YES

We show the termination of the TRS R:

  active(f(a(),X,X)) -> mark(f(X,b(),b()))
  active(b()) -> mark(a())
  mark(f(X1,X2,X3)) -> active(f(X1,mark(X2),X3))
  mark(a()) -> active(a())
  mark(b()) -> active(b())
  f(mark(X1),X2,X3) -> f(X1,X2,X3)
  f(X1,mark(X2),X3) -> f(X1,X2,X3)
  f(X1,X2,mark(X3)) -> f(X1,X2,X3)
  f(active(X1),X2,X3) -> f(X1,X2,X3)
  f(X1,active(X2),X3) -> f(X1,X2,X3)
  f(X1,X2,active(X3)) -> f(X1,X2,X3)

-- SCC decomposition.

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

p1: active#(f(a(),X,X)) -> mark#(f(X,b(),b()))
p2: active#(f(a(),X,X)) -> f#(X,b(),b())
p3: active#(b()) -> mark#(a())
p4: mark#(f(X1,X2,X3)) -> active#(f(X1,mark(X2),X3))
p5: mark#(f(X1,X2,X3)) -> f#(X1,mark(X2),X3)
p6: mark#(f(X1,X2,X3)) -> mark#(X2)
p7: mark#(a()) -> active#(a())
p8: mark#(b()) -> active#(b())
p9: f#(mark(X1),X2,X3) -> f#(X1,X2,X3)
p10: f#(X1,mark(X2),X3) -> f#(X1,X2,X3)
p11: f#(X1,X2,mark(X3)) -> f#(X1,X2,X3)
p12: f#(active(X1),X2,X3) -> f#(X1,X2,X3)
p13: f#(X1,active(X2),X3) -> f#(X1,X2,X3)
p14: f#(X1,X2,active(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: mark(f(X1,X2,X3)) -> active(f(X1,mark(X2),X3))
r4: mark(a()) -> active(a())
r5: mark(b()) -> active(b())
r6: f(mark(X1),X2,X3) -> f(X1,X2,X3)
r7: f(X1,mark(X2),X3) -> f(X1,X2,X3)
r8: f(X1,X2,mark(X3)) -> f(X1,X2,X3)
r9: f(active(X1),X2,X3) -> f(X1,X2,X3)
r10: f(X1,active(X2),X3) -> f(X1,X2,X3)
r11: f(X1,X2,active(X3)) -> f(X1,X2,X3)

The estimated dependency graph contains the following SCCs:

  {p1, p4, p6}
  {p9, p10, p11, p12, p13, p14}


-- Reduction pair.

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

p1: active#(f(a(),X,X)) -> mark#(f(X,b(),b()))
p2: mark#(f(X1,X2,X3)) -> mark#(X2)
p3: mark#(f(X1,X2,X3)) -> active#(f(X1,mark(X2),X3))

and R consists of:

r1: active(f(a(),X,X)) -> mark(f(X,b(),b()))
r2: active(b()) -> mark(a())
r3: mark(f(X1,X2,X3)) -> active(f(X1,mark(X2),X3))
r4: mark(a()) -> active(a())
r5: mark(b()) -> active(b())
r6: f(mark(X1),X2,X3) -> f(X1,X2,X3)
r7: f(X1,mark(X2),X3) -> f(X1,X2,X3)
r8: f(X1,X2,mark(X3)) -> f(X1,X2,X3)
r9: f(active(X1),X2,X3) -> f(X1,X2,X3)
r10: f(X1,active(X2),X3) -> f(X1,X2,X3)
r11: f(X1,X2,active(X3)) -> f(X1,X2,X3)

The set of usable rules consists of

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

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          active#_A(x1) = max{23, x1 + 11}
          f_A(x1,x2,x3) = max{x1 - 1, x2 + 9, x3 + 5}
          a_A = 0
          mark#_A(x1) = x1 + 14
          b_A = 0
          mark_A(x1) = x1
          active_A(x1) = x1
    
      precedence:
      
        active# = f = mark# = b > mark = active > a
      
      partial status:
    
        pi(active#) = [1]
        pi(f) = []
        pi(a) = []
        pi(mark#) = [1]
        pi(b) = []
        pi(mark) = [1]
        pi(active) = [1]
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          active#_A(x1) = 29
          f_A(x1,x2,x3) = 28
          a_A = 9
          mark#_A(x1) = max{29, x1 - 28}
          b_A = 11
          mark_A(x1) = 10
          active_A(x1) = 10
    
      precedence:
      
        f > active# = a = mark# = b > mark = active
      
      partial status:
    
        pi(active#) = []
        pi(f) = []
        pi(a) = []
        pi(mark#) = []
        pi(b) = []
        pi(mark) = []
        pi(active) = []
    

