YES

We show the termination of the TRS R:

  active(f(X)) -> mark(g(h(f(X))))
  mark(f(X)) -> active(f(mark(X)))
  mark(g(X)) -> active(g(X))
  mark(h(X)) -> active(h(mark(X)))
  f(mark(X)) -> f(X)
  f(active(X)) -> f(X)
  g(mark(X)) -> g(X)
  g(active(X)) -> g(X)
  h(mark(X)) -> h(X)
  h(active(X)) -> h(X)

-- SCC decomposition.

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

p1: active#(f(X)) -> mark#(g(h(f(X))))
p2: active#(f(X)) -> g#(h(f(X)))
p3: active#(f(X)) -> h#(f(X))
p4: mark#(f(X)) -> active#(f(mark(X)))
p5: mark#(f(X)) -> f#(mark(X))
p6: mark#(f(X)) -> mark#(X)
p7: mark#(g(X)) -> active#(g(X))
p8: mark#(h(X)) -> active#(h(mark(X)))
p9: mark#(h(X)) -> h#(mark(X))
p10: mark#(h(X)) -> mark#(X)
p11: f#(mark(X)) -> f#(X)
p12: f#(active(X)) -> f#(X)
p13: g#(mark(X)) -> g#(X)
p14: g#(active(X)) -> g#(X)
p15: h#(mark(X)) -> h#(X)
p16: h#(active(X)) -> h#(X)

and R consists of:

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

The estimated dependency graph contains the following SCCs:

  {p1, p4, p6, p7, p8, p10}
  {p13, p14}
  {p15, p16}
  {p11, p12}


-- Reduction pair.

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

p1: active#(f(X)) -> mark#(g(h(f(X))))
p2: mark#(h(X)) -> mark#(X)
p3: mark#(h(X)) -> active#(h(mark(X)))
p4: mark#(g(X)) -> active#(g(X))
p5: mark#(f(X)) -> mark#(X)
p6: mark#(f(X)) -> active#(f(mark(X)))

and R consists of:

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

The set of usable rules consists of

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

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

The next rules are strictly ordered:

  p5

We remove them from the problem.

-- SCC decomposition.

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

p1: active#(f(X)) -> mark#(g(h(f(X))))
p2: mark#(h(X)) -> mark#(X)
p3: mark#(h(X)) -> active#(h(mark(X)))
p4: mark#(g(X)) -> active#(g(X))
p5: mark#(f(X)) -> active#(f(mark(X)))

and R consists of:

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

The estimated dependency graph contains the following SCCs:

  {p1, p2, p3, p4, p5}


-- Reduction pair.

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

p1: active#(f(X)) -> mark#(g(h(f(X))))
p2: mark#(f(X)) -> active#(f(mark(X)))
p3: mark#(g(X)) -> active#(g(X))
p4: mark#(h(X)) -> active#(h(mark(X)))
p5: mark#(h(X)) -> mark#(X)

and R consists of:

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

The set of usable rules consists of

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

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) = ((1,0),(1,1)) x1 + (1,1)
            f_A(x1) = ((1,0),(1,1)) x1 + (6,24)
            mark#_A(x1) = ((1,0),(1,1)) x1 + (4,9)
            g_A(x1) = ((0,0),(1,0)) x1 + (1,2)
            h_A(x1) = x1 + (5,8)
            mark_A(x1) = ((1,0),(1,1)) x1 + (2,7)
            active_A(x1) = x1 + (0,1)
    
      precedence:
      
        active# = f = mark# = g = h = mark = active
      
      partial status:
    
        pi(active#) = []
        pi(f) = []
        pi(mark#) = []
        pi(g) = []
        pi(h) = []
        pi(mark) = []
        pi(active) = []
    
