YES

We show the termination of the TRS R:

  active(f(f(a()))) -> mark(c(f(g(f(a())))))
  active(f(X)) -> f(active(X))
  active(g(X)) -> g(active(X))
  f(mark(X)) -> mark(f(X))
  g(mark(X)) -> mark(g(X))
  proper(f(X)) -> f(proper(X))
  proper(a()) -> ok(a())
  proper(c(X)) -> c(proper(X))
  proper(g(X)) -> g(proper(X))
  f(ok(X)) -> ok(f(X))
  c(ok(X)) -> ok(c(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()))) -> c#(f(g(f(a()))))
p2: active#(f(f(a()))) -> f#(g(f(a())))
p3: active#(f(f(a()))) -> g#(f(a()))
p4: active#(f(X)) -> f#(active(X))
p5: active#(f(X)) -> active#(X)
p6: active#(g(X)) -> g#(active(X))
p7: active#(g(X)) -> active#(X)
p8: f#(mark(X)) -> f#(X)
p9: g#(mark(X)) -> g#(X)
p10: proper#(f(X)) -> f#(proper(X))
p11: proper#(f(X)) -> proper#(X)
p12: proper#(c(X)) -> c#(proper(X))
p13: proper#(c(X)) -> proper#(X)
p14: proper#(g(X)) -> g#(proper(X))
p15: proper#(g(X)) -> proper#(X)
p16: f#(ok(X)) -> f#(X)
p17: c#(ok(X)) -> c#(X)
p18: g#(ok(X)) -> g#(X)
p19: top#(mark(X)) -> top#(proper(X))
p20: top#(mark(X)) -> proper#(X)
p21: top#(ok(X)) -> top#(active(X))
p22: top#(ok(X)) -> active#(X)

and R consists of:

r1: active(f(f(a()))) -> mark(c(f(g(f(a())))))
r2: active(f(X)) -> f(active(X))
r3: active(g(X)) -> g(active(X))
r4: f(mark(X)) -> mark(f(X))
r5: g(mark(X)) -> mark(g(X))
r6: proper(f(X)) -> f(proper(X))
r7: proper(a()) -> ok(a())
r8: proper(c(X)) -> c(proper(X))
r9: proper(g(X)) -> g(proper(X))
r10: f(ok(X)) -> ok(f(X))
r11: c(ok(X)) -> ok(c(X))
r12: g(ok(X)) -> ok(g(X))
r13: top(mark(X)) -> top(proper(X))
r14: top(ok(X)) -> top(active(X))

The estimated dependency graph contains the following SCCs:

  {p19, p21}
  {p5, p7}
  {p11, p13, p15}
  {p8, p16}
  {p9, p18}
  {p17}


-- 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(c(f(g(f(a())))))
r2: active(f(X)) -> f(active(X))
r3: active(g(X)) -> g(active(X))
r4: f(mark(X)) -> mark(f(X))
r5: g(mark(X)) -> mark(g(X))
r6: proper(f(X)) -> f(proper(X))
r7: proper(a()) -> ok(a())
r8: proper(c(X)) -> c(proper(X))
r9: proper(g(X)) -> g(proper(X))
r10: f(ok(X)) -> ok(f(X))
r11: c(ok(X)) -> ok(c(X))
r12: g(ok(X)) -> ok(g(X))
r13: top(mark(X)) -> top(proper(X))
r14: top(ok(X)) -> top(active(X))

The set of usable rules consists of

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

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{2, x1 - 20}
          ok_A(x1) = max{21, x1}
          active_A(x1) = x1
          mark_A(x1) = max{40, x1 + 3}
          proper_A(x1) = max{21, x1}
          f_A(x1) = max{32, x1 + 11}
          g_A(x1) = max{43, x1 + 6}
          c_A(x1) = max{40, x1 - 21}
          a_A = 5
    
      precedence:
      
        top# > a > active > f > mark = proper = g = c > ok
      
      partial status:
    
        pi(top#) = []
        pi(ok) = []
        pi(active) = []
        pi(mark) = []
        pi(proper) = [1]
        pi(f) = []
        pi(g) = [1]
        pi(c) = []
        pi(a) = []
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          top#_A(x1) = 67
          ok_A(x1) = 9
          active_A(x1) = 5
          mark_A(x1) = 66
          proper_A(x1) = x1
          f_A(x1) = 8
          g_A(x1) = x1 + 6
          c_A(x1) = 17
          a_A = 3
    
      precedence:
      
        top# = ok = active > mark = proper = f = g = c = a
      
      partial status:
    
        pi(top#) = []
        pi(ok) = []
        pi(active) = []
        pi(mark) = []
        pi(proper) = [1]
        pi(f) = []
        pi(g) = [1]
        pi(c) = []
        pi(a) = []
    

