YES

We show the termination of the TRS R:

  active(f(f(a()))) -> mark(f(g(f(a()))))
  active(f(X)) -> f(active(X))
  f(mark(X)) -> mark(f(X))
  proper(f(X)) -> f(proper(X))
  proper(a()) -> ok(a())
  proper(g(X)) -> g(proper(X))
  f(ok(X)) -> ok(f(X))
  g(ok(X)) -> ok(g(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(f(a()))) -> f#(g(f(a())))
p2: active#(f(f(a()))) -> g#(f(a()))
p3: active#(f(X)) -> f#(active(X))
p4: active#(f(X)) -> active#(X)
p5: f#(mark(X)) -> f#(X)
p6: proper#(f(X)) -> f#(proper(X))
p7: proper#(f(X)) -> proper#(X)
p8: proper#(g(X)) -> g#(proper(X))
p9: proper#(g(X)) -> proper#(X)
p10: f#(ok(X)) -> f#(X)
p11: g#(ok(X)) -> g#(X)
p12: top#(mark(X)) -> top#(proper(X))
p13: top#(mark(X)) -> proper#(X)
p14: top#(ok(X)) -> top#(active(X))
p15: top#(ok(X)) -> active#(X)

and R consists of:

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

The estimated dependency graph contains the following SCCs:

  {p12, p14}
  {p4}
  {p7, p9}
  {p5, p10}
  {p11}


-- 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(f(a()))) -> mark(f(g(f(a()))))
r2: active(f(X)) -> f(active(X))
r3: f(mark(X)) -> mark(f(X))
r4: proper(f(X)) -> f(proper(X))
r5: proper(a()) -> ok(a())
r6: proper(g(X)) -> g(proper(X))
r7: f(ok(X)) -> ok(f(X))
r8: g(ok(X)) -> ok(g(X))
r9: top(mark(X)) -> top(proper(X))
r10: top(ok(X)) -> top(active(X))

The set of usable rules consists of

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

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          top#_A(x1) = x1 + 8
          ok_A(x1) = max{13, x1 - 1}
          active_A(x1) = max{4, x1 - 10}
          mark_A(x1) = max{5, x1 + 3}
          proper_A(x1) = max{1, x1}
          f_A(x1) = max{2, x1}
          g_A(x1) = max{13, x1 - 24}
          a_A = 27
    
      precedence:
      
        active > g > top# = ok = f = a > mark > proper
      
      partial status:
    
        pi(top#) = []
        pi(ok) = []
        pi(active) = []
        pi(mark) = [1]
        pi(proper) = [1]
        pi(f) = []
        pi(g) = []
        pi(a) = []
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          top#_A(x1) = 0
          ok_A(x1) = 24
          active_A(x1) = 26
          mark_A(x1) = max{37, x1}
          proper_A(x1) = 33
          f_A(x1) = 38
          g_A(x1) = 25
          a_A = 23
    
      precedence:
      
        active = proper > f = g > ok > top# > mark > a
      
      partial status:
    
        pi(top#) = []
        pi(ok) = []
        pi(active) = []
        pi(mark) = []
        pi(proper) = []
        pi(f) = []
        pi(g) = []
        pi(a) = []
    

The next rules are strictly ordered:

  p1

We remove them from the problem.

-- SCC decomposition.

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

p1: top#(mark(X)) -> top#(proper(X))

and R consists of:

r1: active(f(f(a()))) -> mark(f(g(f(a()))))
r2: active(f(X)) -> f(active(X))
r3: f(mark(X)) -> mark(f(X))
r4: proper(f(X)) -> f(proper(X))
r5: proper(a()) -> ok(a())
r6: proper(g(X)) -> g(proper(X))
r7: f(ok(X)) -> ok(f(X))
r8: g(ok(X)) -> ok(g(X))
r9: top(mark(X)) -> top(proper(X))
r10: 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#(mark(X)) -> top#(proper(X))

and R consists of:

