YES

We show the termination of the TRS R:

  active(f(f(a()))) -> mark(f(g(f(a()))))
  active(f(X)) -> f(active(X))
  f(mark(X)) -> mark(f(X))
  proper(f(X)) -> f(proper(X))
  proper(a()) -> ok(a())
  proper(g(X)) -> g(proper(X))
  f(ok(X)) -> ok(f(X))
  g(ok(X)) -> ok(g(X))
  top(mark(X)) -> top(proper(X))
  top(ok(X)) -> top(active(X))

-- SCC decomposition.

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

p1: active#(f(f(a()))) -> f#(g(f(a())))
p2: active#(f(f(a()))) -> g#(f(a()))
p3: active#(f(X)) -> f#(active(X))
p4: active#(f(X)) -> active#(X)
p5: f#(mark(X)) -> f#(X)
p6: proper#(f(X)) -> f#(proper(X))
p7: proper#(f(X)) -> proper#(X)
p8: proper#(g(X)) -> g#(proper(X))
p9: proper#(g(X)) -> proper#(X)
p10: f#(ok(X)) -> f#(X)
p11: g#(ok(X)) -> g#(X)
p12: top#(mark(X)) -> top#(proper(X))
p13: top#(mark(X)) -> proper#(X)
p14: top#(ok(X)) -> top#(active(X))
p15: top#(ok(X)) -> active#(X)

and R consists of:

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

The estimated dependency graph contains the following SCCs:

  {p12, p14}
  {p4}
  {p7, p9}
  {p5, p10}
  {p11}


-- Reduction pair.

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

p1: top#(ok(X)) -> top#(active(X))
p2: top#(mark(X)) -> top#(proper(X))

and R consists of:

r1: active(f(f(a()))) -> mark(f(g(f(a()))))
r2: active(f(X)) -> f(active(X))
r3: f(mark(X)) -> mark(f(X))
r4: proper(f(X)) -> f(proper(X))
r5: proper(a()) -> ok(a())
r6: proper(g(X)) -> g(proper(X))
r7: f(ok(X)) -> ok(f(X))
r8: g(ok(X)) -> ok(g(X))
r9: top(mark(X)) -> top(proper(X))
r10: top(ok(X)) -> top(active(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:
      top#_A(x1) = x1
      ok_A(x1) = ((1,0),(1,0)) x1 + (0,2)
      active_A(x1) = ((1,0),(1,0)) x1 + (0,1)
      mark_A(x1) = x1 + (2,0)
      proper_A(x1) = ((1,0),(1,1)) x1 + (1,0)
      f_A(x1) = x1 + (1,1)
      g_A(x1) = ((0,0),(1,0)) x1 + (2,5)
      a_A() = (4,3)

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(X)) -> active#(X)

and R consists of:

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

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  matrix interpretations:
  
    carrier: N^2
    order: lexicographic order
    interpretations:
      active#_A(x1) = ((0,0),(1,0)) x1
      f_A(x1) = ((1,0),(1,1)) x1 + (1,1)

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: proper#(g(X)) -> proper#(X)
p2: proper#(f(X)) -> proper#(X)

and R consists of:

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

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  matrix interpretations:
  
    carrier: N^2
    order: lexicographic order
    interpretations:
      proper#_A(x1) = x1
      g_A(x1) = ((1,0),(1,1)) x1 + (1,1)
      f_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: f#(mark(X)) -> f#(X)
p2: f#(ok(X)) -> f#(X)

and R consists of:

r1: active(f(f(a()))) -> mark(f(g(f(a()))))
r2: active(f(X)) -> f(active(X))
r3: f(mark(X)) -> mark(f(X))
r4: proper(f(X)) -> f(proper(X))
r5: proper(a()) -> ok(a())
r6: proper(g(X)) -> g(proper(X))
r7: f(ok(X)) -> ok(f(X))
r8: g(ok(X)) -> ok(g(X))
r9: top(mark(X)) -> top(proper(X))
r10: top(ok(X)) -> top(active(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) = x1
      mark_A(x1) = ((1,0),(1,1)) x1 + (1,1)
      ok_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#(ok(X)) -> g#(X)

and R consists of:

r1: active(f(f(a()))) -> mark(f(g(f(a()))))
r2: active(f(X)) -> f(active(X))
r3: f(mark(X)) -> mark(f(X))
r4: proper(f(X)) -> f(proper(X))
r5: proper(a()) -> ok(a())
r6: proper(g(X)) -> g(proper(X))
r7: f(ok(X)) -> ok(f(X))
r8: g(ok(X)) -> ok(g(X))
r9: top(mark(X)) -> top(proper(X))
r10: top(ok(X)) -> top(active(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) = ((0,0),(1,0)) x1
      ok_A(x1) = ((1,0),(1,1)) x1 + (1,1)

The next rules are strictly ordered:

  p1

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