YES

We show the termination of the TRS R:

  a__2nd(cons1(X,cons(Y,Z))) -> mark(Y)
  a__2nd(cons(X,X1)) -> a__2nd(cons1(mark(X),mark(X1)))
  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))
  mark(cons1(X1,X2)) -> cons1(mark(X1),mark(X2))
  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#(cons1(X,cons(Y,Z))) -> mark#(Y)
p2: a__2nd#(cons(X,X1)) -> a__2nd#(cons1(mark(X),mark(X1)))
p3: a__2nd#(cons(X,X1)) -> mark#(X)
p4: a__2nd#(cons(X,X1)) -> mark#(X1)
p5: a__from#(X) -> mark#(X)
p6: mark#(|2nd|(X)) -> a__2nd#(mark(X))
p7: mark#(|2nd|(X)) -> mark#(X)
p8: mark#(from(X)) -> a__from#(mark(X))
p9: mark#(from(X)) -> mark#(X)
p10: mark#(cons(X1,X2)) -> mark#(X1)
p11: mark#(s(X)) -> mark#(X)
p12: mark#(cons1(X1,X2)) -> mark#(X1)
p13: mark#(cons1(X1,X2)) -> mark#(X2)

and R consists of:

r1: a__2nd(cons1(X,cons(Y,Z))) -> mark(Y)
r2: a__2nd(cons(X,X1)) -> a__2nd(cons1(mark(X),mark(X1)))
r3: a__from(X) -> cons(mark(X),from(s(X)))
r4: mark(|2nd|(X)) -> a__2nd(mark(X))
r5: mark(from(X)) -> a__from(mark(X))
r6: mark(cons(X1,X2)) -> cons(mark(X1),X2)
r7: mark(s(X)) -> s(mark(X))
r8: mark(cons1(X1,X2)) -> cons1(mark(X1),mark(X2))
r9: a__2nd(X) -> |2nd|(X)
r10: a__from(X) -> from(X)

The estimated dependency graph contains the following SCCs:

  {p1, p2, p3, p4, p5, p6, p7, p8, p9, p10, p11, p12, p13}


-- Reduction pair.

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

p1: a__2nd#(cons1(X,cons(Y,Z))) -> mark#(Y)
p2: mark#(cons1(X1,X2)) -> mark#(X2)
p3: mark#(cons1(X1,X2)) -> mark#(X1)
p4: mark#(s(X)) -> mark#(X)
p5: mark#(cons(X1,X2)) -> mark#(X1)
p6: mark#(from(X)) -> mark#(X)
p7: mark#(from(X)) -> a__from#(mark(X))
p8: a__from#(X) -> mark#(X)
p9: mark#(|2nd|(X)) -> mark#(X)
p10: mark#(|2nd|(X)) -> a__2nd#(mark(X))
p11: a__2nd#(cons(X,X1)) -> mark#(X1)
p12: a__2nd#(cons(X,X1)) -> mark#(X)
p13: a__2nd#(cons(X,X1)) -> a__2nd#(cons1(mark(X),mark(X1)))

and R consists of:

r1: a__2nd(cons1(X,cons(Y,Z))) -> mark(Y)
r2: a__2nd(cons(X,X1)) -> a__2nd(cons1(mark(X),mark(X1)))
r3: a__from(X) -> cons(mark(X),from(s(X)))
r4: mark(|2nd|(X)) -> a__2nd(mark(X))
r5: mark(from(X)) -> a__from(mark(X))
r6: mark(cons(X1,X2)) -> cons(mark(X1),X2)
r7: mark(s(X)) -> s(mark(X))
r8: mark(cons1(X1,X2)) -> cons1(mark(X1),mark(X2))
r9: a__2nd(X) -> |2nd|(X)
r10: a__from(X) -> from(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:
    
        max/plus interpretations on natural numbers:
        
          a__2nd#_A(x1) = max{4, x1 - 1}
          cons1_A(x1,x2) = max{x1, x2}
          cons_A(x1,x2) = max{x1 + 7, x2}
          mark#_A(x1) = max{0, x1 - 2}
          s_A(x1) = max{1, x1}
          from_A(x1) = max{9, x1 + 8}
          a__from#_A(x1) = max{5, x1 + 4}
          mark_A(x1) = x1
          |2nd|_A(x1) = max{6, x1 + 1}
          a__2nd_A(x1) = max{6, x1 + 1}
          a__from_A(x1) = max{9, x1 + 8}
    
      precedence:
      
        cons1 = s = mark = a__2nd = a__from > mark# > from = |2nd| > a__from# > a__2nd# = cons
      
      partial status:
    
