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. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        a__f#_A(x1) = x1 + (1,0)
        s_A(x1) = x1 + (4,0)
        |0|_A() = (1,1)
        a__p_A(x1) = x1 + (0,1)
        a__p#_A(x1) = x1
        mark#_A(x1) = x1 + (3,0)
        cons_A(x1,x2) = x1 + (0,1)
        p_A(x1) = x1 + (0,1)
        mark_A(x1) = x1 + (1,1)
        f_A(x1) = x1 + (0,1)
        a__f_A(x1) = x1 + (0,1)
    
    2. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        a__f#_A(x1) = x1 + (5,0)
        s_A(x1) = (3,4)
        |0|_A() = (1,8)
        a__p_A(x1) = ((0,0),(1,0)) x1 + (2,1)
        a__p#_A(x1) = ((0,1),(0,0)) x1 + (0,8)
        mark#_A(x1) = (6,7)
        cons_A(x1,x2) = x1
        p_A(x1) = (1,1)
        mark_A(x1) = (4,7)
        f_A(x1) = x1
        a__f_A(x1) = x1
    
    3. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        a__f#_A(x1) = (1,0)
        s_A(x1) = (1,1)
        |0|_A() = (4,1)
        a__p_A(x1) = (2,4)
        a__p#_A(x1) = (0,0)
        mark#_A(x1) = (2,0)
        cons_A(x1,x2) = ((0,1),(1,0)) x1
        p_A(x1) = (3,5)
        mark_A(x1) = (3,3)
        f_A(x1) = ((0,0),(1,0)) x1 + (1,0)
        a__f_A(x1) = ((0,0),(1,0)) x1 + (2,0)
    

The next rules are strictly ordered:

  p2, p3, p4, p7, 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: mark#(cons(X1,X2)) -> mark#(X1)
p3: mark#(p(X)) -> mark#(X)
p4: mark#(f(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}


-- 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. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        a__f#_A(x1) = x1
        s_A(x1) = x1 + (6,7)
        |0|_A() = (5,1)
        a__p_A(x1) = ((0,1),(0,1)) x1 + (2,0)
        a__f_A(x1) = (8,3)
        cons_A(x1,x2) = x1 + ((0,1),(0,1)) x2 + (1,1)
        f_A(x1) = (1,1)
        mark_A(x1) = ((0,1),(0,1)) x1 + (8,2)
        p_A(x1) = x1 + (1,0)
    
    2. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        a__f#_A(x1) = x1
        s_A(x1) = x1 + (2,1)
        |0|_A() = (2,2)
        a__p_A(x1) = (2,4)
        a__f_A(x1) = (3,0)
        cons_A(x1,x2) = x1 + (0,1)
        f_A(x1) = (4,1)
        mark_A(x1) = (1,1)
        p_A(x1) = (3,0)
    
    3. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        a__f#_A(x1) = (0,0)
        s_A(x1) = (1,6)
        |0|_A() = (6,6)
        a__p_A(x1) = (5,6)
        a__f_A(x1) = (2,6)
        cons_A(x1,x2) = ((0,0),(1,0)) x1 + (3,1)
        f_A(x1) = (1,7)
        mark_A(x1) = (4,5)
        p_A(x1) = (6,7)
    

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: mark#(cons(X1,X2)) -> mark#(X1)
p2: mark#(f(X)) -> mark#(X)
p3: 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 monotone reduction pair:

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

The next rules are strictly ordered:

  p1, p2, p3
  r1, r2, r3, r4, r5, r6, r7, r8, r9, r10

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