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

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

We have the following SCCs.
  { 1, 3, 4, 5 }

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

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

This gives the following inequalities:
  1 <= I0  ==>  I0 - 1 > I0 - 2  with  I0 - 1 >= 0
    ==>  I4 - 2 >= I4 - 2
  1 <= I7  ==>  I6 - 2 >= I6 - 2
  I9 <= 0  ==>  I8 - 2 >= (-1 + I8) - 1

We remove all the strictly oriented dependency pairs.

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

The dependency graph for this problem is:
  3 -> 4, 5
  4 -> 3
  5 -> 
Where:
  3) f3#(I4, I5) -> f1#(I4, I5) 
  4) f1#(I6, I7) -> f3#(I6, -1 + I7) [1 <= I7]
  5) f1#(I8, I9) -> f2#(-1 + I8, 1 + I9) [I9 <= 0]

We have the following SCCs.
  { 3, 4 }

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

We use the basic value criterion with the projection function NU:
  NU[f1#(z1,z2)] = z2
  NU[f3#(z1,z2)] = z2

This gives the following inequalities:
    ==>  I5 (>! \union =) I5
  1 <= I7  ==>  I7 >! -1 + I7

We remove all the strictly oriented dependency pairs.

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

The dependency graph for this problem is:
  3 -> 
Where:
  3) f3#(I4, I5) -> f1#(I4, I5) 

We have the following SCCs.