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:
 Arctic Interpretation Processor:
  dimension: 1
  interpretation:
   [h](x0) = x0,
   
   [c](x0) = x0,
   
   [mark](x0) = 2x0,
   
   [a__h](x0) = x0,
   
   [d](x0) = x0,
   
   [a__c](x0) = x0,
   
   [g](x0) = x0,
   
   [a__f](x0) = x0,
   
   [f](x0) = x0
  orientation:
   a__f(f(X)) = X >= X = a__c(f(g(f(X))))
   
   a__c(X) = X >= X = d(X)
   
   a__h(X) = X >= X = a__c(d(X))
   
   mark(f(X)) = 2X >= 2X = a__f(mark(X))
   
   mark(c(X)) = 2X >= X = a__c(X)
   
   mark(h(X)) = 2X >= 2X = a__h(mark(X))
   
   mark(g(X)) = 2X >= X = g(X)
   
   mark(d(X)) = 2X >= X = d(X)
   
   a__f(X) = X >= X = f(X)
   
   a__c(X) = X >= X = c(X)
   
   a__h(X) = X >= X = h(X)
  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(h(X)) -> a__h(mark(X))
   a__f(X) -> f(X)
   a__c(X) -> c(X)
   a__h(X) -> h(X)
  Arctic Interpretation Processor:
   dimension: 1
   interpretation:
    [h](x0) = 1x0,
    
    [c](x0) = x0,
    
    [mark](x0) = x0,
    
    [a__h](x0) = 1x0,
    
    [d](x0) = x0,
    
    [a__c](x0) = x0,
    
    [g](x0) = x0,
    
    [a__f](x0) = 8x0,
    
    [f](x0) = 8x0
   orientation:
    a__f(f(X)) = 16X >= 16X = a__c(f(g(f(X))))
    
    a__c(X) = X >= X = d(X)
    
    a__h(X) = 1X >= X = a__c(d(X))
    
    mark(f(X)) = 8X >= 8X = a__f(mark(X))
    
    mark(h(X)) = 1X >= 1X = a__h(mark(X))
    
    a__f(X) = 8X >= 8X = f(X)
    
    a__c(X) = X >= X = c(X)
    
    a__h(X) = 1X >= 1X = h(X)
   problem:
    a__f(f(X)) -> a__c(f(g(f(X))))
    a__c(X) -> d(X)
    mark(f(X)) -> a__f(mark(X))
    mark(h(X)) -> a__h(mark(X))
    a__f(X) -> f(X)
    a__c(X) -> c(X)
    a__h(X) -> h(X)
   String Reversal Processor:
    f(a__f(X)) -> f(g(f(a__c(X))))
    a__c(X) -> d(X)
    f(mark(X)) -> mark(a__f(X))
    h(mark(X)) -> mark(a__h(X))
    a__f(X) -> f(X)
    a__c(X) -> c(X)
    a__h(X) -> h(X)
    Bounds Processor:
     bound: 1
     enrichment: match
     automaton:
      final states: {13,12,11,9,7,6,1}
      transitions:
       h1(28) -> 29*
       c1(22) -> 23*
       f1(20) -> 21*
       d1(14) -> 15*
       f90() -> 2*
       f0(5) -> 1*
       f0(2) -> 11*
       f0(3) -> 4*
       g0(4) -> 5*
       a__c0(2) -> 3*
       d0(2) -> 6*
       mark0(10) -> 9*
       mark0(8) -> 7*
       a__f0(2) -> 8*
       a__h0(2) -> 10*
       c0(2) -> 12*
       h0(2) -> 13*
       1 -> 21,11,8
       2 -> 28,22,20,14
       7 -> 21,11,8
       9 -> 29,13,10
       15 -> 3*
       21 -> 8*
       23 -> 3*
       29 -> 10*
     problem:
      
     Qed