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 combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            top#_A(x1) = ((1,0),(0,0)) x1 + (0,2)
            ok_A(x1) = ((1,0),(0,0)) x1 + (26,3)
            active_A(x1) = ((1,0),(1,1)) x1 + (26,1)
            mark_A(x1) = ((1,0),(0,0)) x1 + (8,3)
            proper_A(x1) = ((1,0),(1,0)) x1 + (7,0)
            f_A(x1) = x1 + (5,4)
            h_A(x1) = ((1,0),(0,0)) x1 + (6,5)
            g_A(x1) = ((1,0),(1,0)) x1 + (11,4)
    
      precedence:
      
        active > h > mark > proper > ok = g > f > top#
      
      partial status:
    
        pi(top#) = []
        pi(ok) = []
        pi(active) = [1]
        pi(mark) = []
        pi(proper) = []
        pi(f) = [1]
        pi(h) = []
        pi(g) = []
    
    2. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            top#_A(x1) = (0,0)
            ok_A(x1) = (0,0)
            active_A(x1) = (1,2)
            mark_A(x1) = (0,0)
            proper_A(x1) = (0,4)
            f_A(x1) = x1
            h_A(x1) = (0,1)
            g_A(x1) = (0,3)
    
      precedence:
      
        top# = ok = proper = g > active = f = h > mark
      
      partial status:
    
        pi(top#) = []
        pi(ok) = []
        pi(active) = []
        pi(mark) = []
        pi(proper) = []
        pi(f) = []
        pi(h) = []
        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 reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            top#_A(x1) = ((1,0),(0,0)) x1 + (1,5)
            ok_A(x1) = ((1,0),(0,0)) x1 + (3,4)
            active_A(x1) = ((1,0),(1,1)) x1 + (1,3)
            f_A(x1) = ((1,0),(0,0)) x1 + (4,5)
            mark_A(x1) = (2,1)
            h_A(x1) = x1 + (1,2)
            g_A(x1) = ((1,0),(0,0)) x1 + (1,4)
    
      precedence:
      
        active > mark > top# > f = h > ok = g
      
      partial status:
    
        pi(top#) = []
        pi(ok) = []
        pi(active) = [1]
        pi(f) = []
        pi(mark) = []
        pi(h) = [1]
        pi(g) = []
    
    2. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            top#_A(x1) = (1,1)
            ok_A(x1) = (3,0)
            active_A(x1) = ((1,0),(1,1)) x1 + (2,0)
            f_A(x1) = (5,2)
            mark_A(x1) = (0,0)
            h_A(x1) = (5,1)
            g_A(x1) = (4,0)
    
      precedence:
      
        g > active = f = h > top# > ok = mark
      
      partial status:
    
        pi(top#) = []
        pi(ok) = []
        pi(active) = [1]
        pi(f) = []
        pi(mark) = []
        pi(h) = []
        pi(g) = []
    

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#(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 reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            active#_A(x1) = x1 + (1,2)
            h_A(x1) = x1 + (2,1)
            f_A(x1) = ((1,0),(1,1)) x1 + (2,1)
    
      precedence:
      
        active# = h > f
      
      partial status:
    
        pi(active#) = [1]
        pi(h) = [1]
        pi(f) = [1]
    
    2. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            active#_A(x1) = ((1,0),(1,0)) x1 + (0,2)
            h_A(x1) = ((1,0),(0,0)) x1 + (1,1)
            f_A(x1) = ((1,0),(0,0)) x1 + (1,1)
    
      precedence:
      
        active# = h = f
      
      partial status:
    
        pi(active#) = []
        pi(h) = []
        pi(f) = []
    

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#(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 reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            proper#_A(x1) = x1
            h_A(x1) = ((1,0),(0,0)) x1 + (1,1)
            g_A(x1) = x1 + (1,1)
            f_A(x1) = ((1,0),(0,0)) x1 + (1,1)
    
      precedence:
      
        g > proper# > h = f
      
      partial status:
    
        pi(proper#) = [1]
        pi(h) = []
        pi(g) = [1]
        pi(f) = []
    
    2. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            proper#_A(x1) = ((0,0),(1,0)) x1
            h_A(x1) = (1,1)
            g_A(x1) = ((1,0),(0,0)) x1 + (1,1)
            f_A(x1) = (1,1)
    
      precedence:
      
        proper# = h = g = f
      
      partial status:
    
        pi(proper#) = []
        pi(h) = []
        pi(g) = []
        pi(f) = []
    

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: 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 reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            g#_A(x1) = ((1,0),(1,1)) x1 + (2,2)
            ok_A(x1) = ((1,0),(0,0)) x1 + (1,1)
    
      precedence:
      
        g# = ok
      
      partial status:
    
        pi(g#) = []
        pi(ok) = []
    
    2. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            g#_A(x1) = ((1,0),(0,0)) x1
            ok_A(x1) = (1,1)
    
      precedence:
      
        ok > g#
      
      partial status:
    
        pi(g#) = []
        pi(ok) = []
    

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: 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 reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            h#_A(x1) = x1 + (1,2)
            mark_A(x1) = x1 + (2,1)
            ok_A(x1) = ((1,0),(1,1)) x1 + (2,1)
    
      precedence:
      
        h# = mark = ok
      
      partial status:
    
        pi(h#) = [1]
        pi(mark) = [1]
        pi(ok) = [1]
    
    2. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            h#_A(x1) = ((1,0),(1,0)) x1 + (0,2)
            mark_A(x1) = ((1,0),(0,0)) x1 + (1,1)
            ok_A(x1) = ((1,0),(0,0)) x1 + (1,1)
    
      precedence:
      
        h# = mark = ok
      
      partial status:
    
        pi(h#) = []
        pi(mark) = []
        pi(ok) = []
    

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(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 reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            f#_A(x1) = x1 + (1,2)
            mark_A(x1) = x1 + (2,1)
            ok_A(x1) = ((1,0),(1,1)) x1 + (2,1)
    
      precedence:
      
        f# = mark = ok
      
      partial status:
    
        pi(f#) = [1]
        pi(mark) = [1]
        pi(ok) = [1]
    
    2. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            f#_A(x1) = ((1,0),(1,0)) x1 + (0,2)
            mark_A(x1) = ((1,0),(0,0)) x1 + (1,1)
            ok_A(x1) = ((1,0),(0,0)) x1 + (1,1)
    
      precedence:
      
        f# = mark = ok
      
      partial status:
    
        pi(f#) = []
        pi(mark) = []
        pi(ok) = []
    

The next rules are strictly ordered:

  p1, p2

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