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(a(),X,X)) -> mark#(f(X,b(),b()))
p2: mark#(f(X1,X2,X3)) -> active#(f(X1,mark(X2),X3))

and R consists of:

r1: active(f(a(),X,X)) -> mark(f(X,b(),b()))
r2: active(b()) -> mark(a())
r3: mark(f(X1,X2,X3)) -> active(f(X1,mark(X2),X3))
r4: mark(a()) -> active(a())
r5: mark(b()) -> active(b())
r6: f(mark(X1),X2,X3) -> f(X1,X2,X3)
r7: f(X1,mark(X2),X3) -> f(X1,X2,X3)
r8: f(X1,X2,mark(X3)) -> f(X1,X2,X3)
r9: f(active(X1),X2,X3) -> f(X1,X2,X3)
r10: f(X1,active(X2),X3) -> f(X1,X2,X3)
r11: f(X1,X2,active(X3)) -> f(X1,X2,X3)

The estimated dependency graph contains the following SCCs:

  {p1, p2}


-- Reduction pair.

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

p1: active#(f(a(),X,X)) -> mark#(f(X,b(),b()))
p2: mark#(f(X1,X2,X3)) -> active#(f(X1,mark(X2),X3))

and R consists of:

r1: active(f(a(),X,X)) -> mark(f(X,b(),b()))
r2: active(b()) -> mark(a())
r3: mark(f(X1,X2,X3)) -> active(f(X1,mark(X2),X3))
r4: mark(a()) -> active(a())
r5: mark(b()) -> active(b())
r6: f(mark(X1),X2,X3) -> f(X1,X2,X3)
r7: f(X1,mark(X2),X3) -> f(X1,X2,X3)
r8: f(X1,X2,mark(X3)) -> f(X1,X2,X3)
r9: f(active(X1),X2,X3) -> f(X1,X2,X3)
r10: f(X1,active(X2),X3) -> f(X1,X2,X3)
r11: f(X1,X2,active(X3)) -> f(X1,X2,X3)

The set of usable rules consists of

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

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          active#_A(x1) = max{26, x1 + 25}
          f_A(x1,x2,x3) = max{x1 + 39, x3 + 48}
          a_A = 19
          mark#_A(x1) = x1 + 32
          b_A = 1
          mark_A(x1) = max{28, x1 + 8}
          active_A(x1) = max{28, x1}
    
      precedence:
      
        mark = active > active# = a = mark# > b > f
      
      partial status:
    
        pi(active#) = [1]
        pi(f) = [1]
        pi(a) = []
        pi(mark#) = []
        pi(b) = []
        pi(mark) = []
        pi(active) = []
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          active#_A(x1) = 11
          f_A(x1,x2,x3) = 15
          a_A = 6
          mark#_A(x1) = 0
          b_A = 10
          mark_A(x1) = 2
          active_A(x1) = 2
    
      precedence:
      
        f > active# = a = mark# = b = mark = active
      
      partial status:
    
        pi(active#) = []
        pi(f) = []
        pi(a) = []
        pi(mark#) = []
        pi(b) = []
        pi(mark) = []
        pi(active) = []
    

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(a(),X,X)) -> mark#(f(X,b(),b()))

and R consists of:

r1: active(f(a(),X,X)) -> mark(f(X,b(),b()))
r2: active(b()) -> mark(a())
r3: mark(f(X1,X2,X3)) -> active(f(X1,mark(X2),X3))
r4: mark(a()) -> active(a())
r5: mark(b()) -> active(b())
r6: f(mark(X1),X2,X3) -> f(X1,X2,X3)
r7: f(X1,mark(X2),X3) -> f(X1,X2,X3)
r8: f(X1,X2,mark(X3)) -> f(X1,X2,X3)
r9: f(active(X1),X2,X3) -> f(X1,X2,X3)
r10: f(X1,active(X2),X3) -> f(X1,X2,X3)
r11: f(X1,X2,active(X3)) -> f(X1,X2,X3)

The estimated dependency graph contains the following SCCs:

  (no SCCs)

-- Reduction pair.