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(f(a()))) -> mark(c(f(g(f(a())))))
r2: active(f(X)) -> f(active(X))
r3: active(g(X)) -> g(active(X))
r4: f(mark(X)) -> mark(f(X))
r5: g(mark(X)) -> mark(g(X))
r6: proper(f(X)) -> f(proper(X))
r7: proper(a()) -> ok(a())
r8: proper(c(X)) -> c(proper(X))
r9: proper(g(X)) -> g(proper(X))
r10: f(ok(X)) -> ok(f(X))
r11: c(ok(X)) -> ok(c(X))
r12: g(ok(X)) -> ok(g(X))
r13: top(mark(X)) -> top(proper(X))
r14: 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(f(a()))) -> mark(c(f(g(f(a())))))
r2: active(f(X)) -> f(active(X))
r3: active(g(X)) -> g(active(X))
r4: f(mark(X)) -> mark(f(X))
r5: g(mark(X)) -> mark(g(X))
r6: proper(f(X)) -> f(proper(X))
r7: proper(a()) -> ok(a())
r8: proper(c(X)) -> c(proper(X))
r9: proper(g(X)) -> g(proper(X))
r10: f(ok(X)) -> ok(f(X))
r11: c(ok(X)) -> ok(c(X))
r12: g(ok(X)) -> ok(g(X))
r13: top(mark(X)) -> top(proper(X))
r14: top(ok(X)) -> top(active(X))

The set of usable rules consists of

  r1, r2, r3, r4, r5, r10, r12

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{11, x1 + 10}
          ok_A(x1) = x1 + 11
          active_A(x1) = x1 + 8
          f_A(x1) = x1 + 10
          mark_A(x1) = max{35, x1 + 28}
          g_A(x1) = x1 + 9
          a_A = 14
          c_A(x1) = max{7, x1 - 29}
    
      precedence:
      
        top# = active = f = mark = g > ok = a = c
      
      partial status:
    
        pi(top#) = [1]
        pi(ok) = [1]
        pi(active) = [1]
        pi(f) = [1]
        pi(mark) = [1]
        pi(g) = [1]
        pi(a) = []
        pi(c) = []
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          top#_A(x1) = x1 + 25
          ok_A(x1) = x1 + 23
          active_A(x1) = x1
          f_A(x1) = max{21, x1 + 15}
          mark_A(x1) = max{37, x1 + 18}
          g_A(x1) = x1 + 13
          a_A = 5
          c_A(x1) = 4
    
      precedence:
      
        top# = ok = active = f = mark = g = a = c
      
      partial status:
    
        pi(top#) = []
        pi(ok) = [1]
        pi(active) = [1]
        pi(f) = [1]
        pi(mark) = [1]
        pi(g) = [1]
        pi(a) = []
        pi(c) = []
    

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

and R consists of:

r1: active(f(f(a()))) -> mark(c(f(g(f(a())))))
r2: active(f(X)) -> f(active(X))
r3: active(g(X)) -> g(active(X))
r4: f(mark(X)) -> mark(f(X))
r5: g(mark(X)) -> mark(g(X))
r6: proper(f(X)) -> f(proper(X))
r7: proper(a()) -> ok(a())
r8: proper(c(X)) -> c(proper(X))
r9: proper(g(X)) -> g(proper(X))
r10: f(ok(X)) -> ok(f(X))
r11: c(ok(X)) -> ok(c(X))
r12: g(ok(X)) -> ok(g(X))
r13: top(mark(X)) -> top(proper(X))
r14: 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:
        
          active#_A(x1) = x1 + 1
          g_A(x1) = x1
          f_A(x1) = x1
    
      precedence:
      
        active# = g = f
      
      partial status:
    
        pi(active#) = []
        pi(g) = []
        pi(f) = []
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          active#_A(x1) = x1 + 1
          g_A(x1) = x1
          f_A(x1) = x1
    
      precedence:
      
        active# = g = f
      
      partial status:
    
