YES

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

Proof:
 Arctic Interpretation Processor:
  dimension: 1
  interpretation:
   [J](x0) = x0,
   
   [I](x0) = 1x0,
   
   [W](x0) = x0,
   
   [B](x0) = x0
  orientation:
   W(B(x)) = x >= x = W(x)
   
   B(I(x)) = 1x >= x = J(x)
   
   W(I(x)) = 1x >= x = W(J(x))
  problem:
   W(B(x)) -> W(x)
  Arctic Interpretation Processor:
   dimension: 2
   interpretation:
                
    [W](x0) = x0,
    
              [1 0]  
    [B](x0) = [1 3]x0
   orientation:
              [1 0]             
    W(B(x)) = [1 3]x >= x = W(x)
   problem:
    
   Qed