YES

Problem:
 a__f(f(X)) -> a__c(f(g(f(X))))
 a__c(X) -> d(X)
 a__h(X) -> a__c(d(X))
 mark(f(X)) -> a__f(mark(X))
 mark(c(X)) -> a__c(X)
 mark(h(X)) -> a__h(mark(X))
 mark(g(X)) -> g(X)
 mark(d(X)) -> d(X)
 a__f(X) -> f(X)
 a__c(X) -> c(X)
 a__h(X) -> h(X)

Proof:
 String Reversal Processor:
  f(a__f(X)) -> f(g(f(a__c(X))))
  a__c(X) -> d(X)
  a__h(X) -> d(a__c(X))
  f(mark(X)) -> mark(a__f(X))
  c(mark(X)) -> a__c(X)
  h(mark(X)) -> mark(a__h(X))
  g(mark(X)) -> g(X)
  d(mark(X)) -> d(X)
  a__f(X) -> f(X)
  a__c(X) -> c(X)
  a__h(X) -> h(X)
  WPO Processor:
   algebra: Sum
   weight function:
    w0 = 0
    w(a__f) = w(f) = 2
    w(mark) = 1
    w(h) = w(c) = w(a__h) = w(d) = w(a__c) = w(g) = 0
   status function:
    st(h) = st(c) = st(mark) = st(a__h) = st(d) = st(a__c) = st(g) = st(a__f) = st(f) = [0]
   precedence:
    a__h > a__c > a__f > f > h > c ~ mark ~ d ~ g
   problem:
    
   Qed