YES

Problem:
 a(b(x)) -> b(b(a(x)))

Proof:
 DP Processor:
  DPs:
   a#(b(x)) -> a#(x)
  TRS:
   a(b(x)) -> b(b(a(x)))
  Matrix Interpretation Processor: dim=4
   
   interpretation:
    [a#](x0) = [1 1 1 1]x0,
    
              [0 0 0 0]  
              [0 1 0 0]  
    [a](x0) = [0 0 1 1]x0
              [0 0 1 1]  ,
    
              [1 0 0 0]     [0]
              [0 1 0 0]     [0]
    [b](x0) = [0 0 0 1]x0 + [1]
              [0 0 1 0]     [0]
   orientation:
    a#(b(x)) = [1 1 1 1]x + [1] >= [1 1 1 1]x = a#(x)
    
              [0 0 0 0]    [0]    [0 0 0 0]    [0]             
              [0 1 0 0]    [0]    [0 1 0 0]    [0]             
    a(b(x)) = [0 0 1 1]x + [1] >= [0 0 1 1]x + [1] = b(b(a(x)))
              [0 0 1 1]    [1]    [0 0 1 1]    [1]             
   problem:
    DPs:
     
    TRS:
     a(b(x)) -> b(b(a(x)))
   Qed