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:
    
   Restore Modifier:
    DPs:
     a#(b(x)) -> a#(c(b(x)))
    TRS:
     a(b(x)) -> a(c(b(x)))
    SCC Processor:
     #sccs: 1
     #rules: 1
     #arcs: 1/1
     DPs:
      a#(b(x)) -> a#(c(b(x)))
     TRS:
      a(b(x)) -> a(c(b(x)))
     Matrix Interpretation Processor:
      dimension: 1
      interpretation:
       [a#](x0) = x0,
       
       [c](x0) = 0,
       
       [a](x0) = 0,
       
       [b](x0) = 1
      orientation:
       a#(b(x)) = 1 >= 0 = a#(c(b(x)))
       
       a(b(x)) = 0 >= 0 = a(c(b(x)))
      problem:
       DPs:
        
       TRS:
        a(b(x)) -> a(c(b(x)))
      Qed