YES

We show the termination of the TRS R:

  a__f(|0|()) -> cons(|0|(),f(s(|0|())))
  a__f(s(|0|())) -> a__f(a__p(s(|0|())))
  a__p(s(X)) -> mark(X)
  mark(f(X)) -> a__f(mark(X))
  mark(p(X)) -> a__p(mark(X))
  mark(|0|()) -> |0|()
  mark(cons(X1,X2)) -> cons(mark(X1),X2)
  mark(s(X)) -> s(mark(X))
  a__f(X) -> f(X)
  a__p(X) -> p(X)

-- SCC decomposition.

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

p1: a__f#(s(|0|())) -> a__f#(a__p(s(|0|())))
p2: a__f#(s(|0|())) -> a__p#(s(|0|()))
p3: a__p#(s(X)) -> mark#(X)
p4: mark#(f(X)) -> a__f#(mark(X))
p5: mark#(f(X)) -> mark#(X)
p6: mark#(p(X)) -> a__p#(mark(X))
p7: mark#(p(X)) -> mark#(X)
p8: mark#(cons(X1,X2)) -> mark#(X1)
p9: mark#(s(X)) -> mark#(X)

and R consists of:

r1: a__f(|0|()) -> cons(|0|(),f(s(|0|())))
r2: a__f(s(|0|())) -> a__f(a__p(s(|0|())))
r3: a__p(s(X)) -> mark(X)
r4: mark(f(X)) -> a__f(mark(X))
r5: mark(p(X)) -> a__p(mark(X))
r6: mark(|0|()) -> |0|()
r7: mark(cons(X1,X2)) -> cons(mark(X1),X2)
r8: mark(s(X)) -> s(mark(X))
r9: a__f(X) -> f(X)
r10: a__p(X) -> p(X)

The estimated dependency graph contains the following SCCs:

  {p1, p2, p3, p4, p5, p6, p7, p8, p9}


-- Reduction pair.

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

p1: a__f#(s(|0|())) -> a__f#(a__p(s(|0|())))
p2: a__f#(s(|0|())) -> a__p#(s(|0|()))
p3: a__p#(s(X)) -> mark#(X)
p4: mark#(s(X)) -> mark#(X)
p5: mark#(cons(X1,X2)) -> mark#(X1)
p6: mark#(p(X)) -> mark#(X)
p7: mark#(p(X)) -> a__p#(mark(X))
p8: mark#(f(X)) -> mark#(X)
p9: mark#(f(X)) -> a__f#(mark(X))

and R consists of:

r1: a__f(|0|()) -> cons(|0|(),f(s(|0|())))
r2: a__f(s(|0|())) -> a__f(a__p(s(|0|())))
r3: a__p(s(X)) -> mark(X)
r4: mark(f(X)) -> a__f(mark(X))
r5: mark(p(X)) -> a__p(mark(X))
r6: mark(|0|()) -> |0|()
r7: mark(cons(X1,X2)) -> cons(mark(X1),X2)
r8: mark(s(X)) -> s(mark(X))
r9: a__f(X) -> f(X)
r10: a__p(X) -> p(X)

The set of usable rules consists of

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

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          a__f#_A(x1) = x1 + 15
          s_A(x1) = x1 + 9
          |0|_A = 5
          a__p_A(x1) = x1
          a__p#_A(x1) = max{0, x1 - 8}
          mark#_A(x1) = x1
          cons_A(x1,x2) = max{27, x1 + 22, x2 - 12}
          p_A(x1) = x1
          mark_A(x1) = x1 + 8
          f_A(x1) = max{25, x1 + 23}
          a__f_A(x1) = max{26, x1 + 23}
    
      precedence:
      
        a__f# = s = |0| = a__p = a__p# = mark# = cons = p = mark = f = a__f
      
      partial status:
    
        pi(a__f#) = []
        pi(s) = []
        pi(|0|) = []
        pi(a__p) = []
        pi(a__p#) = []
        pi(mark#) = []
        pi(cons) = []
        pi(p) = []
        pi(mark) = []
        pi(f) = []
        pi(a__f) = []
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          a__f#_A(x1) = max{12, x1}
          s_A(x1) = max{17, x1 - 6}
          |0|_A = 17
          a__p_A(x1) = max{10, x1}
          a__p#_A(x1) = 13
          mark#_A(x1) = max{24, x1 + 5}
          cons_A(x1,x2) = 23
          p_A(x1) = max{9, x1}
          mark_A(x1) = 24
          f_A(x1) = 18
          a__f_A(x1) = 24
    
      precedence:
      
        mark# > mark > |0| = a__f > f > s > a__f# = a__p = a__p# > cons = p
      
      partial status:
    
        pi(a__f#) = [1]
        pi(s) = []
        pi(|0|) = []
        pi(a__p) = []
        pi(a__p#) = []
        pi(mark#) = []
        pi(cons) = []
        pi(p) = []
        pi(mark) = []
        pi(f) = []
        pi(a__f) = []
    

The next rules are strictly ordered:

  p1, p3, p4, p5, p8

We remove them from the problem.

-- SCC decomposition.

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

p1: a__f#(s(|0|())) -> a__p#(s(|0|()))
p2: mark#(p(X)) -> mark#(X)
p3: mark#(p(X)) -> a__p#(mark(X))
p4: mark#(f(X)) -> a__f#(mark(X))

and R consists of:

r1: a__f(|0|()) -> cons(|0|(),f(s(|0|())))
r2: a__f(s(|0|())) -> a__f(a__p(s(|0|())))
r3: a__p(s(X)) -> mark(X)
r4: mark(f(X)) -> a__f(mark(X))
r5: mark(p(X)) -> a__p(mark(X))
r6: mark(|0|()) -> |0|()
r7: mark(cons(X1,X2)) -> cons(mark(X1),X2)
r8: mark(s(X)) -> s(mark(X))
r9: a__f(X) -> f(X)
r10: a__p(X) -> p(X)

The estimated dependency graph contains the following SCCs:

  {p2}


-- Reduction pair.

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

p1: mark#(p(X)) -> mark#(X)

and R consists of:

r1: a__f(|0|()) -> cons(|0|(),f(s(|0|())))
r2: a__f(s(|0|())) -> a__f(a__p(s(|0|())))
r3: a__p(s(X)) -> mark(X)
r4: mark(f(X)) -> a__f(mark(X))
r5: mark(p(X)) -> a__p(mark(X))
r6: mark(|0|()) -> |0|()
r7: mark(cons(X1,X2)) -> cons(mark(X1),X2)
r8: mark(s(X)) -> s(mark(X))
r9: a__f(X) -> f(X)
r10: a__p(X) -> p(X)

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:
        
          mark#_A(x1) = max{4, x1 + 3}
          p_A(x1) = max{3, x1 + 2}
    
      precedence:
      
        mark# = p
      
      partial status:
    
        pi(mark#) = [1]
        pi(p) = [1]
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          mark#_A(x1) = max{1, x1 - 1}
          p_A(x1) = x1
    
      precedence:
      
        mark# = p
      
      partial status:
    
        pi(mark#) = []
        pi(p) = [1]
    

The next rules are strictly ordered:

  p1

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