    2. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            active#_A(x1) = ((1,0),(1,1)) x1 + (12,15)
            f_A(x1) = ((1,0),(1,1)) x1 + (1,14)
            mark#_A(x1) = ((0,0),(1,0)) x1 + (10,1)
            g_A(x1) = (9,9)
            h_A(x1) = ((1,0),(1,1)) x1 + (11,14)
            mark_A(x1) = ((1,0),(1,1)) x1 + (0,13)
            active_A(x1) = ((1,0),(1,1)) x1 + (12,15)
    
      precedence:
      
        active# > f = mark# = h = mark = active > g
      
      partial status:
    
        pi(active#) = []
        pi(f) = [1]
        pi(mark#) = []
        pi(g) = []
        pi(h) = []
        pi(mark) = [1]
        pi(active) = [1]
    

The next rules are strictly ordered:

  p1, p2, p3, p4, p5

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#(active(X)) -> g#(X)

and R consists of:

r1: active(f(X)) -> mark(g(h(f(X))))
r2: mark(f(X)) -> active(f(mark(X)))
r3: mark(g(X)) -> active(g(X))
r4: mark(h(X)) -> active(h(mark(X)))
r5: f(mark(X)) -> f(X)
r6: f(active(X)) -> f(X)
r7: g(mark(X)) -> g(X)
r8: g(active(X)) -> g(X)
r9: h(mark(X)) -> h(X)
r10: h(active(X)) -> h(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) = x1 + (1,2)
            mark_A(x1) = x1 + (2,1)
            active_A(x1) = ((1,0),(1,1)) x1 + (2,1)
    
      precedence:
      
        g# = mark = active
      
      partial status:
    
        pi(g#) = [1]
        pi(mark) = [1]
        pi(active) = [1]
    
    2. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            g#_A(x1) = ((1,0),(1,0)) x1 + (0,2)
            mark_A(x1) = ((1,0),(0,0)) x1 + (1,1)
            active_A(x1) = ((1,0),(0,0)) x1 + (1,1)
    
      precedence:
      
        g# = mark = active
      
      partial status:
    
        pi(g#) = []
        pi(mark) = []
        pi(active) = []
    

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: h#(mark(X)) -> h#(X)
p2: h#(active(X)) -> h#(X)

and R consists of:

r1: active(f(X)) -> mark(g(h(f(X))))
r2: mark(f(X)) -> active(f(mark(X)))
r3: mark(g(X)) -> active(g(X))
r4: mark(h(X)) -> active(h(mark(X)))
r5: f(mark(X)) -> f(X)
r6: f(active(X)) -> f(X)
r7: g(mark(X)) -> g(X)
r8: g(active(X)) -> g(X)
r9: h(mark(X)) -> h(X)
r10: h(active(X)) -> h(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)
            active_A(x1) = ((1,0),(1,1)) x1 + (2,1)
    
      precedence:
      
        h# = mark = active
      
      partial status:
    
        pi(h#) = [1]
        pi(mark) = [1]
        pi(active) = [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)
            active_A(x1) = ((1,0),(0,0)) x1 + (1,1)
    
      precedence:
      
        h# = mark = active
      
      partial status:
    
        pi(h#) = []
        pi(mark) = []
        pi(active) = []
    

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

and R consists of:

r1: active(f(X)) -> mark(g(h(f(X))))
r2: mark(f(X)) -> active(f(mark(X)))
r3: mark(g(X)) -> active(g(X))
r4: mark(h(X)) -> active(h(mark(X)))
r5: f(mark(X)) -> f(X)
r6: f(active(X)) -> f(X)
r7: g(mark(X)) -> g(X)
r8: g(active(X)) -> g(X)
r9: h(mark(X)) -> h(X)
r10: h(active(X)) -> h(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)
            active_A(x1) = ((1,0),(1,1)) x1 + (2,1)
    
      precedence:
      
        f# = mark = active
      
      partial status:
    
        pi(f#) = [1]
        pi(mark) = [1]
        pi(active) = [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)
            active_A(x1) = ((1,0),(0,0)) x1 + (1,1)
    
      precedence:
      
        f# = mark = active
      
      partial status:
    
        pi(f#) = []
        pi(mark) = []
        pi(active) = []
    

The next rules are strictly ordered:

  p1, p2

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