YES

We show the termination of the TRS R:

  active(f(f(X))) -> mark(c(f(g(f(X)))))
  active(c(X)) -> mark(d(X))
  active(h(X)) -> mark(c(d(X)))
  mark(f(X)) -> active(f(mark(X)))
  mark(c(X)) -> active(c(X))
  mark(g(X)) -> active(g(X))
  mark(d(X)) -> active(d(X))
  mark(h(X)) -> active(h(mark(X)))
  f(mark(X)) -> f(X)
  f(active(X)) -> f(X)
  c(mark(X)) -> c(X)
  c(active(X)) -> c(X)
  g(mark(X)) -> g(X)
  g(active(X)) -> g(X)
  d(mark(X)) -> d(X)
  d(active(X)) -> d(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(f(X))) -> mark#(c(f(g(f(X)))))
p2: active#(f(f(X))) -> c#(f(g(f(X))))
p3: active#(f(f(X))) -> f#(g(f(X)))
p4: active#(f(f(X))) -> g#(f(X))
p5: active#(c(X)) -> mark#(d(X))
p6: active#(c(X)) -> d#(X)
p7: active#(h(X)) -> mark#(c(d(X)))
p8: active#(h(X)) -> c#(d(X))
p9: active#(h(X)) -> d#(X)
p10: mark#(f(X)) -> active#(f(mark(X)))
p11: mark#(f(X)) -> f#(mark(X))
p12: mark#(f(X)) -> mark#(X)
p13: mark#(c(X)) -> active#(c(X))
p14: mark#(g(X)) -> active#(g(X))
p15: mark#(d(X)) -> active#(d(X))
p16: mark#(h(X)) -> active#(h(mark(X)))
p17: mark#(h(X)) -> h#(mark(X))
p18: mark#(h(X)) -> mark#(X)
p19: f#(mark(X)) -> f#(X)
p20: f#(active(X)) -> f#(X)
p21: c#(mark(X)) -> c#(X)
p22: c#(active(X)) -> c#(X)
p23: g#(mark(X)) -> g#(X)
p24: g#(active(X)) -> g#(X)
p25: d#(mark(X)) -> d#(X)
p26: d#(active(X)) -> d#(X)
p27: h#(mark(X)) -> h#(X)
p28: h#(active(X)) -> h#(X)

and R consists of:

r1: active(f(f(X))) -> mark(c(f(g(f(X)))))
r2: active(c(X)) -> mark(d(X))
r3: active(h(X)) -> mark(c(d(X)))
r4: mark(f(X)) -> active(f(mark(X)))
r5: mark(c(X)) -> active(c(X))
r6: mark(g(X)) -> active(g(X))
r7: mark(d(X)) -> active(d(X))
r8: mark(h(X)) -> active(h(mark(X)))
r9: f(mark(X)) -> f(X)
r10: f(active(X)) -> f(X)
r11: c(mark(X)) -> c(X)
r12: c(active(X)) -> c(X)
r13: g(mark(X)) -> g(X)
r14: g(active(X)) -> g(X)
r15: d(mark(X)) -> d(X)
r16: d(active(X)) -> d(X)
r17: h(mark(X)) -> h(X)
r18: h(active(X)) -> h(X)

The estimated dependency graph contains the following SCCs:

  {p1, p5, p7, p10, p12, p13, p14, p15, p16, p18}
  {p21, p22}
  {p19, p20}
  {p23, p24}
  {p25, p26}
  {p27, p28}


-- Reduction pair.

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

p1: active#(f(f(X))) -> mark#(c(f(g(f(X)))))
p2: mark#(h(X)) -> mark#(X)
p3: mark#(h(X)) -> active#(h(mark(X)))
p4: active#(h(X)) -> mark#(c(d(X)))
p5: mark#(d(X)) -> active#(d(X))
p6: active#(c(X)) -> mark#(d(X))
p7: mark#(g(X)) -> active#(g(X))
p8: mark#(c(X)) -> active#(c(X))
p9: mark#(f(X)) -> mark#(X)
p10: mark#(f(X)) -> active#(f(mark(X)))

and R consists of:

r1: active(f(f(X))) -> mark(c(f(g(f(X)))))
r2: active(c(X)) -> mark(d(X))
r3: active(h(X)) -> mark(c(d(X)))
r4: mark(f(X)) -> active(f(mark(X)))
r5: mark(c(X)) -> active(c(X))
r6: mark(g(X)) -> active(g(X))
r7: mark(d(X)) -> active(d(X))
r8: mark(h(X)) -> active(h(mark(X)))
r9: f(mark(X)) -> f(X)
r10: f(active(X)) -> f(X)
r11: c(mark(X)) -> c(X)
r12: c(active(X)) -> c(X)
r13: g(mark(X)) -> g(X)
r14: g(active(X)) -> g(X)
r15: d(mark(X)) -> d(X)
r16: d(active(X)) -> d(X)
r17: h(mark(X)) -> h(X)
r18: h(active(X)) -> h(X)

