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

The next rules are strictly ordered:

  p1, p2, p3, p4, p5, p6

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. 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, r9, r10

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. lexicographic path order with precedence:
    
      precedence:
      
        h# > active > mark
      
      argument filter:
    
        pi(h#) = 1
        pi(mark) = [1]
        pi(active) = 1
    
    2. matrix interpretations:
    
      carrier: N^1
      order: standard order
      interpretations:
        h#_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, r9, r10

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. 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, r9, r10

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