YES

We show the termination of the TRS R:

  active(f(a(),X,X)) -> mark(f(X,b(),b()))
  active(b()) -> mark(a())
  active(f(X1,X2,X3)) -> f(X1,active(X2),X3)
  f(X1,mark(X2),X3) -> mark(f(X1,X2,X3))
  proper(f(X1,X2,X3)) -> f(proper(X1),proper(X2),proper(X3))
  proper(a()) -> ok(a())
  proper(b()) -> ok(b())
  f(ok(X1),ok(X2),ok(X3)) -> ok(f(X1,X2,X3))
  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(a(),X,X)) -> f#(X,b(),b())
p2: active#(f(X1,X2,X3)) -> f#(X1,active(X2),X3)
p3: active#(f(X1,X2,X3)) -> active#(X2)
p4: f#(X1,mark(X2),X3) -> f#(X1,X2,X3)
p5: proper#(f(X1,X2,X3)) -> f#(proper(X1),proper(X2),proper(X3))
p6: proper#(f(X1,X2,X3)) -> proper#(X1)
p7: proper#(f(X1,X2,X3)) -> proper#(X2)
p8: proper#(f(X1,X2,X3)) -> proper#(X3)
p9: f#(ok(X1),ok(X2),ok(X3)) -> f#(X1,X2,X3)
p10: top#(mark(X)) -> top#(proper(X))
p11: top#(mark(X)) -> proper#(X)
p12: top#(ok(X)) -> top#(active(X))
p13: top#(ok(X)) -> active#(X)

and R consists of:

r1: active(f(a(),X,X)) -> mark(f(X,b(),b()))
r2: active(b()) -> mark(a())
r3: active(f(X1,X2,X3)) -> f(X1,active(X2),X3)
r4: f(X1,mark(X2),X3) -> mark(f(X1,X2,X3))
r5: proper(f(X1,X2,X3)) -> f(proper(X1),proper(X2),proper(X3))
r6: proper(a()) -> ok(a())
r7: proper(b()) -> ok(b())
r8: f(ok(X1),ok(X2),ok(X3)) -> ok(f(X1,X2,X3))
r9: top(mark(X)) -> top(proper(X))
r10: top(ok(X)) -> top(active(X))

The estimated dependency graph contains the following SCCs:

  {p10, p12}
  {p3}
  {p6, p7, p8}
  {p4, p9}


-- 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(a(),X,X)) -> mark(f(X,b(),b()))
r2: active(b()) -> mark(a())
r3: active(f(X1,X2,X3)) -> f(X1,active(X2),X3)
r4: f(X1,mark(X2),X3) -> mark(f(X1,X2,X3))
r5: proper(f(X1,X2,X3)) -> f(proper(X1),proper(X2),proper(X3))
r6: proper(a()) -> ok(a())
r7: proper(b()) -> ok(b())
r8: f(ok(X1),ok(X2),ok(X3)) -> ok(f(X1,X2,X3))
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:

  lexicographic combination of reduction pairs:
  
    1. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        top#_A(x1) = ((0,1),(0,0)) x1
        ok_A(x1) = x1
        active_A(x1) = x1 + (2,0)
        mark_A(x1) = x1 + (1,2)
        proper_A(x1) = x1
        f_A(x1,x2,x3) = ((0,0),(1,0)) x1 + ((0,1),(0,1)) x2 + ((0,0),(1,0)) x3 + (0,1)
        a_A() = (6,1)
        b_A() = (1,3)
    
    2. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        top#_A(x1) = (0,0)
        ok_A(x1) = ((0,1),(0,1)) x1
        active_A(x1) = x1 + (0,1)
        mark_A(x1) = (0,3)
        proper_A(x1) = ((1,1),(0,1)) x1 + (1,1)
        f_A(x1,x2,x3) = (3,2)
        a_A() = (1,1)
        b_A() = (1,1)
    

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(a(),X,X)) -> mark(f(X,b(),b()))
r2: active(b()) -> mark(a())
r3: active(f(X1,X2,X3)) -> f(X1,active(X2),X3)
r4: f(X1,mark(X2),X3) -> mark(f(X1,X2,X3))
r5: proper(f(X1,X2,X3)) -> f(proper(X1),proper(X2),proper(X3))
r6: proper(a()) -> ok(a())
r7: proper(b()) -> ok(b())
r8: f(ok(X1),ok(X2),ok(X3)) -> ok(f(X1,X2,X3))
r9: top(mark(X)) -> top(proper(X))
r10: 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(a(),X,X)) -> mark(f(X,b(),b()))
r2: active(b()) -> mark(a())
r3: active(f(X1,X2,X3)) -> f(X1,active(X2),X3)
r4: f(X1,mark(X2),X3) -> mark(f(X1,X2,X3))
r5: proper(f(X1,X2,X3)) -> f(proper(X1),proper(X2),proper(X3))
r6: proper(a()) -> ok(a())
r7: proper(b()) -> ok(b())
r8: f(ok(X1),ok(X2),ok(X3)) -> ok(f(X1,X2,X3))
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, r8

