YES

We show the termination of the TRS R:

  active(f(X)) -> mark(g(h(f(X))))
  active(f(X)) -> f(active(X))
  active(h(X)) -> h(active(X))
  f(mark(X)) -> mark(f(X))
  h(mark(X)) -> mark(h(X))
  proper(f(X)) -> f(proper(X))
  proper(g(X)) -> g(proper(X))
  proper(h(X)) -> h(proper(X))
  f(ok(X)) -> ok(f(X))
  g(ok(X)) -> ok(g(X))
  h(ok(X)) -> ok(h(X))
  top(mark(X)) -> top(proper(X))
  top(ok(X)) -> top(active(X))

-- SCC decomposition.

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

p1: active#(f(X)) -> g#(h(f(X)))
p2: active#(f(X)) -> h#(f(X))
p3: active#(f(X)) -> f#(active(X))
p4: active#(f(X)) -> active#(X)
p5: active#(h(X)) -> h#(active(X))
p6: active#(h(X)) -> active#(X)
p7: f#(mark(X)) -> f#(X)
p8: h#(mark(X)) -> h#(X)
p9: proper#(f(X)) -> f#(proper(X))
p10: proper#(f(X)) -> proper#(X)
p11: proper#(g(X)) -> g#(proper(X))
p12: proper#(g(X)) -> proper#(X)
p13: proper#(h(X)) -> h#(proper(X))
p14: proper#(h(X)) -> proper#(X)
p15: f#(ok(X)) -> f#(X)
p16: g#(ok(X)) -> g#(X)
p17: h#(ok(X)) -> h#(X)
p18: top#(mark(X)) -> top#(proper(X))
p19: top#(mark(X)) -> proper#(X)
p20: top#(ok(X)) -> top#(active(X))
p21: top#(ok(X)) -> active#(X)

and R consists of:

r1: active(f(X)) -> mark(g(h(f(X))))
r2: active(f(X)) -> f(active(X))
r3: active(h(X)) -> h(active(X))
r4: f(mark(X)) -> mark(f(X))
r5: h(mark(X)) -> mark(h(X))
r6: proper(f(X)) -> f(proper(X))
r7: proper(g(X)) -> g(proper(X))
r8: proper(h(X)) -> h(proper(X))
r9: f(ok(X)) -> ok(f(X))
r10: g(ok(X)) -> ok(g(X))
r11: h(ok(X)) -> ok(h(X))
r12: top(mark(X)) -> top(proper(X))
r13: top(ok(X)) -> top(active(X))

The estimated dependency graph contains the following SCCs:

  {p18, p20}
  {p4, p6}
  {p10, p12, p14}
  {p16}
  {p8, p17}
  {p7, p15}


-- Reduction pair.

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

p1: top#(ok(X)) -> top#(active(X))
p2: top#(mark(X)) -> top#(proper(X))

and R consists of:

r1: active(f(X)) -> mark(g(h(f(X))))
r2: active(f(X)) -> f(active(X))
r3: active(h(X)) -> h(active(X))
r4: f(mark(X)) -> mark(f(X))
r5: h(mark(X)) -> mark(h(X))
r6: proper(f(X)) -> f(proper(X))
r7: proper(g(X)) -> g(proper(X))
r8: proper(h(X)) -> h(proper(X))
r9: f(ok(X)) -> ok(f(X))
r10: g(ok(X)) -> ok(g(X))
r11: h(ok(X)) -> ok(h(X))
r12: top(mark(X)) -> top(proper(X))
r13: top(ok(X)) -> top(active(X))

The set of usable rules consists of

  r1, r2, r3, r4, r5, r6, r7, r8, r9, r10, r11

Take the reduction pair:

  lexicographic path order with precedence:
  
    precedence:
    
      f > g > active > h > proper > mark > ok > top#
    
    argument filter:
  
      pi(top#) = 1
      pi(ok) = 1
      pi(active) = 1
      pi(mark) = [1]
      pi(proper) = 1
      pi(f) = [1]
      pi(h) = [1]
      pi(g) = []

The next rules are strictly ordered:

  p2

We remove them from the problem.

