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()
  Arctic Interpretation Processor:
   dimension: 1
   interpretation:
    [mark#](x0) = x0 + 1,
    
    [a__c#] = 0,
    
    [a__h#](x0) = x0,
    
    [a__g#](x0) = 4x0 + 0,
    
    [h](x0) = 1x0,
    
    [mark](x0) = 2x0 + 15,
    
    [g](x0) = 5x0 + 5,
    
    [c] = 0,
    
    [d] = 5,
    
    [a__c] = 10,
    
    [a__h](x0) = 2x0,
    
    [a__g](x0) = 6x0 + 6
   orientation:
    a__g#(X) = 4X + 0 >= X = a__h#(X)
    
    a__h#(d()) = 5 >= 4 = a__g#(c())
    
    mark#(g(X)) = 5X + 5 >= 4X + 0 = a__g#(X)
    
    mark#(h(X)) = 1X + 1 >= X = a__h#(X)
    
    mark#(c()) = 1 >= 0 = a__c#()
    
    a__g(X) = 6X + 6 >= 2X = a__h(X)
    
    a__c() = 10 >= 5 = d()
    
    a__h(d()) = 7 >= 6 = a__g(c())
    
    mark(g(X)) = 7X + 15 >= 6X + 6 = a__g(X)
    
    mark(h(X)) = 3X + 15 >= 2X = a__h(X)
    
    mark(c()) = 15 >= 10 = a__c()
    
    mark(d()) = 15 >= 5 = d()
    
    a__g(X) = 6X + 6 >= 5X + 5 = g(X)
    
    a__h(X) = 2X >= 1X = h(X)
    
    a__c() = 10 >= 0 = c()
   problem:
    DPs:
     
    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()
   Qed