YES

We show the termination of the TRS R:

  active(f(f(a()))) -> mark(c(f(g(f(a())))))
  mark(f(X)) -> active(f(mark(X)))
  mark(a()) -> active(a())
  mark(c(X)) -> active(c(X))
  mark(g(X)) -> active(g(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)

-- SCC decomposition.

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

p1: active#(f(f(a()))) -> mark#(c(f(g(f(a())))))
p2: active#(f(f(a()))) -> c#(f(g(f(a()))))
p3: active#(f(f(a()))) -> f#(g(f(a())))
p4: active#(f(f(a()))) -> g#(f(a()))
p5: mark#(f(X)) -> active#(f(mark(X)))
p6: mark#(f(X)) -> f#(mark(X))
p7: mark#(f(X)) -> mark#(X)
p8: mark#(a()) -> active#(a())
p9: mark#(c(X)) -> active#(c(X))
p10: mark#(g(X)) -> active#(g(mark(X)))
p11: mark#(g(X)) -> g#(mark(X))
p12: mark#(g(X)) -> mark#(X)
p13: f#(mark(X)) -> f#(X)
p14: f#(active(X)) -> f#(X)
p15: c#(mark(X)) -> c#(X)
p16: c#(active(X)) -> c#(X)
p17: g#(mark(X)) -> g#(X)
p18: g#(active(X)) -> g#(X)

and R consists of:

r1: active(f(f(a()))) -> mark(c(f(g(f(a())))))
r2: mark(f(X)) -> active(f(mark(X)))
r3: mark(a()) -> active(a())
r4: mark(c(X)) -> active(c(X))
r5: mark(g(X)) -> active(g(mark(X)))
r6: f(mark(X)) -> f(X)
r7: f(active(X)) -> f(X)
r8: c(mark(X)) -> c(X)
r9: c(active(X)) -> c(X)
r10: g(mark(X)) -> g(X)
r11: g(active(X)) -> g(X)

The estimated dependency graph contains the following SCCs:

  {p7, p12}
  {p1, p9}
  {p13, p14}
  {p17, p18}
  {p15, p16}


-- Reduction pair.

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

p1: mark#(g(X)) -> mark#(X)
p2: mark#(f(X)) -> mark#(X)

and R consists of:

r1: active(f(f(a()))) -> mark(c(f(g(f(a())))))
r2: mark(f(X)) -> active(f(mark(X)))
r3: mark(a()) -> active(a())
r4: mark(c(X)) -> active(c(X))
r5: mark(g(X)) -> active(g(mark(X)))
r6: f(mark(X)) -> f(X)
r7: f(active(X)) -> f(X)
r8: c(mark(X)) -> c(X)
r9: c(active(X)) -> c(X)
r10: g(mark(X)) -> g(X)
r11: g(active(X)) -> g(X)

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. matrix interpretations:
    
      carrier: N^1
      order: standard order
      interpretations:
        mark#_A(x1) = x1
        g_A(x1) = x1 + 1
        f_A(x1) = x1 + 1
    
    2. matrix interpretations:
    
      carrier: N^1
      order: standard order
      interpretations:
        mark#_A(x1) = 0
        g_A(x1) = x1 + 1
        f_A(x1) = x1 + 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: active#(f(f(a()))) -> mark#(c(f(g(f(a())))))
p2: mark#(c(X)) -> active#(c(X))

and R consists of:

r1: active(f(f(a()))) -> mark(c(f(g(f(a())))))
r2: mark(f(X)) -> active(f(mark(X)))
r3: mark(a()) -> active(a())
r4: mark(c(X)) -> active(c(X))
r5: mark(g(X)) -> active(g(mark(X)))
r6: f(mark(X)) -> f(X)
r7: f(active(X)) -> f(X)
r8: c(mark(X)) -> c(X)
r9: c(active(X)) -> c(X)
r10: g(mark(X)) -> g(X)
r11: g(active(X)) -> g(X)

The set of usable rules consists of

  r8, r9

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) = 3
        a_A() = 0
        mark#_A(x1) = 2
        c_A(x1) = 1
        g_A(x1) = 1
        mark_A(x1) = x1 + 1
        active_A(x1) = 1
    
    2. matrix interpretations:
    
      carrier: N^1
      order: standard order
      interpretations:
        active#_A(x1) = 1
        f_A(x1) = 1
        a_A() = 0
        mark#_A(x1) = 0
        c_A(x1) = 1
        g_A(x1) = 1
        mark_A(x1) = x1
        active_A(x1) = 0
    

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(f(a()))) -> mark(c(f(g(f(a())))))
r2: mark(f(X)) -> active(f(mark(X)))
r3: mark(a()) -> active(a())
r4: mark(c(X)) -> active(c(X))
r5: mark(g(X)) -> active(g(mark(X)))
r6: f(mark(X)) -> f(X)
r7: f(active(X)) -> f(X)
r8: c(mark(X)) -> c(X)
r9: c(active(X)) -> c(X)
r10: g(mark(X)) -> g(X)
r11: g(active(X)) -> g(X)

The set of usable rules consists of

  (no rules)

Take the 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. matrix interpretations:
    
      carrier: N^1
      order: standard order
      interpretations:
        f#_A(x1) = x1
        mark_A(x1) = x1 + 1
        active_A(x1) = 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#(active(X)) -> g#(X)

and R consists of:

r1: active(f(f(a()))) -> mark(c(f(g(f(a())))))
r2: mark(f(X)) -> active(f(mark(X)))
r3: mark(a()) -> active(a())
r4: mark(c(X)) -> active(c(X))
r5: mark(g(X)) -> active(g(mark(X)))
r6: f(mark(X)) -> f(X)
r7: f(active(X)) -> f(X)
r8: c(mark(X)) -> c(X)
r9: c(active(X)) -> c(X)
r10: g(mark(X)) -> g(X)
r11: g(active(X)) -> g(X)

The set of usable rules consists of

  (no rules)

Take the 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. matrix interpretations:
    
      carrier: N^1
      order: standard order
      interpretations:
        g#_A(x1) = 0
        mark_A(x1) = x1 + 1
        active_A(x1) = x1 + 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#(mark(X)) -> c#(X)
p2: c#(active(X)) -> c#(X)

and R consists of:

r1: active(f(f(a()))) -> mark(c(f(g(f(a())))))
r2: mark(f(X)) -> active(f(mark(X)))
r3: mark(a()) -> active(a())
r4: mark(c(X)) -> active(c(X))
r5: mark(g(X)) -> active(g(mark(X)))
r6: f(mark(X)) -> f(X)
r7: f(active(X)) -> f(X)
r8: c(mark(X)) -> c(X)
r9: c(active(X)) -> c(X)
r10: g(mark(X)) -> g(X)
r11: g(active(X)) -> g(X)

The set of usable rules consists of

  (no rules)

Take the 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. matrix interpretations:
    
      carrier: N^1
      order: standard order
      interpretations:
        c#_A(x1) = 0
        mark_A(x1) = x1 + 1
        active_A(x1) = x1 + 1
    

The next rules are strictly ordered:

  p1, p2

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