-- SCC decomposition.

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

p1: top#(ok(X)) -> top#(active(X))

and R consists of:

r1: active(f(X)) -> mark(g(h(f(X))))
r2: active(f(X)) -> f(active(X))
r3: active(h(X)) -> h(active(X))
r4: f(mark(X)) -> mark(f(X))
r5: h(mark(X)) -> mark(h(X))
r6: proper(f(X)) -> f(proper(X))
r7: proper(g(X)) -> g(proper(X))
r8: proper(h(X)) -> h(proper(X))
r9: f(ok(X)) -> ok(f(X))
r10: g(ok(X)) -> ok(g(X))
r11: h(ok(X)) -> ok(h(X))
r12: top(mark(X)) -> top(proper(X))
r13: top(ok(X)) -> top(active(X))

The estimated dependency graph contains the following SCCs:

  {p1}


-- Reduction pair.

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

p1: top#(ok(X)) -> top#(active(X))

and R consists of:

r1: active(f(X)) -> mark(g(h(f(X))))
r2: active(f(X)) -> f(active(X))
r3: active(h(X)) -> h(active(X))
r4: f(mark(X)) -> mark(f(X))
r5: h(mark(X)) -> mark(h(X))
r6: proper(f(X)) -> f(proper(X))
r7: proper(g(X)) -> g(proper(X))
r8: proper(h(X)) -> h(proper(X))
r9: f(ok(X)) -> ok(f(X))
r10: g(ok(X)) -> ok(g(X))
r11: h(ok(X)) -> ok(h(X))
r12: top(mark(X)) -> top(proper(X))
r13: top(ok(X)) -> top(active(X))

The set of usable rules consists of

  r1, r2, r3, r4, r5, r9, r10, r11

Take the monotone reduction pair:

  lexicographic path order with precedence:
  
    precedence:
    
      g > ok > active > h > f > mark > top#
    
    argument filter:
  
      pi(top#) = 1
      pi(ok) = [1]
      pi(active) = 1
      pi(f) = 1
      pi(mark) = 1
      pi(h) = 1
      pi(g) = 1

The next rules are strictly ordered:

  p1
  r6, r7, r8, r12, r13

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#(h(X)) -> active#(X)
p2: active#(f(X)) -> active#(X)

and R consists of:

r1: active(f(X)) -> mark(g(h(f(X))))
r2: active(f(X)) -> f(active(X))
r3: active(h(X)) -> h(active(X))
r4: f(mark(X)) -> mark(f(X))
r5: h(mark(X)) -> mark(h(X))
r6: proper(f(X)) -> f(proper(X))
r7: proper(g(X)) -> g(proper(X))
r8: proper(h(X)) -> h(proper(X))
r9: f(ok(X)) -> ok(f(X))
r10: g(ok(X)) -> ok(g(X))
r11: h(ok(X)) -> ok(h(X))
r12: top(mark(X)) -> top(proper(X))
r13: top(ok(X)) -> top(active(X))

The set of usable rules consists of

  (no rules)

Take the monotone reduction pair:

  lexicographic path order with precedence:
  
    precedence:
    
      active# > f > h
    
    argument filter:
  
      pi(active#) = 1
      pi(h) = [1]
      pi(f) = [1]

The next rules are strictly ordered:

  p1, p2
  r1, r2, r3, r4, r5, r6, r7, r8, r9, r10, r11, r12, r13

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: proper#(h(X)) -> proper#(X)
p2: proper#(g(X)) -> proper#(X)
p3: proper#(f(X)) -> proper#(X)

and R consists of:

r1: active(f(X)) -> mark(g(h(f(X))))
r2: active(f(X)) -> f(active(X))
r3: active(h(X)) -> h(active(X))
r4: f(mark(X)) -> mark(f(X))
r5: h(mark(X)) -> mark(h(X))
r6: proper(f(X)) -> f(proper(X))
r7: proper(g(X)) -> g(proper(X))
r8: proper(h(X)) -> h(proper(X))
r9: f(ok(X)) -> ok(f(X))
r10: g(ok(X)) -> ok(g(X))
r11: h(ok(X)) -> ok(h(X))
r12: top(mark(X)) -> top(proper(X))
r13: top(ok(X)) -> top(active(X))