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

p1: f#(mark(X1),X2,X3) -> f#(X1,X2,X3)
p2: f#(X1,X2,active(X3)) -> f#(X1,X2,X3)
p3: f#(X1,active(X2),X3) -> f#(X1,X2,X3)
p4: f#(active(X1),X2,X3) -> f#(X1,X2,X3)
p5: f#(X1,X2,mark(X3)) -> f#(X1,X2,X3)
p6: f#(X1,mark(X2),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: mark(f(X1,X2,X3)) -> active(f(X1,mark(X2),X3))
r4: mark(a()) -> active(a())
r5: mark(b()) -> active(b())
r6: f(mark(X1),X2,X3) -> f(X1,X2,X3)
r7: f(X1,mark(X2),X3) -> f(X1,X2,X3)
r8: f(X1,X2,mark(X3)) -> f(X1,X2,X3)
r9: f(active(X1),X2,X3) -> f(X1,X2,X3)
r10: f(X1,active(X2),X3) -> f(X1,X2,X3)
r11: f(X1,X2,active(X3)) -> f(X1,X2,X3)

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          f#_A(x1,x2,x3) = max{3, x1 + 1, x2 + 1, x3 + 1}
          mark_A(x1) = max{2, x1 + 1}
          active_A(x1) = max{1, x1}
    
      precedence:
      
        f# = mark = active
      
      partial status:
    
        pi(f#) = [1, 2]
        pi(mark) = [1]
        pi(active) = [1]
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          f#_A(x1,x2,x3) = max{2, x1 + 1, x2 + 1}
          mark_A(x1) = max{1, x1}
          active_A(x1) = x1 + 1
    
      precedence:
      
        f# = mark = active
      
      partial status:
    
        pi(f#) = [1, 2]
        pi(mark) = [1]
        pi(active) = []
    

The next rules are strictly ordered:

  p1, p3, p4, p6

We remove them from the problem.

-- SCC decomposition.

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

p1: f#(X1,X2,active(X3)) -> f#(X1,X2,X3)
p2: f#(X1,X2,mark(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: mark(f(X1,X2,X3)) -> active(f(X1,mark(X2),X3))
r4: mark(a()) -> active(a())
r5: mark(b()) -> active(b())
r6: f(mark(X1),X2,X3) -> f(X1,X2,X3)
r7: f(X1,mark(X2),X3) -> f(X1,X2,X3)
r8: f(X1,X2,mark(X3)) -> f(X1,X2,X3)
r9: f(active(X1),X2,X3) -> f(X1,X2,X3)
r10: f(X1,active(X2),X3) -> f(X1,X2,X3)
r11: f(X1,X2,active(X3)) -> f(X1,X2,X3)

The estimated dependency graph contains the following SCCs:

  {p1, p2}


-- Reduction pair.

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

p1: f#(X1,X2,active(X3)) -> f#(X1,X2,X3)
p2: f#(X1,X2,mark(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: mark(f(X1,X2,X3)) -> active(f(X1,mark(X2),X3))
r4: mark(a()) -> active(a())
r5: mark(b()) -> active(b())
r6: f(mark(X1),X2,X3) -> f(X1,X2,X3)
r7: f(X1,mark(X2),X3) -> f(X1,X2,X3)
r8: f(X1,X2,mark(X3)) -> f(X1,X2,X3)
r9: f(active(X1),X2,X3) -> f(X1,X2,X3)
r10: f(X1,active(X2),X3) -> f(X1,X2,X3)
r11: f(X1,X2,active(X3)) -> f(X1,X2,X3)

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          f#_A(x1,x2,x3) = max{2, x1, x2 + 1, x3 + 1}
          active_A(x1) = max{1, x1}
          mark_A(x1) = max{1, x1}
    
      precedence:
      
        f# = active = mark
      
      partial status:
    
        pi(f#) = [1, 3]
        pi(active) = [1]
        pi(mark) = [1]
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          f#_A(x1,x2,x3) = max{0, x3 - 2}
          active_A(x1) = max{3, x1 + 1}
          mark_A(x1) = x1
    
      precedence:
      
        f# = active = mark
      
      partial status:
    
        pi(f#) = []
        pi(active) = [1]
        pi(mark) = [1]
    

The next rules are strictly ordered:

  p1, p2

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