YES

We show the termination of the TRS R:

  a__2nd(cons(X,cons(Y,Z))) -> mark(Y)
  a__from(X) -> cons(mark(X),from(s(X)))
  mark(|2nd|(X)) -> a__2nd(mark(X))
  mark(from(X)) -> a__from(mark(X))
  mark(cons(X1,X2)) -> cons(mark(X1),X2)
  mark(s(X)) -> s(mark(X))
  a__2nd(X) -> |2nd|(X)
  a__from(X) -> from(X)

-- SCC decomposition.

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

p1: a__2nd#(cons(X,cons(Y,Z))) -> mark#(Y)
p2: a__from#(X) -> mark#(X)
p3: mark#(|2nd|(X)) -> a__2nd#(mark(X))
p4: mark#(|2nd|(X)) -> mark#(X)
p5: mark#(from(X)) -> a__from#(mark(X))
p6: mark#(from(X)) -> mark#(X)
p7: mark#(cons(X1,X2)) -> mark#(X1)
p8: mark#(s(X)) -> mark#(X)

and R consists of:

r1: a__2nd(cons(X,cons(Y,Z))) -> mark(Y)
r2: a__from(X) -> cons(mark(X),from(s(X)))
r3: mark(|2nd|(X)) -> a__2nd(mark(X))
r4: mark(from(X)) -> a__from(mark(X))
r5: mark(cons(X1,X2)) -> cons(mark(X1),X2)
r6: mark(s(X)) -> s(mark(X))
r7: a__2nd(X) -> |2nd|(X)
r8: a__from(X) -> from(X)

The estimated dependency graph contains the following SCCs:

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


-- Reduction pair.

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

p1: a__2nd#(cons(X,cons(Y,Z))) -> mark#(Y)
p2: mark#(s(X)) -> mark#(X)
p3: mark#(cons(X1,X2)) -> mark#(X1)
p4: mark#(from(X)) -> mark#(X)
p5: mark#(from(X)) -> a__from#(mark(X))
p6: a__from#(X) -> mark#(X)
p7: mark#(|2nd|(X)) -> mark#(X)
p8: mark#(|2nd|(X)) -> a__2nd#(mark(X))

and R consists of:

r1: a__2nd(cons(X,cons(Y,Z))) -> mark(Y)
r2: a__from(X) -> cons(mark(X),from(s(X)))
r3: mark(|2nd|(X)) -> a__2nd(mark(X))
r4: mark(from(X)) -> a__from(mark(X))
r5: mark(cons(X1,X2)) -> cons(mark(X1),X2)
r6: mark(s(X)) -> s(mark(X))
r7: a__2nd(X) -> |2nd|(X)
r8: a__from(X) -> from(X)

The set of usable rules consists of

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

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          a__2nd#_A(x1) = max{7, x1 + 1}
          cons_A(x1,x2) = max{7, x1 + 5, x2}
          mark#_A(x1) = x1
          s_A(x1) = max{1, x1}
          from_A(x1) = max{14, x1 + 12}
          a__from#_A(x1) = max{14, x1 + 6}
          mark_A(x1) = max{6, x1 + 4}
          |2nd|_A(x1) = max{10, x1 + 8}
          a__2nd_A(x1) = max{11, x1 + 8}
          a__from_A(x1) = max{15, x1 + 12}
    
      precedence:
      
        a__2nd# = mark = a__2nd > s = |2nd| = a__from > from > cons = mark# = a__from#
      
      partial status:
    
        pi(a__2nd#) = [1]
        pi(cons) = [1, 2]
        pi(mark#) = [1]
        pi(s) = [1]
        pi(from) = [1]
        pi(a__from#) = [1]
        pi(mark) = [1]
        pi(|2nd|) = [1]
        pi(a__2nd) = []
        pi(a__from) = []
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          a__2nd#_A(x1) = max{8, x1 - 1}
          cons_A(x1,x2) = max{x1 + 28, x2 + 13}
          mark#_A(x1) = x1 + 21
          s_A(x1) = max{56, x1 + 28}
          from_A(x1) = max{29, x1 + 10}
          a__from#_A(x1) = max{30, x1 + 22}
          mark_A(x1) = max{27, x1}
          |2nd|_A(x1) = max{5, x1 + 4}
          a__2nd_A(x1) = 6
          a__from_A(x1) = 11
    
      precedence:
      
        a__2nd# = cons = mark# = s = from = a__from# = mark = |2nd| = a__2nd = a__from
      
      partial status:
    
        pi(a__2nd#) = []
        pi(cons) = [1, 2]
        pi(mark#) = [1]
        pi(s) = [1]
        pi(from) = [1]
        pi(a__from#) = [1]
        pi(mark) = [1]
        pi(|2nd|) = [1]
        pi(a__2nd) = []
        pi(a__from) = []
    

The next rules are strictly ordered:

  p1, p2, p3, p4, p5, p6, p7, p8

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