        pi(active#) = [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: proper#(g(X)) -> proper#(X)
p2: proper#(c(X)) -> proper#(X)
p3: proper#(f(X)) -> proper#(X)

and R consists of:

r1: active(f(f(a()))) -> mark(c(f(g(f(a())))))
r2: active(f(X)) -> f(active(X))
r3: active(g(X)) -> g(active(X))
r4: f(mark(X)) -> mark(f(X))
r5: g(mark(X)) -> mark(g(X))
r6: proper(f(X)) -> f(proper(X))
r7: proper(a()) -> ok(a())
r8: proper(c(X)) -> c(proper(X))
r9: proper(g(X)) -> g(proper(X))
r10: f(ok(X)) -> ok(f(X))
r11: c(ok(X)) -> ok(c(X))
r12: g(ok(X)) -> ok(g(X))
r13: top(mark(X)) -> top(proper(X))
r14: 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) = x1 + 1
          g_A(x1) = x1
          c_A(x1) = x1
          f_A(x1) = x1
    
      precedence:
      
        proper# = g = c = f
      
      partial status:
    
        pi(proper#) = []
        pi(g) = []
        pi(c) = []
        pi(f) = []
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          proper#_A(x1) = x1 + 1
          g_A(x1) = x1
          c_A(x1) = x1
          f_A(x1) = x1
    
      precedence:
      
        proper# = g = c = f
      
      partial status:
    
        pi(proper#) = [1]
        pi(g) = [1]
        pi(c) = [1]
        pi(f) = [1]
    

The next rules are strictly ordered:

  p1, p2, p3

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(c(f(g(f(a())))))
r2: active(f(X)) -> f(active(X))
r3: active(g(X)) -> g(active(X))
r4: f(mark(X)) -> mark(f(X))
r5: g(mark(X)) -> mark(g(X))
r6: proper(f(X)) -> f(proper(X))
r7: proper(a()) -> ok(a())
r8: proper(c(X)) -> c(proper(X))
r9: proper(g(X)) -> g(proper(X))
r10: f(ok(X)) -> ok(f(X))
r11: c(ok(X)) -> ok(c(X))
r12: g(ok(X)) -> ok(g(X))
r13: top(mark(X)) -> top(proper(X))
r14: 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#(mark(X)) -> g#(X)
p2: g#(ok(X)) -> g#(X)

and R consists of:

r1: active(f(f(a()))) -> mark(c(f(g(f(a())))))
r2: active(f(X)) -> f(active(X))
r3: active(g(X)) -> g(active(X))
r4: f(mark(X)) -> mark(f(X))
r5: g(mark(X)) -> mark(g(X))
r6: proper(f(X)) -> f(proper(X))
r7: proper(a()) -> ok(a())
r8: proper(c(X)) -> c(proper(X))
r9: proper(g(X)) -> g(proper(X))
r10: f(ok(X)) -> ok(f(X))
r11: c(ok(X)) -> ok(c(X))
r12: g(ok(X)) -> ok(g(X))
r13: top(mark(X)) -> top(proper(X))
r14: 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: c#(ok(X)) -> c#(X)

and R consists of:

r1: active(f(f(a()))) -> mark(c(f(g(f(a())))))
r2: active(f(X)) -> f(active(X))
r3: active(g(X)) -> g(active(X))
r4: f(mark(X)) -> mark(f(X))
r5: g(mark(X)) -> mark(g(X))
r6: proper(f(X)) -> f(proper(X))
r7: proper(a()) -> ok(a())
r8: proper(c(X)) -> c(proper(X))
r9: proper(g(X)) -> g(proper(X))
r10: f(ok(X)) -> ok(f(X))
r11: c(ok(X)) -> ok(c(X))
r12: g(ok(X)) -> ok(g(X))
r13: top(mark(X)) -> top(proper(X))
r14: 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:
        
          c#_A(x1) = max{4, x1 + 3}
          ok_A(x1) = max{3, x1 + 2}
    
      precedence:
      
        c# = ok
      
      partial status:
    
        pi(c#) = [1]
        pi(ok) = [1]
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          c#_A(x1) = max{1, x1 - 1}
          ok_A(x1) = x1
    
      precedence:
      
        c# = ok
      
      partial status:
    
        pi(c#) = []
        pi(ok) = [1]
    

The next rules are strictly ordered:

  p1

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