YES

We show the termination of the TRS R:

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

and R consists of:

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

The estimated dependency graph contains the following SCCs:

  {p9, p11}
  {p4, p6}
  {p7}
  {p8}


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

The set of usable rules consists of

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

Take the reduction pair:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        top#_A(x1) = max{1, x1 - 13}
        ok_A(x1) = max{6, x1 - 5}
        active_A(x1) = max{7, x1 - 17}
        mark_A(x1) = x1 + 17
        proper_A(x1) = x1 + 14
        f_A(x1) = max{46, x1 + 15}
        g_A(x1) = 12
        c_A = 80
  
    precedence:
    
      active = proper = c > top# = mark = f > ok = g
    
    partial status:
  
      pi(top#) = []
      pi(ok) = []
      pi(active) = []
      pi(mark) = [1]
      pi(proper) = [1]
      pi(f) = [1]
      pi(g) = []
      pi(c) = []

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: top#(ok(X)) -> top#(active(X))

and R consists of:

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

The estimated dependency graph contains the following SCCs:

  {p1}


-- Reduction pair.

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

p1: top#(ok(X)) -> top#(active(X))

and R consists of:

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

The set of usable rules consists of

  r1, r2, r7

Take the reduction pair:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        top#_A(x1) = max{1, x1 - 5}
        ok_A(x1) = max{23, x1 + 13}
        active_A(x1) = x1 + 9
        g_A(x1) = max{12, x1 + 2}
        c_A = 6
        mark_A(x1) = max{15, x1 + 3}
        f_A(x1) = max{2, x1}
  
    precedence:
    
      top# = ok = active = g = c = mark = f
    
    partial status:
  
      pi(top#) = []
      pi(ok) = [1]
      pi(active) = [1]
      pi(g) = [1]
      pi(c) = []
      pi(mark) = []
      pi(f) = [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(c()) -> mark(f(g(c())))
r2: active(f(g(X))) -> mark(g(X))
r3: proper(c()) -> ok(c())
r4: proper(f(X)) -> f(proper(X))
r5: proper(g(X)) -> g(proper(X))
r6: f(ok(X)) -> ok(f(X))
r7: g(ok(X)) -> ok(g(X))
r8: top(mark(X)) -> top(proper(X))
r9: top(ok(X)) -> top(active(X))

The set of usable rules consists of

  (no rules)

Take the monotone reduction pair:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        proper#_A(x1) = max{3, x1 + 2}
        g_A(x1) = x1
        f_A(x1) = x1 + 1
  
    precedence:
    
      proper# = g = f
    
    partial status:
  
      pi(proper#) = [1]
      pi(g) = [1]
      pi(f) = [1]

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

and R consists of:

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

The estimated dependency graph contains the following SCCs:

  {p1}


-- Reduction pair.

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

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

and R consists of:

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

The set of usable rules consists of

  (no rules)

Take the monotone reduction pair:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        proper#_A(x1) = x1 + 1
        f_A(x1) = x1
  
    precedence:
    
      proper# = f
    
    partial status:
  
      pi(proper#) = [1]
      pi(f) = [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: f#(ok(X)) -> f#(X)

and R consists of:

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

The set of usable rules consists of

  (no rules)

Take the monotone reduction pair:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        f#_A(x1) = x1 + 2
        ok_A(x1) = x1 + 2
  
    precedence:
    
      f# = ok
    
    partial status:
  
      pi(f#) = [1]
      pi(ok) = [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: g#(ok(X)) -> g#(X)

and R consists of:

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

The set of usable rules consists of

  (no rules)

Take the monotone reduction pair:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        g#_A(x1) = x1 + 2
        ok_A(x1) = x1 + 2
  
    precedence:
    
      g# = ok
    
    partial status:
  
      pi(g#) = [1]
      pi(ok) = [1]

The next rules are strictly ordered:

  p1

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