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:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            a__f#_A(x1) = (10,1)
            s_A(x1) = x1 + (8,0)
            |0|_A() = (1,5)
            a__p_A(x1) = ((1,0),(0,0)) x1 + (0,2)
            a__p#_A(x1) = ((1,0),(1,1)) x1 + (0,1)
            mark#_A(x1) = ((1,0),(0,0)) x1 + (8,8)
            cons_A(x1,x2) = x1 + (1,1)
            p_A(x1) = ((1,0),(0,0)) x1 + (0,1)
            mark_A(x1) = ((1,0),(1,1)) x1 + (3,3)
            f_A(x1) = ((1,0),(0,0)) x1 + (2,4)
            a__f_A(x1) = ((1,0),(0,0)) x1 + (2,7)
    
      precedence:
      
        |0| = a__p = mark > a__f > cons > a__f# > s = a__p# = mark# = p = f
      
      partial status:
    
        pi(a__f#) = []
        pi(s) = [1]
        pi(|0|) = []
        pi(a__p) = []
        pi(a__p#) = [1]
        pi(mark#) = []
        pi(cons) = [1]
        pi(p) = []
        pi(mark) = [1]
        pi(f) = []
        pi(a__f) = []
    
    2. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            a__f#_A(x1) = (4,2)
            s_A(x1) = (1,7)
            |0|_A() = (2,10)
            a__p_A(x1) = (3,11)
            a__p#_A(x1) = (5,1)
            mark#_A(x1) = (6,3)
            cons_A(x1,x2) = (1,1)
            p_A(x1) = (0,6)
            mark_A(x1) = x1 + (3,5)
            f_A(x1) = (7,4)
            a__f_A(x1) = (7,8)
    
      precedence:
      
        a__p = mark > a__f > s = |0| = cons > mark# = f > a__f# > a__p# = p
      
      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) = []
    

The next rules are strictly ordered:

  p2, p3, 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#(s(X)) -> mark#(X)
p3: mark#(cons(X1,X2)) -> mark#(X1)
p4: mark#(p(X)) -> mark#(X)
p5: 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, p5}


-- 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. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            a__f#_A(x1) = x1
            s_A(x1) = ((1,0),(0,0)) x1 + (3,3)
            |0|_A() = (2,2)
            a__p_A(x1) = ((1,0),(0,0)) x1 + (0,3)
            a__f_A(x1) = x1
            cons_A(x1,x2) = (1,1)
            f_A(x1) = x1
            mark_A(x1) = ((1,0),(1,1)) x1 + (2,4)
            p_A(x1) = ((1,0),(0,0)) x1 + (0,2)
    
      precedence:
      
        s > |0| > a__f# = a__p = cons = mark > a__f > f = p
      
      partial status:
    
        pi(a__f#) = [1]
        pi(s) = []
        pi(|0|) = []
        pi(a__p) = []
        pi(a__f) = [1]
        pi(cons) = []
        pi(f) = [1]
        pi(mark) = [1]
        pi(p) = []
    
    2. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            a__f#_A(x1) = (10,4)
            s_A(x1) = (9,3)
            |0|_A() = (11,7)
            a__p_A(x1) = (8,2)
            a__f_A(x1) = (7,6)
            cons_A(x1,x2) = (7,3)
            f_A(x1) = (6,5)
            mark_A(x1) = ((1,0),(0,0)) x1 + (5,4)
            p_A(x1) = (4,1)
    
      precedence:
      
        mark > cons > a__p > p > |0| > a__f# = s = a__f > f
      
      partial status:
    
        pi(a__f#) = []
        pi(s) = []
        pi(|0|) = []
        pi(a__p) = []
        pi(a__f) = []
        pi(cons) = []
        pi(f) = []
        pi(mark) = []
        pi(p) = []
    

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

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:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            mark#_A(x1) = ((1,0),(1,0)) x1 + (1,2)
            s_A(x1) = ((1,0),(1,1)) x1 + (2,3)
            f_A(x1) = x1 + (2,1)
            p_A(x1) = x1
            cons_A(x1,x2) = x1 + x2 + (2,1)
    
      precedence:
      
        f = p = cons > mark# = s
      
      partial status:
    
        pi(mark#) = []
        pi(s) = [1]
        pi(f) = [1]
        pi(p) = [1]
        pi(cons) = [2]
    
    2. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            mark#_A(x1) = (0,0)
            s_A(x1) = (1,1)
            f_A(x1) = ((1,0),(0,0)) x1 + (1,1)
            p_A(x1) = ((1,0),(0,0)) x1 + (1,1)
            cons_A(x1,x2) = x2 + (1,1)
    
      precedence:
      
        mark# = s = f = p = cons
      
      partial status:
    
        pi(mark#) = []
        pi(s) = []
        pi(f) = []
        pi(p) = []
        pi(cons) = []
    

The next rules are strictly ordered:

  p1, p2, p4

We remove them from the problem.

-- SCC decomposition.

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 estimated dependency graph contains the following SCCs:

  {p1}


-- 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:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            mark#_A(x1) = ((1,0),(1,1)) x1 + (2,2)
            p_A(x1) = ((1,0),(0,0)) x1 + (1,1)
    
      precedence:
      
        mark# = p
      
      partial status:
    
        pi(mark#) = []
        pi(p) = []
    
    2. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            mark#_A(x1) = ((1,0),(0,0)) x1
            p_A(x1) = (1,1)
    
      precedence:
      
        p > mark#
      
      partial status:
    
        pi(mark#) = []
        pi(p) = []
    

The next rules are strictly ordered:

  p1

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