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