YES

We show the termination of the TRS R:

  a__f(g(X),Y) -> a__f(mark(X),f(g(X),Y))
  mark(f(X1,X2)) -> a__f(mark(X1),X2)
  mark(g(X)) -> g(mark(X))
  a__f(X1,X2) -> f(X1,X2)

-- SCC decomposition.

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

p1: a__f#(g(X),Y) -> a__f#(mark(X),f(g(X),Y))
p2: a__f#(g(X),Y) -> mark#(X)
p3: mark#(f(X1,X2)) -> a__f#(mark(X1),X2)
p4: mark#(f(X1,X2)) -> mark#(X1)
p5: mark#(g(X)) -> mark#(X)

and R consists of:

r1: a__f(g(X),Y) -> a__f(mark(X),f(g(X),Y))
r2: mark(f(X1,X2)) -> a__f(mark(X1),X2)
r3: mark(g(X)) -> g(mark(X))
r4: a__f(X1,X2) -> f(X1,X2)

The estimated dependency graph contains the following SCCs:

  {p1, p2, p3, p4, p5}


-- Reduction pair.

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

p1: a__f#(g(X),Y) -> a__f#(mark(X),f(g(X),Y))
p2: a__f#(g(X),Y) -> mark#(X)
p3: mark#(g(X)) -> mark#(X)
p4: mark#(f(X1,X2)) -> mark#(X1)
p5: mark#(f(X1,X2)) -> a__f#(mark(X1),X2)

and R consists of:

r1: a__f(g(X),Y) -> a__f(mark(X),f(g(X),Y))
r2: mark(f(X1,X2)) -> a__f(mark(X1),X2)
r3: mark(g(X)) -> g(mark(X))
r4: a__f(X1,X2) -> f(X1,X2)

The set of usable rules consists of

  r1, r2, r3, r4

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. matrix interpretations:
    
      carrier: N^1
      order: standard order
      interpretations:
        a__f#_A(x1,x2) = x1
        g_A(x1) = x1 + 3
        mark_A(x1) = x1 + 1
        f_A(x1,x2) = x1
        mark#_A(x1) = x1 + 2
        a__f_A(x1,x2) = x1
    
    2. matrix interpretations:
    
      carrier: N^1
      order: standard order
      interpretations:
        a__f#_A(x1,x2) = 0
        g_A(x1) = 1
        mark_A(x1) = 3
        f_A(x1,x2) = 1
        mark#_A(x1) = 1
        a__f_A(x1,x2) = 2
    

The next rules are strictly ordered:

  p1, p2, p3, p5

We remove them from the problem.

-- SCC decomposition.

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

p1: mark#(f(X1,X2)) -> mark#(X1)

and R consists of:

r1: a__f(g(X),Y) -> a__f(mark(X),f(g(X),Y))
r2: mark(f(X1,X2)) -> a__f(mark(X1),X2)
r3: mark(g(X)) -> g(mark(X))
r4: a__f(X1,X2) -> f(X1,X2)

The estimated dependency graph contains the following SCCs:

  {p1}


-- Reduction pair.

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

p1: mark#(f(X1,X2)) -> mark#(X1)

and R consists of:

r1: a__f(g(X),Y) -> a__f(mark(X),f(g(X),Y))
r2: mark(f(X1,X2)) -> a__f(mark(X1),X2)
r3: mark(g(X)) -> g(mark(X))
r4: a__f(X1,X2) -> f(X1,X2)

The set of usable rules consists of

  (no rules)

Take the monotone reduction pair:

  lexicographic combination of reduction pairs:
  
    1. matrix interpretations:
    
      carrier: N^1
      order: standard order
      interpretations:
        mark#_A(x1) = x1
        f_A(x1,x2) = x1 + x2 + 1
    
    2. matrix interpretations:
    
      carrier: N^1
      order: standard order
      interpretations:
        mark#_A(x1) = x1
        f_A(x1,x2) = x1 + x2 + 1
    

The next rules are strictly ordered:

  p1
  r1, r2, r3, r4

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