YES

We show the termination of the TRS R:

  active(f(f(a()))) -> mark(f(g(f(a()))))
  mark(f(X)) -> active(f(X))
  mark(a()) -> active(a())
  mark(g(X)) -> active(g(mark(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(X))
p5: mark#(a()) -> active#(a())
p6: mark#(g(X)) -> active#(g(mark(X)))
p7: mark#(g(X)) -> g#(mark(X))
p8: mark#(g(X)) -> mark#(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(X))
r3: mark(a()) -> active(a())
r4: mark(g(X)) -> active(g(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)

The estimated dependency graph contains the following SCCs:

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


-- Reduction pair.

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

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

and R consists of:

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

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^2
        order: lexicographic order
        interpretations:
          mark#_A(x1) = ((1,0),(1,0)) x1 + (2,2)
          g_A(x1) = ((1,0),(0,0)) x1 + (1,1)
  
    precedence:
    
      mark# = g
    
    partial status:
  
      pi(mark#) = []
      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#(f(f(a()))) -> mark#(f(g(f(a()))))
p2: mark#(f(X)) -> active#(f(X))

and R consists of:

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

The set of usable rules consists of

  r5, r6

Take the reduction pair:

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

The next rules are strictly ordered:

  p1

We remove them from the problem.

-- SCC decomposition.

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

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

and R consists of:

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

The estimated dependency graph contains the following SCCs:

  (no SCCs)

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

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^2
        order: lexicographic order
        interpretations:
          g#_A(x1) = ((1,0),(0,0)) x1 + (1,1)
          mark_A(x1) = x1 + (2,2)
          active_A(x1) = ((1,0),(0,0)) x1 + (2,2)
  
    precedence:
    
      g# = mark = active
    
    partial status:
  
      pi(g#) = []
      pi(mark) = [1]
      pi(active) = []

The next rules are strictly ordered:

  p1

We remove them from the problem.

-- SCC decomposition.

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

p1: 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(X))
r3: mark(a()) -> active(a())
r4: mark(g(X)) -> active(g(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)

The estimated dependency graph contains the following SCCs:

  {p1}


-- Reduction pair.

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

p1: 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(X))
r3: mark(a()) -> active(a())
r4: mark(g(X)) -> active(g(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)

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^2
        order: lexicographic order
        interpretations:
          g#_A(x1) = ((1,0),(1,0)) x1 + (2,2)
          active_A(x1) = ((1,0),(0,0)) x1 + (1,1)
  
    precedence:
    
      g# = active
    
    partial status:
  
      pi(g#) = []
      pi(active) = []

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

The set of usable rules consists of

  (no rules)

Take the reduction pair:

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

The next rules are strictly ordered:

  p1

We remove them from the problem.

-- SCC decomposition.

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

p1: 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(X))
r3: mark(a()) -> active(a())
r4: mark(g(X)) -> active(g(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)

The estimated dependency graph contains the following SCCs:

  {p1}


-- Reduction pair.

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

p1: 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(X))
r3: mark(a()) -> active(a())
r4: mark(g(X)) -> active(g(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)

The set of usable rules consists of

  (no rules)

Take the reduction pair:

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

The next rules are strictly ordered:

  p1

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