YES

Problem:
 W(B(x)) -> I(x)
 B(S(x)) -> S(x)
 W(x) -> I(x)

Proof:
 Arctic Interpretation Processor:
  dimension: 1
  interpretation:
   [S](x0) = 14x0,
   
   [I](x0) = x0,
   
   [W](x0) = 1x0,
   
   [B](x0) = 2x0
  orientation:
   W(B(x)) = 3x >= x = I(x)
   
   B(S(x)) = 16x >= 14x = S(x)
   
   W(x) = 1x >= x = I(x)
  problem:
   
  Qed