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^2
      order: standard order
      interpretations:
        a__f#_A(x1,x2) = ((1,0),(1,0)) x1
        g_A(x1) = ((1,1),(1,1)) x1 + (2,2)
        mark_A(x1) = ((1,1),(1,1)) x1 + (1,1)
        f_A(x1,x2) = ((1,1),(1,1)) x1 + (1,3)
        mark#_A(x1) = ((1,1),(1,1)) x1
        a__f_A(x1,x2) = ((1,1),(1,1)) x1 + (2,3)
    
    2. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        a__f#_A(x1,x2) = ((0,1),(1,1)) x1
        g_A(x1) = x1 + (4,1)
        mark_A(x1) = ((0,0),(1,1)) x1 + (3,5)
        f_A(x1,x2) = x1 + (1,1)
        mark#_A(x1) = ((0,1),(0,1)) x1 + (5,6)
        a__f_A(x1,x2) = ((0,0),(1,1)) x1 + (4,0)
    

The next rules are strictly ordered:

  p1, p2, p3, p4, p5

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