YES

We show the termination of the TRS R:

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

and R consists of:

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

The estimated dependency graph contains the following SCCs:

  {p9, p11}
  {p4, p6}
  {p7}
  {p8}


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

The set of usable rules consists of

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

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{0, x1 - 9}
          ok_A(x1) = max{6, x1 - 1}
          active_A(x1) = max{7, x1 - 2}
          mark_A(x1) = x1 + 19
          proper_A(x1) = max{18, x1 + 17}
          f_A(x1) = max{32, x1 + 16}
          g_A(x1) = 11
          c_A = 54
    
      precedence:
      
        active = c > mark = proper > g > f > ok > top#
      
      partial status:
    
        pi(top#) = []
        pi(ok) = []
        pi(active) = []
        pi(mark) = [1]
        pi(proper) = []
        pi(f) = []
        pi(g) = []
        pi(c) = []
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          top#_A(x1) = 0
          ok_A(x1) = 13
          active_A(x1) = 15
          mark_A(x1) = 15
          proper_A(x1) = 14
          f_A(x1) = 12
          g_A(x1) = 14
          c_A = 25
    
      precedence:
      
        c > g > ok = f > active > mark > top# = proper
      
      partial status:
    
        pi(top#) = []
        pi(ok) = []
        pi(active) = []
        pi(mark) = []
        pi(proper) = []
        pi(f) = []
        pi(g) = []
        pi(c) = []
    

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

The set of usable rules consists of

  r1, r2, r7

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
          ok_A(x1) = x1
          active_A(x1) = x1
          g_A(x1) = x1
          c_A = 1
          mark_A(x1) = x1
          f_A(x1) = x1
    
      precedence:
      
        top# = ok = active = g = c = mark = f
      
      partial status:
    
        pi(top#) = []
        pi(ok) = []
        pi(active) = []
        pi(g) = []
        pi(c) = []
        pi(mark) = []
        pi(f) = []
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          top#_A(x1) = x1 + 6
          ok_A(x1) = max{10, x1 + 6}
          active_A(x1) = max{5, x1 + 1}
          g_A(x1) = max{4, x1}
          c_A = 0
          mark_A(x1) = max{5, x1 + 1}
          f_A(x1) = max{2, x1}
    
      precedence:
      
        top# = ok = active = g = c = mark = f
      
      partial status:
    
        pi(top#) = []
        pi(ok) = [1]
        pi(active) = [1]
        pi(g) = [1]
        pi(c) = []
        pi(mark) = []
        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(c()) -> mark(f(g(c())))
r2: active(f(g(X))) -> mark(g(X))
r3: proper(c()) -> ok(c())
r4: proper(f(X)) -> f(proper(X))
r5: proper(g(X)) -> g(proper(X))
r6: f(ok(X)) -> ok(f(X))
r7: g(ok(X)) -> ok(g(X))
r8: top(mark(X)) -> top(proper(X))
r9: 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#(ok(X)) -> f#(X)

and R consists of:

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

and R consists of:

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