YES

We show the termination of the TRS R:

  active(f(f(a()))) -> mark(f(g(f(a()))))
  mark(f(X)) -> active(f(mark(X)))
  mark(a()) -> active(a())
  mark(g(X)) -> active(g(X))
  f(mark(X)) -> f(X)
  f(active(X)) -> f(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#(f(g(f(a()))))
p2: active#(f(f(a()))) -> f#(g(f(a())))
p3: active#(f(f(a()))) -> g#(f(a()))
p4: mark#(f(X)) -> active#(f(mark(X)))
p5: mark#(f(X)) -> f#(mark(X))
p6: mark#(f(X)) -> mark#(X)
p7: mark#(a()) -> active#(a())
p8: mark#(g(X)) -> active#(g(X))
p9: f#(mark(X)) -> f#(X)
p10: f#(active(X)) -> f#(X)
p11: g#(mark(X)) -> g#(X)
p12: g#(active(X)) -> g#(X)

and R consists of:

r1: active(f(f(a()))) -> mark(f(g(f(a()))))
r2: mark(f(X)) -> active(f(mark(X)))
r3: mark(a()) -> active(a())
r4: mark(g(X)) -> active(g(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)

The estimated dependency graph contains the following SCCs:

  {p1, p4, p6, p8}
  {p9, p10}
  {p11, p12}


-- Reduction pair.

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

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

and R consists of:

r1: active(f(f(a()))) -> mark(f(g(f(a()))))
r2: mark(f(X)) -> active(f(mark(X)))
r3: mark(a()) -> active(a())
r4: mark(g(X)) -> active(g(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)

The set of usable rules consists of

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

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 + (7,1)
            f_A(x1) = ((0,0),(1,0)) x1 + (5,2)
            a_A() = (6,9)
            mark#_A(x1) = (7,6)
            g_A(x1) = (1,1)
            mark_A(x1) = x1 + (0,3)
            active_A(x1) = x1
    
      precedence:
      
        active# = f = a = mark# = g = mark = active
      
      partial status:
    
        pi(active#) = []
        pi(f) = []
        pi(a) = []
        pi(mark#) = []
        pi(g) = []
        pi(mark) = []
        pi(active) = []
    
    2. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            active#_A(x1) = (4,2)
            f_A(x1) = (3,1)
            a_A() = (2,0)
            mark#_A(x1) = (4,2)
            g_A(x1) = (5,2)
            mark_A(x1) = ((1,0),(1,1)) x1 + (1,3)
            active_A(x1) = ((1,0),(0,0)) x1 + (6,11)
    
      precedence:
      
        active > g > f = mark > active# = mark# > a
      
      partial status:
    
        pi(active#) = []
        pi(f) = []
        pi(a) = []
        pi(mark#) = []
        pi(g) = []
        pi(mark) = [1]
        pi(active) = []
    

The next rules are strictly ordered:

  p3

We remove them from the problem.

-- SCC decomposition.

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

p1: active#(f(f(a()))) -> mark#(f(g(f(a()))))
p2: mark#(f(X)) -> mark#(X)
p3: mark#(f(X)) -> active#(f(mark(X)))

and R consists of:

r1: active(f(f(a()))) -> mark(f(g(f(a()))))
r2: mark(f(X)) -> active(f(mark(X)))
r3: mark(a()) -> active(a())
r4: mark(g(X)) -> active(g(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)

The estimated dependency graph contains the following SCCs:

  {p1, p2, p3}


-- Reduction pair.

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

p1: active#(f(f(a()))) -> mark#(f(g(f(a()))))
p2: mark#(f(X)) -> active#(f(mark(X)))
p3: mark#(f(X)) -> mark#(X)

and R consists of:

r1: active(f(f(a()))) -> mark(f(g(f(a()))))
r2: mark(f(X)) -> active(f(mark(X)))
r3: mark(a()) -> active(a())
r4: mark(g(X)) -> active(g(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)

The set of usable rules consists of

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

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
            f_A(x1) = x1 + (0,2)
            a_A() = (4,5)
            mark#_A(x1) = ((0,0),(1,0)) x1 + (0,4)
            g_A(x1) = (0,0)
            mark_A(x1) = x1 + (1,1)
            active_A(x1) = x1
    
      precedence:
      
        active# > mark# = mark > active > g > f > a
      
      partial status:
    
        pi(active#) = []
        pi(f) = [1]
        pi(a) = []
        pi(mark#) = []
        pi(g) = []
        pi(mark) = []
        pi(active) = [1]
    
    2. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            active#_A(x1) = (4,4)
            f_A(x1) = (0,2)
            a_A() = (2,1)
            mark#_A(x1) = (1,0)
            g_A(x1) = (3,3)
            mark_A(x1) = (0,4)
            active_A(x1) = x1
    
      precedence:
      
        active# > mark# > g > f = mark > a = active
      
      partial status:
    
        pi(active#) = []
        pi(f) = []
        pi(a) = []
        pi(mark#) = []
        pi(g) = []
        pi(mark) = []
        pi(active) = [1]
    

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: active#(f(f(a()))) -> mark#(f(g(f(a()))))
p2: mark#(f(X)) -> mark#(X)

and R consists of:

r1: active(f(f(a()))) -> mark(f(g(f(a()))))
r2: mark(f(X)) -> active(f(mark(X)))
r3: mark(a()) -> active(a())
r4: mark(g(X)) -> active(g(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)

The estimated dependency graph contains the following SCCs:

  {p2}


-- Reduction pair.

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

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

and R consists of:

r1: active(f(f(a()))) -> mark(f(g(f(a()))))
r2: mark(f(X)) -> active(f(mark(X)))
r3: mark(a()) -> active(a())
r4: mark(g(X)) -> active(g(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)

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

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

and R consists of:

r1: active(f(f(a()))) -> mark(f(g(f(a()))))
r2: mark(f(X)) -> active(f(mark(X)))
r3: mark(a()) -> active(a())
r4: mark(g(X)) -> active(g(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)

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.

-- 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(f(g(f(a()))))
r2: mark(f(X)) -> active(f(mark(X)))
r3: mark(a()) -> active(a())
r4: mark(g(X)) -> active(g(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)

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.