YES

We show the termination of the TRS R:

  active(f(f(a()))) -> mark(f(g(f(a()))))
  active(g(X)) -> g(active(X))
  g(mark(X)) -> mark(g(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#(g(X)) -> g#(active(X))
p4: active#(g(X)) -> active#(X)
p5: g#(mark(X)) -> g#(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(g(X)) -> g(active(X))
r3: g(mark(X)) -> mark(g(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, p11}
  {p10}


-- 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(g(X)) -> g(active(X))
r3: g(mark(X)) -> mark(g(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:
    
        matrix interpretations:
        
          carrier: N^2
          order: standard order
          interpretations:
            top#_A(x1) = ((0,1),(0,1)) x1 + (2,2)
            ok_A(x1) = x1 + (11,11)
            active_A(x1) = ((0,1),(0,1)) x1 + (12,10)
            mark_A(x1) = ((0,1),(0,1)) x1 + (1,14)
            proper_A(x1) = x1 + (13,13)
            g_A(x1) = ((0,1),(0,1)) x1 + (2,12)
            f_A(x1) = ((0,1),(1,0)) x1 + (0,5)
            a_A() = (0,11)
    
      precedence:
      
        active > proper > g > mark > a > top# > f > ok
      
      partial status:
    
        pi(top#) = []
        pi(ok) = []
        pi(active) = []
        pi(mark) = []
        pi(proper) = []
        pi(g) = []
        pi(f) = []
        pi(a) = []
    
    2. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: standard order
          interpretations:
            top#_A(x1) = (4,3)
            ok_A(x1) = (2,2)
            active_A(x1) = (3,3)
            mark_A(x1) = (1,1)
            proper_A(x1) = (3,4)
            g_A(x1) = (2,2)
            f_A(x1) = (2,2)
            a_A() = (2,2)
    
      precedence:
      
        a > active > ok = proper = g = f > top# = mark
      
      partial status:
    
        pi(top#) = []
        pi(ok) = []
        pi(active) = []
        pi(mark) = []
        pi(proper) = []
        pi(g) = []
        pi(f) = []
        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(g(X)) -> g(active(X))
r3: g(mark(X)) -> mark(g(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(g(X)) -> g(active(X))
r3: g(mark(X)) -> mark(g(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{10, x1 + 3}
          mark_A(x1) = max{9, x1 + 8}
          proper_A(x1) = max{4, x1 + 3}
          g_A(x1) = max{7, x1 + 6}
          f_A(x1) = max{1, x1}
          ok_A(x1) = max{6, x1 + 2}
          a_A = 5
    
      precedence:
      
        top# = mark = proper = g > f = ok = a
      
      partial status:
    
        pi(top#) = [1]
        pi(mark) = [1]
        pi(proper) = [1]
        pi(g) = [1]
        pi(f) = [1]
        pi(ok) = [1]
        pi(a) = []
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          top#_A(x1) = max{3, x1 - 3}
          mark_A(x1) = x1 + 2
          proper_A(x1) = max{11, x1 + 6}
          g_A(x1) = max{5, x1 + 3}
          f_A(x1) = max{16, x1 + 12}
          ok_A(x1) = x1 + 20
          a_A = 12
    
      precedence:
      
        top# = proper > mark > g > f = ok = a
      
      partial status:
    
        pi(top#) = []
        pi(mark) = [1]
        pi(proper) = [1]
        pi(g) = []
        pi(f) = [1]
        pi(ok) = [1]
        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#(g(X)) -> active#(X)

and R consists of:

r1: active(f(f(a()))) -> mark(f(g(f(a()))))
r2: active(g(X)) -> g(active(X))
r3: g(mark(X)) -> mark(g(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}
          g_A(x1) = max{3, x1 + 2}
    
      precedence:
      
        active# = g
      
      partial status:
    
        pi(active#) = [1]
        pi(g) = [1]
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          active#_A(x1) = max{1, x1 - 1}
          g_A(x1) = x1
    
      precedence:
      
        active# = g
      
      partial status:
    
        pi(active#) = []
        pi(g) = [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(g(X)) -> g(active(X))
r3: g(mark(X)) -> mark(g(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: g#(mark(X)) -> g#(X)
p2: g#(ok(X)) -> g#(X)

and R consists of:

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

and R consists of:

r1: active(f(f(a()))) -> mark(f(g(f(a()))))
r2: active(g(X)) -> g(active(X))
r3: g(mark(X)) -> mark(g(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:
        
          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.