YES

We show the termination of the TRS R:

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

-- SCC decomposition.

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

p1: active#(f(X)) -> mark#(g(h(f(X))))
p2: active#(f(X)) -> g#(h(f(X)))
p3: active#(f(X)) -> h#(f(X))
p4: mark#(f(X)) -> active#(f(mark(X)))
p5: mark#(f(X)) -> f#(mark(X))
p6: mark#(f(X)) -> mark#(X)
p7: mark#(g(X)) -> active#(g(X))
p8: mark#(h(X)) -> active#(h(mark(X)))
p9: mark#(h(X)) -> h#(mark(X))
p10: mark#(h(X)) -> mark#(X)
p11: f#(mark(X)) -> f#(X)
p12: f#(active(X)) -> f#(X)
p13: g#(mark(X)) -> g#(X)
p14: g#(active(X)) -> g#(X)
p15: h#(mark(X)) -> h#(X)
p16: h#(active(X)) -> h#(X)

and R consists of:

r1: active(f(X)) -> mark(g(h(f(X))))
r2: mark(f(X)) -> active(f(mark(X)))
r3: mark(g(X)) -> active(g(X))
r4: mark(h(X)) -> active(h(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)
r9: h(mark(X)) -> h(X)
r10: h(active(X)) -> h(X)

The estimated dependency graph contains the following SCCs:

  {p1, p4, p6, p7, p8, p10}
  {p13, p14}
  {p15, p16}
  {p11, p12}


-- Reduction pair.

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

p1: active#(f(X)) -> mark#(g(h(f(X))))
p2: mark#(h(X)) -> mark#(X)
p3: mark#(h(X)) -> active#(h(mark(X)))
p4: mark#(g(X)) -> active#(g(X))
p5: mark#(f(X)) -> mark#(X)
p6: mark#(f(X)) -> active#(f(mark(X)))

and R consists of:

r1: active(f(X)) -> mark(g(h(f(X))))
r2: mark(f(X)) -> active(f(mark(X)))
r3: mark(g(X)) -> active(g(X))
r4: mark(h(X)) -> active(h(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)
r9: h(mark(X)) -> h(X)
r10: h(active(X)) -> h(X)

The set of usable rules consists of

  r1, r2, r3, r4, r5, r6, r7, r8, r9, r10

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{7, x1 - 10}
          f_A(x1) = x1 + 18
          mark#_A(x1) = max{1, x1 - 4}
          g_A(x1) = max{12, x1 - 26}
          h_A(x1) = x1 + 19
          mark_A(x1) = x1 + 5
          active_A(x1) = x1
    
      precedence:
      
        active# = f = mark# = g = h = mark = active
      
      partial status:
    
        pi(active#) = []
        pi(f) = []
        pi(mark#) = []
        pi(g) = []
        pi(h) = []
        pi(mark) = []
        pi(active) = []
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          active#_A(x1) = 31
          f_A(x1) = 30
          mark#_A(x1) = 30
          g_A(x1) = 29
          h_A(x1) = 51
          mark_A(x1) = max{12, x1 + 5}
          active_A(x1) = max{13, x1 + 2}
    
      precedence:
      
        h > f = mark = active > g > mark# > active#
      
      partial status:
    
        pi(active#) = []
        pi(f) = []
        pi(mark#) = []
        pi(g) = []
        pi(h) = []
        pi(mark) = [1]
        pi(active) = [1]
    

The next rules are strictly ordered:

  p1, p3, p4, p6

We remove them from the problem.

-- SCC decomposition.

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

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

and R consists of:

r1: active(f(X)) -> mark(g(h(f(X))))
r2: mark(f(X)) -> active(f(mark(X)))
r3: mark(g(X)) -> active(g(X))
r4: mark(h(X)) -> active(h(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)
r9: h(mark(X)) -> h(X)
r10: h(active(X)) -> h(X)

The estimated dependency graph contains the following SCCs:

  {p1, p2}


-- Reduction pair.

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

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

and R consists of:

r1: active(f(X)) -> mark(g(h(f(X))))
r2: mark(f(X)) -> active(f(mark(X)))
r3: mark(g(X)) -> active(g(X))
r4: mark(h(X)) -> active(h(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)
r9: h(mark(X)) -> h(X)
r10: h(active(X)) -> h(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:
        
          mark#_A(x1) = max{6, x1 + 4}
          h_A(x1) = max{4, x1 + 3}
          f_A(x1) = max{2, x1 + 1}
    
      precedence:
      
        mark# = h = f
      
      partial status:
    
        pi(mark#) = [1]
        pi(h) = [1]
        pi(f) = [1]
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          mark#_A(x1) = x1 + 1
          h_A(x1) = x1
          f_A(x1) = x1
    
      precedence:
      
        mark# = h = f
      
      partial status:
    
        pi(mark#) = [1]
        pi(h) = [1]
        pi(f) = [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(X)) -> mark(g(h(f(X))))
r2: mark(f(X)) -> active(f(mark(X)))
r3: mark(g(X)) -> active(g(X))
r4: mark(h(X)) -> active(h(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)
r9: h(mark(X)) -> h(X)
r10: h(active(X)) -> h(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.

-- Reduction pair.

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

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

and R consists of:

r1: active(f(X)) -> mark(g(h(f(X))))
r2: mark(f(X)) -> active(f(mark(X)))
r3: mark(g(X)) -> active(g(X))
r4: mark(h(X)) -> active(h(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)
r9: h(mark(X)) -> h(X)
r10: h(active(X)) -> h(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:
        
          h#_A(x1) = x1 + 2
          mark_A(x1) = x1
          active_A(x1) = x1 + 2
    
      precedence:
      
        mark > active > h#
      
      partial status:
    
        pi(h#) = [1]
        pi(mark) = [1]
        pi(active) = [1]
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          h#_A(x1) = x1 + 1
          mark_A(x1) = x1
          active_A(x1) = x1
    
      precedence:
      
        h# = mark = active
      
      partial status:
    
        pi(h#) = [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: f#(mark(X)) -> f#(X)
p2: f#(active(X)) -> f#(X)

and R consists of:

r1: active(f(X)) -> mark(g(h(f(X))))
r2: mark(f(X)) -> active(f(mark(X)))
r3: mark(g(X)) -> active(g(X))
r4: mark(h(X)) -> active(h(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)
r9: h(mark(X)) -> h(X)
r10: h(active(X)) -> h(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.