YES

We show the termination of the TRS R:

  active(f(f(a()))) -> mark(c(f(g(f(a())))))
  active(f(X)) -> f(active(X))
  active(g(X)) -> g(active(X))
  f(mark(X)) -> mark(f(X))
  g(mark(X)) -> mark(g(X))
  proper(f(X)) -> f(proper(X))
  proper(a()) -> ok(a())
  proper(c(X)) -> c(proper(X))
  proper(g(X)) -> g(proper(X))
  f(ok(X)) -> ok(f(X))
  c(ok(X)) -> ok(c(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()))) -> c#(f(g(f(a()))))
p2: active#(f(f(a()))) -> f#(g(f(a())))
p3: active#(f(f(a()))) -> g#(f(a()))
p4: active#(f(X)) -> f#(active(X))
p5: active#(f(X)) -> active#(X)
p6: active#(g(X)) -> g#(active(X))
p7: active#(g(X)) -> active#(X)
p8: f#(mark(X)) -> f#(X)
p9: g#(mark(X)) -> g#(X)
p10: proper#(f(X)) -> f#(proper(X))
p11: proper#(f(X)) -> proper#(X)
p12: proper#(c(X)) -> c#(proper(X))
p13: proper#(c(X)) -> proper#(X)
p14: proper#(g(X)) -> g#(proper(X))
p15: proper#(g(X)) -> proper#(X)
p16: f#(ok(X)) -> f#(X)
p17: c#(ok(X)) -> c#(X)
p18: g#(ok(X)) -> g#(X)
p19: top#(mark(X)) -> top#(proper(X))
p20: top#(mark(X)) -> proper#(X)
p21: top#(ok(X)) -> top#(active(X))
p22: top#(ok(X)) -> active#(X)

and R consists of:

r1: active(f(f(a()))) -> mark(c(f(g(f(a())))))
r2: active(f(X)) -> f(active(X))
r3: active(g(X)) -> g(active(X))
r4: f(mark(X)) -> mark(f(X))
r5: g(mark(X)) -> mark(g(X))
r6: proper(f(X)) -> f(proper(X))
r7: proper(a()) -> ok(a())
r8: proper(c(X)) -> c(proper(X))
r9: proper(g(X)) -> g(proper(X))
r10: f(ok(X)) -> ok(f(X))
r11: c(ok(X)) -> ok(c(X))
r12: g(ok(X)) -> ok(g(X))
r13: top(mark(X)) -> top(proper(X))
r14: top(ok(X)) -> top(active(X))

The estimated dependency graph contains the following SCCs:

  {p19, p21}
  {p5, p7}
  {p11, p13, p15}
  {p8, p16}
  {p9, p18}
  {p17}


-- 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(c(f(g(f(a())))))
r2: active(f(X)) -> f(active(X))
r3: active(g(X)) -> g(active(X))
r4: f(mark(X)) -> mark(f(X))
r5: g(mark(X)) -> mark(g(X))
r6: proper(f(X)) -> f(proper(X))
r7: proper(a()) -> ok(a())
r8: proper(c(X)) -> c(proper(X))
r9: proper(g(X)) -> g(proper(X))
r10: f(ok(X)) -> ok(f(X))
r11: c(ok(X)) -> ok(c(X))
r12: g(ok(X)) -> ok(g(X))
r13: top(mark(X)) -> top(proper(X))
r14: top(ok(X)) -> top(active(X))

The set of usable rules consists of

  r1, r2, r3, r4, r5, r6, r7, r8, r9, r10, r11, r12

Take the reduction pair:

  lexicographic path order with precedence:
  
    precedence:
    
      a > f > ok > active > mark > proper > c > g > top#
    
    argument filter:
  
      pi(top#) = [1]
      pi(ok) = 1
      pi(active) = 1
      pi(mark) = [1]
      pi(proper) = [1]
      pi(f) = 1
      pi(g) = 1
      pi(c) = []
      pi(a) = []

