YES

We show the termination of the TRS R:

  active(f(f(a()))) -> mark(f(g(f(a()))))
  mark(f(X)) -> active(f(mark(X)))
  mark(a()) -> active(a())
  mark(g(X)) -> active(g(X))
  f(mark(X)) -> f(X)
  f(active(X)) -> f(X)
  g(mark(X)) -> g(X)
  g(active(X)) -> g(X)

-- SCC decomposition.

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

p1: active#(f(f(a()))) -> mark#(f(g(f(a()))))
p2: active#(f(f(a()))) -> f#(g(f(a())))
p3: active#(f(f(a()))) -> g#(f(a()))
p4: mark#(f(X)) -> active#(f(mark(X)))
p5: mark#(f(X)) -> f#(mark(X))
p6: mark#(f(X)) -> mark#(X)
p7: mark#(a()) -> active#(a())
p8: mark#(g(X)) -> active#(g(X))
p9: f#(mark(X)) -> f#(X)
p10: f#(active(X)) -> f#(X)
p11: g#(mark(X)) -> g#(X)
p12: g#(active(X)) -> g#(X)

and R consists of:

r1: active(f(f(a()))) -> mark(f(g(f(a()))))
r2: mark(f(X)) -> active(f(mark(X)))
r3: mark(a()) -> active(a())
r4: mark(g(X)) -> active(g(X))
r5: f(mark(X)) -> f(X)
r6: f(active(X)) -> f(X)
r7: g(mark(X)) -> g(X)
r8: g(active(X)) -> g(X)

The estimated dependency graph contains the following SCCs:

  {p1, p4, p6, p8}
  {p9, p10}
  {p11, p12}


-- Reduction pair.

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

p1: active#(f(f(a()))) -> mark#(f(g(f(a()))))
p2: mark#(f(X)) -> mark#(X)
p3: mark#(g(X)) -> active#(g(X))
p4: mark#(f(X)) -> active#(f(mark(X)))

and R consists of:

r1: active(f(f(a()))) -> mark(f(g(f(a()))))
r2: mark(f(X)) -> active(f(mark(X)))
r3: mark(a()) -> active(a())
r4: mark(g(X)) -> active(g(X))
r5: f(mark(X)) -> f(X)
r6: f(active(X)) -> f(X)
r7: g(mark(X)) -> g(X)
r8: g(active(X)) -> g(X)

The set of usable rules consists of

  r1, r2, r3, r4, r5, r6, r7, r8

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          active#_A(x1) = max{8, x1}
          f_A(x1) = max{6, x1}
          a_A = 10
          mark#_A(x1) = max{9, x1 + 4}
          g_A(x1) = 0
          mark_A(x1) = max{8, x1}
          active_A(x1) = max{6, x1}
    
      precedence:
      
        mark > active > f = a > mark# > active# > g
      
      partial status:
    
        pi(active#) = [1]
        pi(f) = [1]
        pi(a) = []
        pi(mark#) = [1]
        pi(g) = []
        pi(mark) = [1]
        pi(active) = [1]
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          active#_A(x1) = max{13, x1 + 12}
          f_A(x1) = max{7, x1 + 4}
          a_A = 19
          mark#_A(x1) = x1 + 12
          g_A(x1) = 2
          mark_A(x1) = max{0, x1 - 29}
          active_A(x1) = max{8, x1 - 28}
    
      precedence:
      
        a = mark# > active# = f = g = mark = active
      
      partial status:
    
        pi(active#) = [1]
        pi(f) = [1]
        pi(a) = []
        pi(mark#) = []
        pi(g) = []
        pi(mark) = []
        pi(active) = []
    

The next rules are strictly ordered:

  p1, p2, p3, p4

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

-- Reduction pair.

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

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

and R consists of:

r1: active(f(f(a()))) -> mark(f(g(f(a()))))
r2: mark(f(X)) -> active(f(mark(X)))
r3: mark(a()) -> active(a())
r4: mark(g(X)) -> active(g(X))
r5: f(mark(X)) -> f(X)
r6: f(active(X)) -> f(X)
r7: g(mark(X)) -> g(X)
r8: g(active(X)) -> g(X)

The set of usable rules consists of

  (no rules)

Take the monotone reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          f#_A(x1) = x1 + 2
          mark_A(x1) = x1
          active_A(x1) = x1 + 2
    
      precedence:
      
        mark > active > f#
      
      partial status:
    
        pi(f#) = [1]
        pi(mark) = [1]
        pi(active) = [1]
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          f#_A(x1) = x1 + 1
          mark_A(x1) = x1
          active_A(x1) = x1
    
      precedence:
      
        f# = mark = active
      
      partial status:
    
        pi(f#) = [1]
        pi(mark) = [1]
        pi(active) = [1]
    

The next rules are strictly ordered:

  p1, p2

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

-- Reduction pair.

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

p1: g#(mark(X)) -> g#(X)
p2: g#(active(X)) -> g#(X)

and R consists of:

r1: active(f(f(a()))) -> mark(f(g(f(a()))))
r2: mark(f(X)) -> active(f(mark(X)))
r3: mark(a()) -> active(a())
r4: mark(g(X)) -> active(g(X))
r5: f(mark(X)) -> f(X)
r6: f(active(X)) -> f(X)
r7: g(mark(X)) -> g(X)
r8: g(active(X)) -> g(X)

The set of usable rules consists of

  (no rules)

Take the monotone reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          g#_A(x1) = x1 + 2
          mark_A(x1) = x1
          active_A(x1) = x1 + 2
    
      precedence:
      
        mark > active > g#
      
      partial status:
    
        pi(g#) = [1]
        pi(mark) = [1]
        pi(active) = [1]
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          g#_A(x1) = x1 + 1
          mark_A(x1) = x1
          active_A(x1) = x1
    
      precedence:
      
        g# = mark = active
      
      partial status:
    
        pi(g#) = [1]
        pi(mark) = [1]
        pi(active) = [1]
    

The next rules are strictly ordered:

  p1, p2

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