/export/starexec/sandbox2/solver/bin/starexec_run_Itrs /export/starexec/sandbox2/benchmark/theBenchmark.itrs /export/starexec/sandbox2/output/output_files


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YES

DP problem for innermost termination.
P =
  b15#(sv14_14, sv23_37, sv24_38) -> b10#(sv14_14, sv23_37 - sv14_14, sv24_38 + 1) 
  b14#(I0, I1, I2) -> Cond_b14#(I1 >= I0 && 1 < I0 && 0 < I2 && 1 < I1, I0, I1, I2) 
  Cond_b14#(true, I3, I4, I5) -> b15#(I3, I4, I5) 
  b10#(I6, I7, I8) -> b14#(I6, I7, I8) 
R = 
  b15(sv14_14, sv23_37, sv24_38) -> b10(sv14_14, sv23_37 - sv14_14, sv24_38 + 1) 
  b14(I0, I1, I2) -> Cond_b14(I1 >= I0 && 1 < I0 && 0 < I2 && 1 < I1, I0, I1, I2) 
  Cond_b14(true, I3, I4, I5) -> b15(I3, I4, I5) 
  b10(I6, I7, I8) -> b14(I6, I7, I8) 

This problem is converted using chaining, where edges between chained DPs are removed.

DP problem for innermost termination.
P =
  b15#(sv14_14, sv23_37, sv24_38) -> b10#(sv14_14, sv23_37 - sv14_14, sv24_38 + 1) 
  b14#(I0, I1, I2) -> Cond_b14#(I1 >= I0 && 1 < I0 && 0 < I2 && 1 < I1, I0, I1, I2) 
  Cond_b14#(true, I3, I4, I5) -> b15#(I3, I4, I5) 
  b10#(I6, I7, I8) -> b14#(I6, I7, I8) 
  b14#(I0, I1, I2) -> b15#(I0, I1, I2) [I1 >= I0 && 1 < I0 && 0 < I2 && 1 < I1]
R = 
  b15(sv14_14, sv23_37, sv24_38) -> b10(sv14_14, sv23_37 - sv14_14, sv24_38 + 1) 
  b14(I0, I1, I2) -> Cond_b14(I1 >= I0 && 1 < I0 && 0 < I2 && 1 < I1, I0, I1, I2) 
  Cond_b14(true, I3, I4, I5) -> b15(I3, I4, I5) 
  b10(I6, I7, I8) -> b14(I6, I7, I8) 

The dependency graph for this problem is:
  0 -> 3
  1 -> 
  2 -> 0
  3 -> 4, 1
  4 -> 0
Where:
  0) b15#(sv14_14, sv23_37, sv24_38) -> b10#(sv14_14, sv23_37 - sv14_14, sv24_38 + 1) 
  1) b14#(I0, I1, I2) -> Cond_b14#(I1 >= I0 && 1 < I0 && 0 < I2 && 1 < I1, I0, I1, I2) 
  2) Cond_b14#(true, I3, I4, I5) -> b15#(I3, I4, I5) 
  3) b10#(I6, I7, I8) -> b14#(I6, I7, I8) 
  4) b14#(I0, I1, I2) -> b15#(I0, I1, I2) [I1 >= I0 && 1 < I0 && 0 < I2 && 1 < I1]

We have the following SCCs.
  { 0, 3, 4 }

DP problem for innermost termination.
P =
  b15#(sv14_14, sv23_37, sv24_38) -> b10#(sv14_14, sv23_37 - sv14_14, sv24_38 + 1) 
  b10#(I6, I7, I8) -> b14#(I6, I7, I8) 
  b14#(I0, I1, I2) -> b15#(I0, I1, I2) [I1 >= I0 && 1 < I0 && 0 < I2 && 1 < I1]
R = 
  b15(sv14_14, sv23_37, sv24_38) -> b10(sv14_14, sv23_37 - sv14_14, sv24_38 + 1) 
  b14(I0, I1, I2) -> Cond_b14(I1 >= I0 && 1 < I0 && 0 < I2 && 1 < I1, I0, I1, I2) 
  Cond_b14(true, I3, I4, I5) -> b15(I3, I4, I5) 
  b10(I6, I7, I8) -> b14(I6, I7, I8) 

We use the extended value criterion with the projection function NU:
  NU[b14#(x0,x1,x2)] = x1 - 2
  NU[b10#(x0,x1,x2)] = x1 - 2
  NU[b15#(x0,x1,x2)] = -x0 + x1 - 2

This gives the following inequalities:
    ==>  -sv14_14 + sv23_37 - 2 >= (sv23_37 - sv14_14) - 2
    ==>  I7 - 2 >= I7 - 2
  I1 >= I0 && 1 < I0 && 0 < I2 && 1 < I1  ==>  I1 - 2 > -I0 + I1 - 2  with  I1 - 2 >= 0

We remove all the strictly oriented dependency pairs.

DP problem for innermost termination.
P =
  b15#(sv14_14, sv23_37, sv24_38) -> b10#(sv14_14, sv23_37 - sv14_14, sv24_38 + 1) 
  b10#(I6, I7, I8) -> b14#(I6, I7, I8) 
R = 
  b15(sv14_14, sv23_37, sv24_38) -> b10(sv14_14, sv23_37 - sv14_14, sv24_38 + 1) 
  b14(I0, I1, I2) -> Cond_b14(I1 >= I0 && 1 < I0 && 0 < I2 && 1 < I1, I0, I1, I2) 
  Cond_b14(true, I3, I4, I5) -> b15(I3, I4, I5) 
  b10(I6, I7, I8) -> b14(I6, I7, I8) 

The dependency graph for this problem is:
  0 -> 3
  3 -> 
Where:
  0) b15#(sv14_14, sv23_37, sv24_38) -> b10#(sv14_14, sv23_37 - sv14_14, sv24_38 + 1) 
  3) b10#(I6, I7, I8) -> b14#(I6, I7, I8) 

We have the following SCCs.