/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 =
  rand#(I3, y) -> rand#(I3 + 1, id_dec(y)) [0 > I3]
  rand#(I3, y) -> id_dec#(y) [0 > I3]
  rand#(I4, B0) -> rand#(I4 - 1, id_inc(B0)) [I4 > 0]
  rand#(I4, B0) -> id_inc#(B0) [I4 > 0]
  random#(I6) -> rand#(I6, w(0)) 
R = 
  id_dec(w(x)) -> w(x - 1) 
  id_dec(w(I0)) -> w(I0) 
  id_inc(w(I1)) -> w(I1 + 1) 
  id_inc(w(I2)) -> w(I2) 
  rand(I3, y) -> rand(I3 + 1, id_dec(y)) [0 > I3]
  rand(I4, B0) -> rand(I4 - 1, id_inc(B0)) [I4 > 0]
  rand(I5, B1) -> B1 [I5 = 0]
  random(I6) -> rand(I6, w(0)) 

The dependency graph for this problem is:
  0 -> 0, 1
  1 -> 
  2 -> 2, 3
  3 -> 
  4 -> 0, 1, 2, 3
Where:
  0) rand#(I3, y) -> rand#(I3 + 1, id_dec(y)) [0 > I3]
  1) rand#(I3, y) -> id_dec#(y) [0 > I3]
  2) rand#(I4, B0) -> rand#(I4 - 1, id_inc(B0)) [I4 > 0]
  3) rand#(I4, B0) -> id_inc#(B0) [I4 > 0]
  4) random#(I6) -> rand#(I6, w(0)) 

We have the following SCCs.
  { 2 }
  { 0 }

DP problem for innermost termination.
P =
  rand#(I3, y) -> rand#(I3 + 1, id_dec(y)) [0 > I3]
R = 
  id_dec(w(x)) -> w(x - 1) 
  id_dec(w(I0)) -> w(I0) 
  id_inc(w(I1)) -> w(I1 + 1) 
  id_inc(w(I2)) -> w(I2) 
  rand(I3, y) -> rand(I3 + 1, id_dec(y)) [0 > I3]
  rand(I4, B0) -> rand(I4 - 1, id_inc(B0)) [I4 > 0]
  rand(I5, B1) -> B1 [I5 = 0]
  random(I6) -> rand(I6, w(0)) 

We use the reverse value criterion with the projection function NU:
  NU[rand#(z1,z2)] = 0 + -1 * z1

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

All dependency pairs are strictly oriented, so the entire dependency pair problem may be removed.

DP problem for innermost termination.
P =
  rand#(I4, B0) -> rand#(I4 - 1, id_inc(B0)) [I4 > 0]
R = 
  id_dec(w(x)) -> w(x - 1) 
  id_dec(w(I0)) -> w(I0) 
  id_inc(w(I1)) -> w(I1 + 1) 
  id_inc(w(I2)) -> w(I2) 
  rand(I3, y) -> rand(I3 + 1, id_dec(y)) [0 > I3]
  rand(I4, B0) -> rand(I4 - 1, id_inc(B0)) [I4 > 0]
  rand(I5, B1) -> B1 [I5 = 0]
  random(I6) -> rand(I6, w(0)) 

We use the reverse value criterion with the projection function NU:
  NU[rand#(z1,z2)] = z1

This gives the following inequalities:
  I4 > 0  ==>  I4 > I4 - 1  with  I4 >= 0

All dependency pairs are strictly oriented, so the entire dependency pair problem may be removed.