The set of usable rules consists of

  r1, r2, r3, r4, r5, r6, r7, r8, r9, r10, r11, r12, r13, r14, r15, r16, r17, r18

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. matrix interpretations:
    
      carrier: N^1
      order: standard order
      interpretations:
        active#_A(x1) = x1
        f_A(x1) = x1 + 4
        mark#_A(x1) = x1 + 2
        c_A(x1) = 4
        g_A(x1) = 1
        h_A(x1) = x1 + 7
        mark_A(x1) = x1 + 1
        d_A(x1) = 1
        active_A(x1) = x1
    
    2. lexicographic path order with precedence:
    
      precedence:
      
        mark > f > active# > mark# > c > d > active > g > h
      
      argument filter:
    
        pi(active#) = [1]
        pi(f) = [1]
        pi(mark#) = []
        pi(c) = []
        pi(g) = []
        pi(h) = []
        pi(mark) = 1
        pi(d) = []
        pi(active) = 1
    

The next rules are strictly ordered:

  p1, p2, p3, p4, p5, p6, p7, p8, p9, p10

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

and R consists of:

r1: active(f(f(X))) -> mark(c(f(g(f(X)))))
r2: active(c(X)) -> mark(d(X))
r3: active(h(X)) -> mark(c(d(X)))
r4: mark(f(X)) -> active(f(mark(X)))
r5: mark(c(X)) -> active(c(X))
r6: mark(g(X)) -> active(g(X))
r7: mark(d(X)) -> active(d(X))
r8: mark(h(X)) -> active(h(mark(X)))
r9: f(mark(X)) -> f(X)
r10: f(active(X)) -> f(X)
r11: c(mark(X)) -> c(X)
r12: c(active(X)) -> c(X)
r13: g(mark(X)) -> g(X)
r14: g(active(X)) -> g(X)
r15: d(mark(X)) -> d(X)
r16: d(active(X)) -> d(X)
r17: h(mark(X)) -> h(X)
r18: h(active(X)) -> h(X)

The set of usable rules consists of

  (no rules)

Take the monotone reduction pair:

  lexicographic combination of reduction pairs:
  
    1. matrix interpretations:
    
      carrier: N^1
      order: standard order
      interpretations:
        c#_A(x1) = x1
        mark_A(x1) = x1 + 1
        active_A(x1) = x1 + 1
    
    2. lexicographic path order with precedence:
    
      precedence:
      
        c# > active > mark
      
      argument filter:
    
        pi(c#) = 1
        pi(mark) = [1]
        pi(active) = [1]
    

The next rules are strictly ordered:

  p1, p2
  r1, r2, r3, r4, r5, r6, r7, r8, r9, r10, r11, r12, r13, r14, r15, r16, r17, r18

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(f(X))) -> mark(c(f(g(f(X)))))
r2: active(c(X)) -> mark(d(X))
r3: active(h(X)) -> mark(c(d(X)))
r4: mark(f(X)) -> active(f(mark(X)))
r5: mark(c(X)) -> active(c(X))
r6: mark(g(X)) -> active(g(X))
r7: mark(d(X)) -> active(d(X))
r8: mark(h(X)) -> active(h(mark(X)))
r9: f(mark(X)) -> f(X)
r10: f(active(X)) -> f(X)
r11: c(mark(X)) -> c(X)
r12: c(active(X)) -> c(X)
r13: g(mark(X)) -> g(X)
r14: g(active(X)) -> g(X)
r15: d(mark(X)) -> d(X)
r16: d(active(X)) -> d(X)
r17: h(mark(X)) -> h(X)
r18: h(active(X)) -> h(X)

The set of usable rules consists of

  (no rules)

Take the monotone reduction pair:

  lexicographic combination of reduction pairs:
  
    1. matrix interpretations:
    
      carrier: N^1
      order: standard order
      interpretations:
        f#_A(x1) = x1
        mark_A(x1) = x1 + 1
        active_A(x1) = x1 + 1
    
    2. lexicographic path order with precedence:
    
      precedence:
      
        f# > active > mark
      
      argument filter:
    
        pi(f#) = 1
        pi(mark) = [1]
        pi(active) = [1]
    

The next rules are strictly ordered:

  p1, p2
  r1, r2, r3, r4, r5, r6, r7, r8, r9, r10, r11, r12, r13, r14, r15, r16, r17, r18

