YES

We show the termination of the TRS R:

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

and R consists of:

r1: active(c()) -> mark(f(g(c())))
r2: active(f(g(X))) -> mark(g(X))
r3: mark(c()) -> active(c())
r4: mark(f(X)) -> active(f(X))
r5: mark(g(X)) -> active(g(X))
r6: f(mark(X)) -> f(X)
r7: f(active(X)) -> f(X)
r8: g(mark(X)) -> g(X)
r9: g(active(X)) -> g(X)

The estimated dependency graph contains the following SCCs:

  {p1, p4, p5, p6, p7}
  {p8, p9}
  {p10, p11}


-- Reduction pair.

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

p1: active#(c()) -> mark#(f(g(c())))
p2: mark#(f(X)) -> active#(f(X))
p3: active#(f(g(X))) -> mark#(g(X))
p4: mark#(g(X)) -> active#(g(X))
p5: mark#(c()) -> active#(c())

and R consists of:

r1: active(c()) -> mark(f(g(c())))
r2: active(f(g(X))) -> mark(g(X))
r3: mark(c()) -> active(c())
r4: mark(f(X)) -> active(f(X))
r5: mark(g(X)) -> active(g(X))
r6: f(mark(X)) -> f(X)
r7: f(active(X)) -> f(X)
r8: g(mark(X)) -> g(X)
r9: g(active(X)) -> g(X)

The set of usable rules consists of

  r6, r7, r8, r9

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. lexicographic path order with precedence:
    
      precedence:
      
        c > g > f > active > mark > mark# > active#
      
      argument filter:
    
        pi(active#) = [1]
        pi(c) = []
        pi(mark#) = [1]
        pi(f) = [1]
        pi(g) = []
        pi(mark) = [1]
        pi(active) = [1]
    
    2. lexicographic path order with precedence:
    
      precedence:
      
        g > f > active > mark > mark# > c > active#
      
      argument filter:
    
        pi(active#) = []
        pi(c) = []
        pi(mark#) = [1]
        pi(f) = 1
        pi(g) = []
        pi(mark) = [1]
        pi(active) = [1]
    

The next rules are strictly ordered:

  p1, p2, p3, p4, p5

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(c()) -> mark(f(g(c())))
r2: active(f(g(X))) -> mark(g(X))
r3: mark(c()) -> active(c())
r4: mark(f(X)) -> active(f(X))
r5: mark(g(X)) -> active(g(X))
r6: f(mark(X)) -> f(X)
r7: f(active(X)) -> f(X)
r8: g(mark(X)) -> g(X)
r9: 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. lexicographic path order with precedence:
    
      precedence:
      
        f# > active > mark
      
      argument filter:
    
        pi(f#) = 1
        pi(mark) = [1]
        pi(active) = [1]
    

The next rules are strictly ordered:

  p1, p2
  r1, r2, r3, r4, r5, r6, r7, r8, r9

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(c()) -> mark(f(g(c())))
r2: active(f(g(X))) -> mark(g(X))
r3: mark(c()) -> active(c())
r4: mark(f(X)) -> active(f(X))
r5: mark(g(X)) -> active(g(X))
r6: f(mark(X)) -> f(X)
r7: f(active(X)) -> f(X)
r8: g(mark(X)) -> g(X)
r9: 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. lexicographic path order with precedence:
    
      precedence:
      
        g# > active > mark
      
      argument filter:
    
        pi(g#) = 1
        pi(mark) = [1]
        pi(active) = [1]
    

The next rules are strictly ordered:

  p1, p2
  r1, r2, r3, r4, r5, r6, r7, r8, r9

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