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. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            a__f#_A(x1,x2) = x1 + (0,2)
            g_A(x1) = ((1,0),(1,1)) x1 + (5,2)
            mark_A(x1) = ((1,0),(1,1)) x1 + (4,1)
            f_A(x1,x2) = x1 + (3,0)
            mark#_A(x1) = x1 + (2,0)
            a__f_A(x1,x2) = x1 + (3,2)
    
      precedence:
      
        g = mark > f = a__f > mark# > a__f#
      
      partial status:
    
        pi(a__f#) = [1]
        pi(g) = [1]
        pi(mark) = [1]
        pi(f) = []
        pi(mark#) = [1]
        pi(a__f) = [1]
    
    2. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            a__f#_A(x1,x2) = ((1,0),(0,0)) x1 + (0,1)
            g_A(x1) = ((0,0),(1,0)) x1 + (4,0)
            mark_A(x1) = x1 + (3,4)
            f_A(x1,x2) = (1,0)
            mark#_A(x1) = (0,0)
            a__f_A(x1,x2) = (2,1)
    
      precedence:
      
        mark > g > a__f > a__f# = f = mark#
      
      partial status:
    
        pi(a__f#) = []
        pi(g) = []
        pi(mark) = []
        pi(f) = []
        pi(mark#) = []
        pi(a__f) = []
    

The next rules are strictly ordered:

  p1, p5

We remove them from the problem.

-- SCC decomposition.

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

p1: a__f#(g(X),Y) -> mark#(X)
p2: mark#(g(X)) -> mark#(X)
p3: 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:

  {p2, p3}


-- Reduction pair.

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

p1: mark#(g(X)) -> mark#(X)
p2: 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 reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            mark#_A(x1) = ((1,0),(1,1)) x1 + (2,2)
            g_A(x1) = x1 + (1,3)
            f_A(x1,x2) = x1 + x2 + (3,1)
    
      precedence:
      
        g = f > mark#
      
      partial status:
    
        pi(mark#) = [1]
        pi(g) = [1]
        pi(f) = [1, 2]
    
    2. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            mark#_A(x1) = (0,0)
            g_A(x1) = (1,1)
            f_A(x1,x2) = x2 + (1,1)
    
      precedence:
      
        f > mark# = g
      
      partial status:
    
        pi(mark#) = []
        pi(g) = []
        pi(f) = [2]
    

The next rules are strictly ordered:

  p1

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 reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            mark#_A(x1) = x1 + (1,2)
            f_A(x1,x2) = ((1,0),(0,0)) x1 + (2,1)
    
      precedence:
      
        mark# > f
      
      partial status:
    
        pi(mark#) = [1]
        pi(f) = []
    
    2. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            mark#_A(x1) = (0,0)
            f_A(x1,x2) = (1,1)
    
      precedence:
      
        mark# > f
      
      partial status:
    
        pi(mark#) = []
        pi(f) = []
    

The next rules are strictly ordered:

  p1

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