        pi(a__2nd#) = []
        pi(cons1) = [2]
        pi(cons) = []
        pi(mark#) = []
        pi(s) = [1]
        pi(from) = [1]
        pi(a__from#) = [1]
        pi(mark) = [1]
        pi(|2nd|) = [1]
        pi(a__2nd) = []
        pi(a__from) = [1]
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          a__2nd#_A(x1) = 88
          cons1_A(x1,x2) = 30
          cons_A(x1,x2) = 34
          mark#_A(x1) = 88
          s_A(x1) = 32
          from_A(x1) = max{43, x1 + 30}
          a__from#_A(x1) = max{134, x1 + 104}
          mark_A(x1) = 43
          |2nd|_A(x1) = max{88, x1 + 57}
          a__2nd_A(x1) = 35
          a__from_A(x1) = 61
    
      precedence:
      
        a__2nd# = |2nd| = a__from > mark# = from = a__from# > mark > cons = s = a__2nd > cons1
      
      partial status:
    
        pi(a__2nd#) = []
        pi(cons1) = []
        pi(cons) = []
        pi(mark#) = []
        pi(s) = []
        pi(from) = [1]
        pi(a__from#) = [1]
        pi(mark) = []
        pi(|2nd|) = [1]
        pi(a__2nd) = []
        pi(a__from) = []
    

The next rules are strictly ordered:

  p1, p7, p8, p10, p11

We remove them from the problem.

-- SCC decomposition.

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

p1: mark#(cons1(X1,X2)) -> mark#(X2)
p2: mark#(cons1(X1,X2)) -> mark#(X1)
p3: mark#(s(X)) -> mark#(X)
p4: mark#(cons(X1,X2)) -> mark#(X1)
p5: mark#(from(X)) -> mark#(X)
p6: mark#(|2nd|(X)) -> mark#(X)
p7: a__2nd#(cons(X,X1)) -> mark#(X)
p8: a__2nd#(cons(X,X1)) -> a__2nd#(cons1(mark(X),mark(X1)))

and R consists of:

r1: a__2nd(cons1(X,cons(Y,Z))) -> mark(Y)
r2: a__2nd(cons(X,X1)) -> a__2nd(cons1(mark(X),mark(X1)))
r3: a__from(X) -> cons(mark(X),from(s(X)))
r4: mark(|2nd|(X)) -> a__2nd(mark(X))
r5: mark(from(X)) -> a__from(mark(X))
r6: mark(cons(X1,X2)) -> cons(mark(X1),X2)
r7: mark(s(X)) -> s(mark(X))
r8: mark(cons1(X1,X2)) -> cons1(mark(X1),mark(X2))
r9: a__2nd(X) -> |2nd|(X)
r10: a__from(X) -> from(X)

The estimated dependency graph contains the following SCCs:

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


-- Reduction pair.

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

p1: mark#(cons1(X1,X2)) -> mark#(X2)
p2: mark#(|2nd|(X)) -> mark#(X)
p3: mark#(from(X)) -> mark#(X)
p4: mark#(cons(X1,X2)) -> mark#(X1)
p5: mark#(s(X)) -> mark#(X)
p6: mark#(cons1(X1,X2)) -> mark#(X1)

and R consists of:

r1: a__2nd(cons1(X,cons(Y,Z))) -> mark(Y)
r2: a__2nd(cons(X,X1)) -> a__2nd(cons1(mark(X),mark(X1)))
r3: a__from(X) -> cons(mark(X),from(s(X)))
r4: mark(|2nd|(X)) -> a__2nd(mark(X))
r5: mark(from(X)) -> a__from(mark(X))
r6: mark(cons(X1,X2)) -> cons(mark(X1),X2)
r7: mark(s(X)) -> s(mark(X))
r8: mark(cons1(X1,X2)) -> cons1(mark(X1),mark(X2))
r9: a__2nd(X) -> |2nd|(X)
r10: a__from(X) -> from(X)

The set of usable rules consists of

  (no rules)

Take the monotone reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          mark#_A(x1) = max{2, x1 + 1}
          cons1_A(x1,x2) = max{x1, x2}
          |2nd|_A(x1) = max{1, x1}
          from_A(x1) = max{1, x1}
          cons_A(x1,x2) = max{x1, x2}
          s_A(x1) = max{1, x1}
    
      precedence:
      
        mark# = cons1 = |2nd| = from = cons = s
      
      partial status:
    
        pi(mark#) = [1]
        pi(cons1) = [1, 2]
        pi(|2nd|) = [1]
        pi(from) = [1]
        pi(cons) = [1, 2]
        pi(s) = [1]
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          mark#_A(x1) = x1 + 1
          cons1_A(x1,x2) = max{x1, x2}
          |2nd|_A(x1) = x1
          from_A(x1) = x1
          cons_A(x1,x2) = max{x1, x2}
          s_A(x1) = x1
    
      precedence:
      
        mark# = cons1 = |2nd| = from = cons = s
      
      partial status:
    
        pi(mark#) = [1]
        pi(cons1) = [1, 2]
        pi(|2nd|) = [1]
        pi(from) = [1]
        pi(cons) = [1, 2]
        pi(s) = [1]
    

The next rules are strictly ordered:

  p1, p2, p3, p4, p5, p6

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