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:

  matrix interpretations:
  
    carrier: N^2
    order: lexicographic order
    interpretations:
      active#_A(x1) = ((0,0),(1,0)) x1
      f_A(x1) = x1 + (1,0)
      a_A() = (1,6)
      mark#_A(x1) = ((0,0),(1,0)) x1
      g_A(x1) = x1 + (0,1)
      mark_A(x1) = ((1,0),(1,1)) x1 + (0,1)
      active_A(x1) = x1

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

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:

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

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(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:

  matrix interpretations:
  
    carrier: N^2
    order: lexicographic order
    interpretations:
      f#_A(x1) = ((1,0),(1,0)) x1
      mark_A(x1) = ((1,0),(1,1)) x1 + (1,1)
      active_A(x1) = ((1,0),(1,1)) x1 + (1,1)

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(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:

  matrix interpretations:
  
    carrier: N^2
    order: lexicographic order
    interpretations:
      g#_A(x1) = ((1,0),(1,0)) x1
      mark_A(x1) = ((1,0),(1,1)) x1 + (1,1)
      active_A(x1) = ((1,0),(1,1)) x1 + (1,1)

The next rules are strictly ordered:

  p1, p2

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