YES

We show the termination of the TRS R:

  active(f(f(a()))) -> mark(f(g(f(a()))))
  mark(f(X)) -> active(f(X))
  mark(a()) -> active(a())
  mark(g(X)) -> active(g(mark(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(X))
p5: mark#(a()) -> active#(a())
p6: mark#(g(X)) -> active#(g(mark(X)))
p7: mark#(g(X)) -> g#(mark(X))
p8: mark#(g(X)) -> mark#(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(X))
r3: mark(a()) -> active(a())
r4: mark(g(X)) -> active(g(mark(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:

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


-- Reduction pair.

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

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

and R consists of:

r1: active(f(f(a()))) -> mark(f(g(f(a()))))
r2: mark(f(X)) -> active(f(X))
r3: mark(a()) -> active(a())
r4: mark(g(X)) -> active(g(mark(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. lexicographic path order with precedence:
    
      precedence:
      
        mark# > g
      
      argument filter:
    
        pi(mark#) = 1
        pi(g) = [1]
    
    2. matrix interpretations:
    
      carrier: N^1
      order: standard order
      interpretations:
        mark#_A(x1) = x1
        g_A(x1) = x1 + 1
    

The next rules are strictly ordered:

  p1
  r1, r2, r3, r4, r5, r6, r7, r8

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

and R consists of:

r1: active(f(f(a()))) -> mark(f(g(f(a()))))
r2: mark(f(X)) -> active(f(X))
r3: mark(a()) -> active(a())
r4: mark(g(X)) -> active(g(mark(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

  r5, r6

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. lexicographic path order with precedence:
    
      precedence:
      
        f > active > mark > a > g > mark# > active#
      
      argument filter:
    
        pi(active#) = [1]
        pi(f) = 1
        pi(a) = []
        pi(mark#) = [1]
        pi(g) = []
        pi(mark) = 1
        pi(active) = [1]
    
    2. matrix interpretations:
    
      carrier: N^1
      order: standard order
      interpretations:
        active#_A(x1) = 1
        f_A(x1) = x1 + 1
        a_A() = 0
        mark#_A(x1) = 0
        g_A(x1) = 1
        mark_A(x1) = x1
        active_A(x1) = x1
    

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(X))
r3: mark(a()) -> active(a())
r4: mark(g(X)) -> active(g(mark(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. lexicographic path order with precedence:
    
      precedence:
      
        g# > active > mark
      
      argument filter:
    
        pi(g#) = 1
        pi(mark) = [1]
        pi(active) = 1
    
    2. matrix interpretations:
    
      carrier: N^1
      order: standard order
      interpretations:
        g#_A(x1) = x1
        mark_A(x1) = x1 + 1
        active_A(x1) = x1 + 1
    

The next rules are strictly ordered:

  p1, p2
  r1, r2, r3, r4, r5, r6, r7, r8

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(X))
r3: mark(a()) -> active(a())
r4: mark(g(X)) -> active(g(mark(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. lexicographic path order with precedence:
    
      precedence:
      
        f# > active > mark
      
      argument filter:
    
        pi(f#) = 1
        pi(mark) = [1]
        pi(active) = 1
    
    2. matrix interpretations:
    
      carrier: N^1
      order: standard order
      interpretations:
        f#_A(x1) = x1
        mark_A(x1) = x1 + 1
        active_A(x1) = x1 + 1
    

The next rules are strictly ordered:

  p1, p2
  r1, r2, r3, r4, r5, r6, r7, r8

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