YES

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

Proof:
 DP Processor:
  DPs:
   a__g#(X) -> a__h#(X)
   a__h#(d()) -> a__g#(c())
   mark#(g(X)) -> a__g#(X)
   mark#(h(X)) -> a__h#(X)
   mark#(c()) -> a__c#()
  TRS:
   a__g(X) -> a__h(X)
   a__c() -> d()
   a__h(d()) -> a__g(c())
   mark(g(X)) -> a__g(X)
   mark(h(X)) -> a__h(X)
   mark(c()) -> a__c()
   mark(d()) -> d()
   a__g(X) -> g(X)
   a__h(X) -> h(X)
   a__c() -> c()
  Usable Rule Processor:
   DPs:
    a__g#(X) -> a__h#(X)
    a__h#(d()) -> a__g#(c())
    mark#(g(X)) -> a__g#(X)
    mark#(h(X)) -> a__h#(X)
    mark#(c()) -> a__c#()
   TRS:
    
   Arctic Interpretation Processor:
    dimension: 1
    usable rules:
     
    interpretation:
     [mark#](x0) = -1x0 + 9,
     
     [a__c#] = 0,
     
     [a__h#](x0) = -8x0 + 0,
     
     [a__g#](x0) = x0 + 1,
     
     [h](x0) = 4x0,
     
     [g](x0) = 8x0 + 8,
     
     [c] = 1,
     
     [d] = 10
    orientation:
     a__g#(X) = X + 1 >= -8X + 0 = a__h#(X)
     
     a__h#(d()) = 2 >= 1 = a__g#(c())
     
     mark#(g(X)) = 7X + 9 >= X + 1 = a__g#(X)
     
     mark#(h(X)) = 3X + 9 >= -8X + 0 = a__h#(X)
     
     mark#(c()) = 9 >= 0 = a__c#()
    problem:
     DPs:
      
     TRS:
      
    Qed