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:
    
   Restore Modifier:
    DPs:
     a#(b(x)) -> a#(x)
    TRS:
     a(b(x)) -> b(a(x))
     a(c(x)) -> x
    SCC Processor:
     #sccs: 1
     #rules: 1
     #arcs: 1/1
     DPs:
      a#(b(x)) -> a#(x)
     TRS:
      a(b(x)) -> b(a(x))
      a(c(x)) -> x
     Matrix Interpretation Processor:
      dimension: 1
      interpretation:
       [a#](x0) = x0,
       
       [c](x0) = x0,
       
       [a](x0) = x0 + 1,
       
       [b](x0) = x0 + 1
      orientation:
       a#(b(x)) = x + 1 >= x = a#(x)
       
       a(b(x)) = x + 2 >= x + 2 = b(a(x))
       
       a(c(x)) = x + 1 >= x = x
      problem:
       DPs:
        
       TRS:
        a(b(x)) -> b(a(x))
        a(c(x)) -> x
      Qed