YES

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

Proof:
 DP Processor:
  DPs:
   a#(b(x)) -> a#(x)
  TRS:
   a(b(x)) -> b(a(x))
   a(c(x)) -> x
  Matrix Interpretation Processor: dim=3
   
   interpretation:
    [a#](x0) = [0 1 0]x0 + [1],
    
              [1 1 0]  
    [c](x0) = [0 0 0]x0
              [1 1 1]  ,
    
              [1 0 0]  
    [a](x0) = [0 0 1]x0
              [0 0 1]  ,
    
              [0 0 0]     [1]
    [b](x0) = [0 1 0]x0 + [1]
              [0 0 1]     [1]
   orientation:
    a#(b(x)) = [0 1 0]x + [2] >= [0 1 0]x + [1] = a#(x)
    
              [0 0 0]    [1]    [0 0 0]    [1]          
    a(b(x)) = [0 0 1]x + [1] >= [0 0 1]x + [1] = b(a(x))
              [0 0 1]    [1]    [0 0 1]    [1]          
    
              [1 1 0]          
    a(c(x)) = [1 1 1]x >= x = x
              [1 1 1]          
   problem:
    DPs:
     
    TRS:
     a(b(x)) -> b(a(x))
     a(c(x)) -> x
   Qed