YES

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

Proof:
 DP Processor:
  DPs:
   a#(b(x)) -> a#(c(b(x)))
  TRS:
   a(b(x)) -> a(c(b(x)))
  Usable Rule Processor:
   DPs:
    a#(b(x)) -> a#(c(b(x)))
   TRS:
    
   Arctic Interpretation Processor:
    dimension: 2
    usable rules:
     
    interpretation:
     [a#](x0) = [0 3]x0 + [0],
     
               [1  -4]     [-4]
     [c](x0) = [-1 -&]x0 + [-&],
     
               [3  -4]     [-4]
     [b](x0) = [3  -3]x0 + [2 ]
    orientation:
     a#(b(x)) = [6 0]x + [5] >= [5  -2]x + [0] = a#(c(b(x)))
    problem:
     DPs:
      
     TRS:
      
    Qed