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(f(X))) -> mark(c(f(g(f(X)))))
r2: active(c(X)) -> mark(d(X))
r3: active(h(X)) -> mark(c(d(X)))
r4: mark(f(X)) -> active(f(mark(X)))
r5: mark(c(X)) -> active(c(X))
r6: mark(g(X)) -> active(g(X))
r7: mark(d(X)) -> active(d(X))
r8: mark(h(X)) -> active(h(mark(X)))
r9: f(mark(X)) -> f(X)
r10: f(active(X)) -> f(X)
r11: c(mark(X)) -> c(X)
r12: c(active(X)) -> c(X)
r13: g(mark(X)) -> g(X)
r14: g(active(X)) -> g(X)
r15: d(mark(X)) -> d(X)
r16: d(active(X)) -> d(X)
r17: h(mark(X)) -> h(X)
r18: h(active(X)) -> h(X)

The set of usable rules consists of

  (no rules)

Take the monotone reduction pair:

  lexicographic combination of reduction pairs:
  
    1. matrix interpretations:
    
      carrier: N^1
      order: standard order
      interpretations:
        g#_A(x1) = x1
        mark_A(x1) = x1 + 1
        active_A(x1) = x1 + 1
    
    2. lexicographic path order with precedence:
    
      precedence:
      
        g# > active > mark
      
      argument filter:
    
        pi(g#) = 1
        pi(mark) = [1]
        pi(active) = [1]
    

The next rules are strictly ordered:

  p1, p2
  r1, r2, r3, r4, r5, r6, r7, r8, r9, r10, r11, r12, r13, r14, r15, r16, r17, r18

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

and R consists of:

r1: active(f(f(X))) -> mark(c(f(g(f(X)))))
r2: active(c(X)) -> mark(d(X))
r3: active(h(X)) -> mark(c(d(X)))
r4: mark(f(X)) -> active(f(mark(X)))
r5: mark(c(X)) -> active(c(X))
r6: mark(g(X)) -> active(g(X))
r7: mark(d(X)) -> active(d(X))
r8: mark(h(X)) -> active(h(mark(X)))
r9: f(mark(X)) -> f(X)
r10: f(active(X)) -> f(X)
r11: c(mark(X)) -> c(X)
r12: c(active(X)) -> c(X)
r13: g(mark(X)) -> g(X)
r14: g(active(X)) -> g(X)
r15: d(mark(X)) -> d(X)
r16: d(active(X)) -> d(X)
r17: h(mark(X)) -> h(X)
r18: h(active(X)) -> h(X)

The set of usable rules consists of

  (no rules)

Take the monotone reduction pair:

  lexicographic combination of reduction pairs:
  
    1. matrix interpretations:
    
      carrier: N^1
      order: standard order
      interpretations:
        d#_A(x1) = x1
        mark_A(x1) = x1 + 1
        active_A(x1) = x1 + 1
    
    2. lexicographic path order with precedence:
    
      precedence:
      
        d# > active > mark
      
      argument filter:
    
        pi(d#) = 1
        pi(mark) = [1]
        pi(active) = [1]
    

The next rules are strictly ordered:

  p1, p2
  r1, r2, r3, r4, r5, r6, r7, r8, r9, r10, r11, r12, r13, r14, r15, r16, r17, r18

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(f(X))) -> mark(c(f(g(f(X)))))
r2: active(c(X)) -> mark(d(X))
r3: active(h(X)) -> mark(c(d(X)))
r4: mark(f(X)) -> active(f(mark(X)))
r5: mark(c(X)) -> active(c(X))
r6: mark(g(X)) -> active(g(X))
r7: mark(d(X)) -> active(d(X))
r8: mark(h(X)) -> active(h(mark(X)))
r9: f(mark(X)) -> f(X)
r10: f(active(X)) -> f(X)
r11: c(mark(X)) -> c(X)
r12: c(active(X)) -> c(X)
r13: g(mark(X)) -> g(X)
r14: g(active(X)) -> g(X)
r15: d(mark(X)) -> d(X)
r16: d(active(X)) -> d(X)
r17: h(mark(X)) -> h(X)
r18: h(active(X)) -> h(X)

The set of usable rules consists of

  (no rules)

Take the monotone reduction pair:

  lexicographic combination of reduction pairs:
  
    1. matrix interpretations:
    
      carrier: N^1
      order: standard order
      interpretations:
        h#_A(x1) = x1
        mark_A(x1) = x1 + 1
        active_A(x1) = x1 + 1
    
    2. lexicographic path order with precedence:
    
      precedence:
      
        h# > active > mark
      
      argument filter:
    
        pi(h#) = 1
        pi(mark) = [1]
        pi(active) = [1]
    

The next rules are strictly ordered:

  p1, p2
  r1, r2, r3, r4, r5, r6, r7, r8, r9, r10, r11, r12, r13, r14, r15, r16, r17, r18

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