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(f(f(a()))) -> mark(c(f(g(f(a())))))
r2: active(f(X)) -> f(active(X))
r3: active(g(X)) -> g(active(X))
r4: f(mark(X)) -> mark(f(X))
r5: g(mark(X)) -> mark(g(X))
r6: proper(f(X)) -> f(proper(X))
r7: proper(a()) -> ok(a())
r8: proper(c(X)) -> c(proper(X))
r9: proper(g(X)) -> g(proper(X))
r10: f(ok(X)) -> ok(f(X))
r11: c(ok(X)) -> ok(c(X))
r12: g(ok(X)) -> ok(g(X))
r13: top(mark(X)) -> top(proper(X))
r14: 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(f(f(a()))) -> mark(c(f(g(f(a())))))
r2: active(f(X)) -> f(active(X))
r3: active(g(X)) -> g(active(X))
r4: f(mark(X)) -> mark(f(X))
r5: g(mark(X)) -> mark(g(X))
r6: proper(f(X)) -> f(proper(X))
r7: proper(a()) -> ok(a())
r8: proper(c(X)) -> c(proper(X))
r9: proper(g(X)) -> g(proper(X))
r10: f(ok(X)) -> ok(f(X))
r11: c(ok(X)) -> ok(c(X))
r12: g(ok(X)) -> ok(g(X))
r13: top(mark(X)) -> top(proper(X))
r14: top(ok(X)) -> top(active(X))

The set of usable rules consists of

  r1, r2, r3, r4, r5, r10, r12

Take the reduction pair:

  lexicographic path order with precedence:
  
    precedence:
    
      c > g > f > mark > a > active > ok > top#
    
    argument filter:
  
      pi(top#) = [1]
      pi(ok) = [1]
      pi(active) = 1
      pi(f) = [1]
      pi(mark) = []
      pi(g) = 1
      pi(a) = []
      pi(c) = 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: active#(g(X)) -> active#(X)
p2: active#(f(X)) -> active#(X)

and R consists of:

r1: active(f(f(a()))) -> mark(c(f(g(f(a())))))
r2: active(f(X)) -> f(active(X))
r3: active(g(X)) -> g(active(X))
r4: f(mark(X)) -> mark(f(X))
r5: g(mark(X)) -> mark(g(X))
r6: proper(f(X)) -> f(proper(X))
r7: proper(a()) -> ok(a())
r8: proper(c(X)) -> c(proper(X))
r9: proper(g(X)) -> g(proper(X))
r10: f(ok(X)) -> ok(f(X))
r11: c(ok(X)) -> ok(c(X))
r12: g(ok(X)) -> ok(g(X))
r13: top(mark(X)) -> top(proper(X))
r14: top(ok(X)) -> top(active(X))

The set of usable rules consists of

  (no rules)

Take the monotone reduction pair:

  lexicographic path order with precedence:
  
    precedence:
    
      active# > f > g
    
    argument filter:
  
      pi(active#) = 1
      pi(g) = [1]
      pi(f) = [1]

The next rules are strictly ordered:

  p1, p2
  r1, r2, r3, r4, r5, r6, r7, r8, r9, r10, r11, r12, r13, r14

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#(c(X)) -> proper#(X)
p3: proper#(f(X)) -> proper#(X)

and R consists of:

r1: active(f(f(a()))) -> mark(c(f(g(f(a())))))
r2: active(f(X)) -> f(active(X))
r3: active(g(X)) -> g(active(X))
r4: f(mark(X)) -> mark(f(X))
r5: g(mark(X)) -> mark(g(X))
r6: proper(f(X)) -> f(proper(X))
r7: proper(a()) -> ok(a())
r8: proper(c(X)) -> c(proper(X))
r9: proper(g(X)) -> g(proper(X))
r10: f(ok(X)) -> ok(f(X))
r11: c(ok(X)) -> ok(c(X))
r12: g(ok(X)) -> ok(g(X))
r13: top(mark(X)) -> top(proper(X))
r14: top(ok(X)) -> top(active(X))

The set of usable rules consists of

  (no rules)

