YES

We show the termination of the TRS R:

  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)

-- SCC decomposition.

Consider the dependency pair problem (P, R), where P consists of

p1: a__f#(f(X)) -> a__c#(f(g(f(X))))
p2: a__h#(X) -> a__c#(d(X))
p3: mark#(f(X)) -> a__f#(mark(X))
p4: mark#(f(X)) -> mark#(X)
p5: mark#(c(X)) -> a__c#(X)
p6: mark#(h(X)) -> a__h#(mark(X))
p7: mark#(h(X)) -> mark#(X)

and R consists of:

r1: a__f(f(X)) -> a__c(f(g(f(X))))
r2: a__c(X) -> d(X)
r3: a__h(X) -> a__c(d(X))
r4: mark(f(X)) -> a__f(mark(X))
r5: mark(c(X)) -> a__c(X)
r6: mark(h(X)) -> a__h(mark(X))
r7: mark(g(X)) -> g(X)
r8: mark(d(X)) -> d(X)
r9: a__f(X) -> f(X)
r10: a__c(X) -> c(X)
r11: a__h(X) -> h(X)

The estimated dependency graph contains the following SCCs:

  {p4, p7}


-- Reduction pair.

Consider the dependency pair problem (P, R), where P consists of

p1: mark#(h(X)) -> mark#(X)
p2: mark#(f(X)) -> mark#(X)

and R consists of:

r1: a__f(f(X)) -> a__c(f(g(f(X))))
r2: a__c(X) -> d(X)
r3: a__h(X) -> a__c(d(X))
r4: mark(f(X)) -> a__f(mark(X))
r5: mark(c(X)) -> a__c(X)
r6: mark(h(X)) -> a__h(mark(X))
r7: mark(g(X)) -> g(X)
r8: mark(d(X)) -> d(X)
r9: a__f(X) -> f(X)
r10: a__c(X) -> c(X)
r11: a__h(X) -> h(X)

The set of usable rules consists of

  (no rules)

Take the monotone reduction pair:

  matrix interpretations:
  
    carrier: N^3
    order: lexicographic order
    interpretations:
      mark#_A(x1) = ((1,0,0),(1,1,0),(1,1,1)) x1
      h_A(x1) = ((1,0,0),(1,1,0),(1,1,1)) x1 + (1,1,1)
      f_A(x1) = ((1,0,0),(1,1,0),(1,1,1)) x1 + (1,1,1)

The next rules are strictly ordered:

  p1, p2
  r1, r2, r3, r4, r5, r6, r7, r8, r9, r10, r11

We remove them from the problem.  Then no dependency pair remains.