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        top#_A(x1) = ((1,1),(1,1)) x1
        ok_A(x1) = ((1,1),(1,1)) x1 + (2,1)
        active_A(x1) = ((1,1),(1,1)) x1 + (1,1)
        f_A(x1,x2,x3) = ((1,1),(1,1)) x1 + ((1,1),(1,1)) x2 + ((1,1),(1,1)) x3 + (1,2)
        mark_A(x1) = (1,1)
        a_A() = (1,1)
        b_A() = (1,1)
    
    2. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        top#_A(x1) = x1
        ok_A(x1) = x1 + (3,1)
        active_A(x1) = x1 + (1,2)
        f_A(x1,x2,x3) = ((0,0),(1,1)) x1 + ((0,1),(1,1)) x2 + (1,1)
        mark_A(x1) = (0,6)
        a_A() = (1,1)
        b_A() = (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: active#(f(X1,X2,X3)) -> active#(X2)

and R consists of:

r1: active(f(a(),X,X)) -> mark(f(X,b(),b()))
r2: active(b()) -> mark(a())
r3: active(f(X1,X2,X3)) -> f(X1,active(X2),X3)
r4: f(X1,mark(X2),X3) -> mark(f(X1,X2,X3))
r5: proper(f(X1,X2,X3)) -> f(proper(X1),proper(X2),proper(X3))
r6: proper(a()) -> ok(a())
r7: proper(b()) -> ok(b())
r8: f(ok(X1),ok(X2),ok(X3)) -> ok(f(X1,X2,X3))
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:

  lexicographic combination of reduction pairs:
  
    1. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        active#_A(x1) = x1
        f_A(x1,x2,x3) = ((0,1),(1,1)) x1 + ((1,1),(1,1)) x2 + ((1,1),(1,1)) x3 + (1,1)
    
    2. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        active#_A(x1) = x1
        f_A(x1,x2,x3) = ((1,1),(0,0)) x2 + ((0,1),(0,0)) x3 + (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#(f(X1,X2,X3)) -> proper#(X3)
p2: proper#(f(X1,X2,X3)) -> proper#(X2)
p3: proper#(f(X1,X2,X3)) -> proper#(X1)

and R consists of:

r1: active(f(a(),X,X)) -> mark(f(X,b(),b()))
r2: active(b()) -> mark(a())
r3: active(f(X1,X2,X3)) -> f(X1,active(X2),X3)
r4: f(X1,mark(X2),X3) -> mark(f(X1,X2,X3))
r5: proper(f(X1,X2,X3)) -> f(proper(X1),proper(X2),proper(X3))
r6: proper(a()) -> ok(a())
r7: proper(b()) -> ok(b())
r8: f(ok(X1),ok(X2),ok(X3)) -> ok(f(X1,X2,X3))
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:

  lexicographic combination of reduction pairs:
  
    1. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        proper#_A(x1) = ((1,1),(1,1)) x1
        f_A(x1,x2,x3) = ((1,1),(1,1)) x1 + ((1,1),(1,1)) x2 + ((1,1),(1,1)) x3 + (1,1)
    
    2. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        proper#_A(x1) = ((1,1),(1,0)) x1
        f_A(x1,x2,x3) = ((0,1),(1,1)) x1 + ((1,1),(1,0)) x2 + ((1,1),(1,1)) x3 + (1,1)
    

The next rules are strictly ordered:

  p1, p2, p3

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#(X1,mark(X2),X3) -> f#(X1,X2,X3)
p2: f#(ok(X1),ok(X2),ok(X3)) -> f#(X1,X2,X3)

and R consists of:

r1: active(f(a(),X,X)) -> mark(f(X,b(),b()))
r2: active(b()) -> mark(a())
r3: active(f(X1,X2,X3)) -> f(X1,active(X2),X3)
r4: f(X1,mark(X2),X3) -> mark(f(X1,X2,X3))
r5: proper(f(X1,X2,X3)) -> f(proper(X1),proper(X2),proper(X3))
r6: proper(a()) -> ok(a())
r7: proper(b()) -> ok(b())
r8: f(ok(X1),ok(X2),ok(X3)) -> ok(f(X1,X2,X3))
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:

  lexicographic combination of reduction pairs:
  
    1. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        f#_A(x1,x2,x3) = ((1,1),(1,1)) x1 + ((1,1),(1,0)) x2 + ((1,1),(0,0)) x3
        mark_A(x1) = ((1,1),(1,1)) x1 + (1,1)
        ok_A(x1) = ((1,1),(1,1)) x1 + (1,1)
    
    2. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        f#_A(x1,x2,x3) = ((0,1),(1,0)) x1 + ((0,1),(1,1)) x2 + ((0,0),(1,1)) x3
        mark_A(x1) = (1,1)
        ok_A(x1) = ((1,1),(1,0)) x1 + (1,1)
    

The next rules are strictly ordered:

  p1, p2

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