The set of usable rules consists of

  (no rules)

Take the monotone reduction pair:

  lexicographic path order with precedence:
  
    precedence:
    
      proper# > f > g > h
    
    argument filter:
  
      pi(proper#) = 1
      pi(h) = [1]
      pi(g) = [1]
      pi(f) = [1]

The next rules are strictly ordered:

  p1, p2, p3
  r1, r2, r3, r4, r5, r6, r7, r8, r9, r10, r11, r12, r13

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#(ok(X)) -> g#(X)

and R consists of:

r1: active(f(X)) -> mark(g(h(f(X))))
r2: active(f(X)) -> f(active(X))
r3: active(h(X)) -> h(active(X))
r4: f(mark(X)) -> mark(f(X))
r5: h(mark(X)) -> mark(h(X))
r6: proper(f(X)) -> f(proper(X))
r7: proper(g(X)) -> g(proper(X))
r8: proper(h(X)) -> h(proper(X))
r9: f(ok(X)) -> ok(f(X))
r10: g(ok(X)) -> ok(g(X))
r11: h(ok(X)) -> ok(h(X))
r12: top(mark(X)) -> top(proper(X))
r13: top(ok(X)) -> top(active(X))

The set of usable rules consists of

  (no rules)

Take the monotone reduction pair:

  lexicographic path order with precedence:
  
    precedence:
    
      ok > g#
    
    argument filter:
  
      pi(g#) = [1]
      pi(ok) = [1]

The next rules are strictly ordered:

  p1
  r1, r2, r3, r4, r5, r6, r7, r8, r9, r10, r11, r12, r13

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#(ok(X)) -> h#(X)

and R consists of:

r1: active(f(X)) -> mark(g(h(f(X))))
r2: active(f(X)) -> f(active(X))
r3: active(h(X)) -> h(active(X))
r4: f(mark(X)) -> mark(f(X))
r5: h(mark(X)) -> mark(h(X))
r6: proper(f(X)) -> f(proper(X))
r7: proper(g(X)) -> g(proper(X))
r8: proper(h(X)) -> h(proper(X))
r9: f(ok(X)) -> ok(f(X))
r10: g(ok(X)) -> ok(g(X))
r11: h(ok(X)) -> ok(h(X))
r12: top(mark(X)) -> top(proper(X))
r13: top(ok(X)) -> top(active(X))

The set of usable rules consists of

  (no rules)

Take the monotone reduction pair:

  lexicographic path order with precedence:
  
    precedence:
    
      h# > ok > mark
    
    argument filter:
  
      pi(h#) = 1
      pi(mark) = [1]
      pi(ok) = [1]

The next rules are strictly ordered:

  p1, p2
  r1, r2, r3, r4, r5, r6, r7, r8, r9, r10, r11, r12, r13

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#(ok(X)) -> f#(X)

and R consists of:

r1: active(f(X)) -> mark(g(h(f(X))))
r2: active(f(X)) -> f(active(X))
r3: active(h(X)) -> h(active(X))
r4: f(mark(X)) -> mark(f(X))
r5: h(mark(X)) -> mark(h(X))
r6: proper(f(X)) -> f(proper(X))
r7: proper(g(X)) -> g(proper(X))
r8: proper(h(X)) -> h(proper(X))
r9: f(ok(X)) -> ok(f(X))
r10: g(ok(X)) -> ok(g(X))
r11: h(ok(X)) -> ok(h(X))
r12: top(mark(X)) -> top(proper(X))
r13: top(ok(X)) -> top(active(X))

The set of usable rules consists of

  (no rules)

Take the monotone reduction pair:

  lexicographic path order with precedence:
  
    precedence:
    
      f# > ok > mark
    
    argument filter:
  
      pi(f#) = 1
      pi(mark) = [1]
      pi(ok) = [1]

The next rules are strictly ordered:

  p1, p2
  r1, r2, r3, r4, r5, r6, r7, r8, r9, r10, r11, r12, r13

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