Take the monotone reduction pair:

  lexicographic path order with precedence:
  
    precedence:
    
      proper# > f > c > g
    
    argument filter:
  
      pi(proper#) = 1
      pi(g) = [1]
      pi(c) = [1]
      pi(f) = [1]

The next rules are strictly ordered:

  p1, p2, p3
  r1, r2, r3, r4, r5, r6, r7, r8, r9, r10, r11, r12, r13, r14

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(c(f(g(f(a())))))
r2: active(f(X)) -> f(active(X))
r3: active(g(X)) -> g(active(X))
r4: f(mark(X)) -> mark(f(X))
r5: g(mark(X)) -> mark(g(X))
r6: proper(f(X)) -> f(proper(X))
r7: proper(a()) -> ok(a())
r8: proper(c(X)) -> c(proper(X))
r9: proper(g(X)) -> g(proper(X))
r10: f(ok(X)) -> ok(f(X))
r11: c(ok(X)) -> ok(c(X))
r12: g(ok(X)) -> ok(g(X))
r13: top(mark(X)) -> top(proper(X))
r14: top(ok(X)) -> top(active(X))

The set of usable rules consists of

  (no rules)

Take the monotone reduction pair:

  lexicographic path order with precedence:
  
    precedence:
    
      f# > ok > mark
    
    argument filter:
  
      pi(f#) = 1
      pi(mark) = [1]
      pi(ok) = [1]

The next rules are strictly ordered:

  p1, p2
  r1, r2, r3, r4, r5, r6, r7, r8, r9, r10, r11, r12, r13, r14

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#(ok(X)) -> g#(X)

and R consists of:

r1: active(f(f(a()))) -> mark(c(f(g(f(a())))))
r2: active(f(X)) -> f(active(X))
r3: active(g(X)) -> g(active(X))
r4: f(mark(X)) -> mark(f(X))
r5: g(mark(X)) -> mark(g(X))
r6: proper(f(X)) -> f(proper(X))
r7: proper(a()) -> ok(a())
r8: proper(c(X)) -> c(proper(X))
r9: proper(g(X)) -> g(proper(X))
r10: f(ok(X)) -> ok(f(X))
r11: c(ok(X)) -> ok(c(X))
r12: g(ok(X)) -> ok(g(X))
r13: top(mark(X)) -> top(proper(X))
r14: top(ok(X)) -> top(active(X))

The set of usable rules consists of

  (no rules)

Take the monotone reduction pair:

  lexicographic path order with precedence:
  
    precedence:
    
      g# > ok > mark
    
    argument filter:
  
      pi(g#) = 1
      pi(mark) = [1]
      pi(ok) = [1]

The next rules are strictly ordered:

  p1, p2
  r1, r2, r3, r4, r5, r6, r7, r8, r9, r10, r11, r12, r13, r14

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: c#(ok(X)) -> c#(X)

and R consists of:

r1: active(f(f(a()))) -> mark(c(f(g(f(a())))))
r2: active(f(X)) -> f(active(X))
r3: active(g(X)) -> g(active(X))
r4: f(mark(X)) -> mark(f(X))
r5: g(mark(X)) -> mark(g(X))
r6: proper(f(X)) -> f(proper(X))
r7: proper(a()) -> ok(a())
r8: proper(c(X)) -> c(proper(X))
r9: proper(g(X)) -> g(proper(X))
r10: f(ok(X)) -> ok(f(X))
r11: c(ok(X)) -> ok(c(X))
r12: g(ok(X)) -> ok(g(X))
r13: top(mark(X)) -> top(proper(X))
r14: top(ok(X)) -> top(active(X))

The set of usable rules consists of

  (no rules)

Take the monotone reduction pair:

  lexicographic path order with precedence:
  
    precedence:
    
      ok > c#
    
    argument filter:
  
      pi(c#) = [1]
      pi(ok) = [1]

The next rules are strictly ordered:

  p1
  r1, r2, r3, r4, r5, r6, r7, r8, r9, r10, r11, r12, r13, r14

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