7 <= 0]

We have the following SCCs.
  { 2, 3 }

DP problem for innermost termination.
P =
  f3#(I2, I3) -> f2#(I2, I3) 
  f2#(I4, I5) -> f3#(I4, -1 + I5) [1 <= I5]
R = 
  f5(x1, x2) -> f4(x1, x2) 
  f4(I0, I1) -> f1(I0, I1) 
  f3(I2, I3) -> f2(I2, I3) 
  f2(I4, I5) -> f3(I4, -1 + I5) [1 <= I5]
  f2(I6, I7) -> f1(-1 + I6, 1 + I7) [I7 <= 0]
  f1(I8, I9) -> f2(I8, I8) [1 <= I8]

We use the basic value criterion with the projection function NU:
  NU[f2#(z1,z2)] = z2
  NU[f3#(z1,z2)] = z2

This gives the following inequalities:
    ==>  I3 (>! \union =) I3
  1 <= I5  ==>  I5 >! -1 + I5

We remove all the strictly oriented dependency pairs.

DP problem for innermost termination.
P =
  f3#(I2, I3) -> f2#(I2, I3) 
R = 
  f5(x1, x2) -> f4(x1, x2) 
  f4(I0, I1) -> f1(I0, I1) 
  f3(I2, I3) -> f2(I2, I3) 
  f2(I4, I5) -> f3(I4, -1 + I5) [1 <= I5]
  f2(I6, I7) -> f1(-1 + I6, 1 + I7) [I7 <= 0]
  f1(I8, I9) -> f2(I8, I8) [1 <= I8]

The dependency graph for this problem is:
  2 -> 
Where:
  2) f3#(I2, I3) -> f2#(I2, I3) 

We have the following SCCs.