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. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        a__2nd#_A(x1) = ((0,1),(0,0)) x1
        cons_A(x1,x2) = ((0,1),(0,1)) x1 + ((1,0),(1,0)) x2 + (2,0)
        mark#_A(x1) = ((0,1),(0,0)) x1 + (1,0)
        s_A(x1) = x1 + (1,1)
        from_A(x1) = x1
        a__from#_A(x1) = ((0,1),(0,0)) x1 + (1,0)
        mark_A(x1) = ((0,1),(0,1)) x1 + (4,0)
        |2nd|_A(x1) = x1 + (1,0)
        a__2nd_A(x1) = ((0,1),(0,1)) x1 + (3,0)
        a__from_A(x1) = ((0,1),(0,1)) x1 + (3,0)
    
    2. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        a__2nd#_A(x1) = (0,0)
        cons_A(x1,x2) = ((1,1),(1,0)) x2
        mark#_A(x1) = (1,1)
        s_A(x1) = (1,0)
        from_A(x1) = (3,3)
        a__from#_A(x1) = (1,1)
        mark_A(x1) = (0,1)
        |2nd|_A(x1) = (2,3)
        a__2nd_A(x1) = (1,2)
        a__from_A(x1) = (2,2)
    
    3. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        a__2nd#_A(x1) = (1,1)
        cons_A(x1,x2) = (1,3)
        mark#_A(x1) = (0,0)
        s_A(x1) = (3,2)
        from_A(x1) = (0,0)
        a__from#_A(x1) = (0,0)
        mark_A(x1) = (2,1)
        |2nd|_A(x1) = (4,3)
        a__2nd_A(x1) = (3,2)
        a__from_A(x1) = (3,2)
    

The next rules are strictly ordered:

  p1, p2, p8

We remove them from the problem.

-- SCC decomposition.

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

p1: mark#(cons(X1,X2)) -> mark#(X1)
p2: mark#(from(X)) -> mark#(X)
p3: mark#(from(X)) -> a__from#(mark(X))
p4: a__from#(X) -> mark#(X)
p5: mark#(|2nd|(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}


-- Reduction pair.

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

p1: mark#(cons(X1,X2)) -> mark#(X1)
p2: mark#(|2nd|(X)) -> mark#(X)
p3: mark#(from(X)) -> a__from#(mark(X))
p4: a__from#(X) -> mark#(X)
p5: mark#(from(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 set of usable rules consists of

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

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        mark#_A(x1) = x1
        cons_A(x1,x2) = x1 + ((1,1),(0,0)) x2 + (0,1)
        |2nd|_A(x1) = x1 + (1,1)
        from_A(x1) = x1 + (2,0)
        a__from#_A(x1) = x1 + (1,0)
        mark_A(x1) = x1 + (0,1)
        a__2nd_A(x1) = x1 + (1,1)
        a__from_A(x1) = x1 + (2,1)
        s_A(x1) = (0,1)
    
    2. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        mark#_A(x1) = ((1,1),(1,1)) x1
        cons_A(x1,x2) = x1 + ((0,1),(1,0)) x2 + (1,1)
        |2nd|_A(x1) = ((0,1),(1,0)) x1 + (1,2)
        from_A(x1) = ((1,1),(1,1)) x1 + (7,5)
        a__from#_A(x1) = ((1,1),(1,1)) x1 + (1,0)
        mark_A(x1) = ((1,1),(1,1)) x1 + (1,1)
        a__2nd_A(x1) = ((0,1),(1,0)) x1 + (2,2)
        a__from_A(x1) = ((1,1),(1,1)) x1 + (10,11)
        s_A(x1) = (1,1)
    
    3. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        mark#_A(x1) = ((1,1),(1,1)) x1
        cons_A(x1,x2) = ((1,1),(0,1)) x1 + (1,1)
        |2nd|_A(x1) = (1,1)
        from_A(x1) = x1 + (1,1)
        a__from#_A(x1) = ((1,1),(1,1)) x1 + (2,2)
        mark_A(x1) = ((1,1),(0,0)) x1 + (1,0)
        a__2nd_A(x1) = (4,1)
        a__from_A(x1) = (0,0)
        s_A(x1) = (1,1)
    

The next rules are strictly ordered:

  p1, p2, p3, p4, p5

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