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

The set of usable rules consists of

  r3, r4, r5, r6, r7, r8

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          top#_A(x1) = max{8, x1 + 3}
          mark_A(x1) = x1 + 8
          proper_A(x1) = x1 + 5
          f_A(x1) = x1 + 2
          ok_A(x1) = x1 + 5
          g_A(x1) = x1 + 6
          a_A = 4
    
      precedence:
      
        proper > f > mark > a > top# = g > ok
      
      partial status:
    
        pi(top#) = [1]
        pi(mark) = [1]
        pi(proper) = [1]
        pi(f) = []
        pi(ok) = [1]
        pi(g) = [1]
        pi(a) = []
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          top#_A(x1) = max{3, x1 + 1}
          mark_A(x1) = max{2, x1 - 2}
          proper_A(x1) = max{11, x1 + 3}
          f_A(x1) = 3
          ok_A(x1) = max{22, x1 + 11}
          g_A(x1) = max{10, x1 - 2}
          a_A = 12
    
      precedence:
      
        mark = ok > top# = proper > f > g > a
      
      partial status:
    
        pi(top#) = []
        pi(mark) = []
        pi(proper) = []
        pi(f) = []
        pi(ok) = []
        pi(g) = []
        pi(a) = []
    

The next rules are strictly ordered:

  p1

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(X)) -> active#(X)

and R consists of:

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

The set of usable rules consists of

  (no rules)

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{4, x1 + 3}
          f_A(x1) = max{3, x1 + 2}
    
      precedence:
      
        active# = f
      
      partial status:
    
        pi(active#) = [1]
        pi(f) = [1]
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          active#_A(x1) = max{1, x1 - 1}
          f_A(x1) = x1
    
      precedence:
      
        active# = f
      
      partial status:
    
        pi(active#) = []
        pi(f) = [1]
    

The next rules are strictly ordered:

  p1

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

and R consists of:

r1: active(f(f(a()))) -> mark(f(g(f(a()))))
r2: active(f(X)) -> f(active(X))
r3: f(mark(X)) -> mark(f(X))
r4: proper(f(X)) -> f(proper(X))
r5: proper(a()) -> ok(a())
r6: proper(g(X)) -> g(proper(X))
r7: f(ok(X)) -> ok(f(X))
r8: g(ok(X)) -> ok(g(X))
r9: top(mark(X)) -> top(proper(X))
r10: top(ok(X)) -> top(active(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:
        
          proper#_A(x1) = max{6, x1 + 4}
          g_A(x1) = max{4, x1 + 3}
          f_A(x1) = max{2, x1 + 1}
    
      precedence:
      
        proper# = g = f
      
      partial status:
    
        pi(proper#) = [1]
        pi(g) = [1]
        pi(f) = [1]
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          proper#_A(x1) = x1 + 1
          g_A(x1) = x1
          f_A(x1) = x1
    
      precedence:
      
        proper# = g = f
      
      partial status:
    
        pi(proper#) = [1]
        pi(g) = [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: f#(mark(X)) -> f#(X)
p2: f#(ok(X)) -> f#(X)

and R consists of:

r1: active(f(f(a()))) -> mark(f(g(f(a()))))
r2: active(f(X)) -> f(active(X))
r3: f(mark(X)) -> mark(f(X))
r4: proper(f(X)) -> f(proper(X))
r5: proper(a()) -> ok(a())
r6: proper(g(X)) -> g(proper(X))
r7: f(ok(X)) -> ok(f(X))
r8: g(ok(X)) -> ok(g(X))
r9: top(mark(X)) -> top(proper(X))
r10: top(ok(X)) -> top(active(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 + 1
          mark_A(x1) = x1
          ok_A(x1) = x1
    
      precedence:
      
        f# = mark = ok
      
      partial status:
    
        pi(f#) = []
        pi(mark) = []
        pi(ok) = []
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          f#_A(x1) = x1 + 1
          mark_A(x1) = x1
          ok_A(x1) = x1
    
      precedence:
      
        f# = mark = ok
      
      partial status:
    
        pi(f#) = [1]
        pi(mark) = [1]
        pi(ok) = [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#(ok(X)) -> g#(X)

and R consists of:

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

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          g#_A(x1) = max{4, x1 + 3}
          ok_A(x1) = max{3, x1 + 2}
    
      precedence:
      
        g# = ok
      
      partial status:
    
        pi(g#) = [1]
        pi(ok) = [1]
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          g#_A(x1) = max{1, x1 - 1}
          ok_A(x1) = x1
    
      precedence:
      
        g# = ok
      
      partial status:
    
        pi(g#) = []
        pi(ok) = [1]
    

The next rules are strictly ordered:

  p1

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