YES(?,O(n^2))

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:
 Matrix Interpretation Processor:
  dimension: 2
  interpretation:
                  [8]
   [h](x0) = x0 + [2],
   
             [1 0]     [0]
   [c](x0) = [0 0]x0 + [2],
   
                [1 1]  
   [mark](x0) = [0 1]x0,
   
                     [9]
   [a__h](x0) = x0 + [2],
   
             [1 0]     [0]
   [d](x0) = [0 0]x0 + [2],
   
                [1 0]     [1]
   [a__c](x0) = [0 0]x0 + [2],
   
             [1 0]     [0]
   [g](x0) = [0 0]x0 + [5],
   
                     [4]
   [a__f](x0) = x0 + [3],
   
                  [2]
   [f](x0) = x0 + [3]
  orientation:
                    [6]    [1 0]    [5]                   
   a__f(f(X)) = X + [6] >= [0 0]X + [2] = a__c(f(g(f(X))))
   
             [1 0]    [1]    [1 0]    [0]       
   a__c(X) = [0 0]X + [2] >= [0 0]X + [2] = d(X)
   
                 [9]    [1 0]    [1]             
   a__h(X) = X + [2] >= [0 0]X + [2] = a__c(d(X))
   
                [1 1]    [5]    [1 1]    [4]                
   mark(f(X)) = [0 1]X + [3] >= [0 1]X + [3] = a__f(mark(X))
   
                [1 0]    [2]    [1 0]    [1]          
   mark(c(X)) = [0 0]X + [2] >= [0 0]X + [2] = a__c(X)
   
                [1 1]    [10]    [1 1]    [9]                
   mark(h(X)) = [0 1]X + [2 ] >= [0 1]X + [2] = a__h(mark(X))
   
                [1 0]    [5]    [1 0]    [0]       
   mark(g(X)) = [0 0]X + [5] >= [0 0]X + [5] = g(X)
   
                [1 0]    [2]    [1 0]    [0]       
   mark(d(X)) = [0 0]X + [2] >= [0 0]X + [2] = d(X)
   
                 [4]        [2]       
   a__f(X) = X + [3] >= X + [3] = f(X)
   
             [1 0]    [1]    [1 0]    [0]       
   a__c(X) = [0 0]X + [2] >= [0 0]X + [2] = c(X)
   
                 [9]        [8]       
   a__h(X) = X + [2] >= X + [2] = h(X)
  problem:
   
  Qed