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. max/plus interpretations on natural numbers:
    
      a__f#_A(x1) = max{14, x1 - 4}
      s_A(x1) = x1 + 7
      |0|_A = 4
      a__p_A(x1) = x1 + 7
      a__p#_A(x1) = max{6, x1 - 8}
      mark#_A(x1) = max{0, x1 - 1}
      cons_A(x1,x2) = max{5, x1 + 2, x2 - 8}
      p_A(x1) = x1 + 7
      mark_A(x1) = x1 + 14
      f_A(x1) = max{21, x1 + 16}
      a__f_A(x1) = max{35, x1 + 16}
    
    2. max/plus interpretations on natural numbers:
    
      a__f#_A(x1) = 0
      s_A(x1) = x1 + 17
      |0|_A = 2
      a__p_A(x1) = max{8, x1 + 6}
      a__p#_A(x1) = 11
      mark#_A(x1) = 11
      cons_A(x1,x2) = 32
      p_A(x1) = max{7, x1 + 6}
      mark_A(x1) = x1 + 22
      f_A(x1) = 10
      a__f_A(x1) = 31
    

The next rules are strictly ordered:

  p2, p4, p5, p6, p8, p9

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__f#(a__p(s(|0|())))
p2: a__p#(s(X)) -> mark#(X)
p3: mark#(p(X)) -> a__p#(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}


-- Reduction pair.

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

p1: a__f#(s(|0|())) -> a__f#(a__p(s(|0|())))

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. max/plus interpretations on natural numbers:
    
      a__f#_A(x1) = max{0, x1 - 18}
      s_A(x1) = x1 + 19
      |0|_A = 5
      a__p_A(x1) = max{3, x1 - 7}
      a__f_A(x1) = max{3, x1 + 1}
      cons_A(x1,x2) = max{4, x1 + 1}
      f_A(x1) = max{2, x1 + 1}
      mark_A(x1) = x1 + 11
      p_A(x1) = max{0, x1 - 7}
    
    2. max/plus interpretations on natural numbers:
    
      a__f#_A(x1) = 0
      s_A(x1) = max{4, x1 - 1}
      |0|_A = 5
      a__p_A(x1) = 1
      a__f_A(x1) = 2
      cons_A(x1,x2) = 0
      f_A(x1) = 0
      mark_A(x1) = x1 + 4
      p_A(x1) = 0
    

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: a__p#(s(X)) -> mark#(X)
p2: mark#(p(X)) -> a__p#(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. max/plus interpretations on natural numbers:
    
      a__p#_A(x1) = x1
      s_A(x1) = x1 + 6
      mark#_A(x1) = x1 + 6
      p_A(x1) = x1 + 6
      mark_A(x1) = x1 + 12
      a__f_A(x1) = 10
      |0|_A = 7
      cons_A(x1,x2) = max{7, x1 + 2, x2 - 3}
      f_A(x1) = 10
      a__p_A(x1) = x1 + 6
    
    2. max/plus interpretations on natural numbers:
    
      a__p#_A(x1) = max{3, x1 + 1}
      s_A(x1) = x1 + 2
      mark#_A(x1) = x1 + 2
      p_A(x1) = max{4, x1}
      mark_A(x1) = x1
      a__f_A(x1) = 4
      |0|_A = 5
      cons_A(x1,x2) = 5
      f_A(x1) = 2
      a__p_A(x1) = max{4, x1}
    

The next rules are strictly ordered:

  p1, p2

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