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:
    
        matrix interpretations:
        
          carrier: N^2
          order: standard order
          interpretations:
            a__2nd#_A(x1) = ((0,1),(0,0)) x1 + (1,6)
            cons_A(x1,x2) = ((0,1),(0,1)) x1 + ((1,0),(1,0)) x2 + (11,6)
            mark#_A(x1) = ((0,1),(0,0)) x1 + (4,6)
            s_A(x1) = ((0,0),(0,1)) x1 + (0,6)
            from_A(x1) = ((0,0),(0,1)) x1 + (11,18)
            a__from#_A(x1) = ((0,1),(0,0)) x1 + (5,6)
            mark_A(x1) = ((0,1),(0,1)) x1 + (10,1)
            |2nd|_A(x1) = ((0,0),(0,1)) x1 + (3,5)
            a__2nd_A(x1) = x1 + (4,5)
            a__from_A(x1) = ((0,1),(0,1)) x1 + (24,18)
    
      precedence:
      
        a__from# > mark > a__2nd > |2nd| > a__from > cons > from > a__2nd# = mark# = s
      
      partial status:
    
        pi(a__2nd#) = []
        pi(cons) = []
        pi(mark#) = []
        pi(s) = []
        pi(from) = []
        pi(a__from#) = []
        pi(mark) = []
        pi(|2nd|) = []
        pi(a__2nd) = [1]
        pi(a__from) = []
    
    2. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: standard order
          interpretations:
            a__2nd#_A(x1) = (2,0)
            cons_A(x1,x2) = (0,0)
            mark#_A(x1) = (2,0)
            s_A(x1) = (0,0)
            from_A(x1) = (1,0)
            a__from#_A(x1) = (0,0)
            mark_A(x1) = (0,0)
            |2nd|_A(x1) = (0,0)
            a__2nd_A(x1) = (0,0)
            a__from_A(x1) = (2,0)
    
      precedence:
      
        a__from# > s = mark = a__2nd > |2nd| > a__from > a__2nd# = cons = mark# = from
      
      partial status:
    
        pi(a__2nd#) = []
        pi(cons) = []
        pi(mark#) = []
        pi(s) = []
        pi(from) = []
        pi(a__from#) = []
        pi(mark) = []
        pi(|2nd|) = []
        pi(a__2nd) = []
        pi(a__from) = []
    

The next rules are strictly ordered:

  p6

We remove them from the problem.

-- SCC decomposition.

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: mark#(|2nd|(X)) -> mark#(X)
p7: 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 estimated dependency graph contains the following SCCs:

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


-- 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#(|2nd|(X)) -> a__2nd#(mark(X))
p3: mark#(|2nd|(X)) -> mark#(X)
p4: mark#(from(X)) -> mark#(X)
p5: mark#(cons(X1,X2)) -> mark#(X1)
p6: 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 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:
    
        matrix interpretations:
        
          carrier: N^2
          order: standard order
          interpretations:
            a__2nd#_A(x1) = x1 + (2,1)
            cons_A(x1,x2) = ((0,1),(0,1)) x1 + ((1,0),(1,0)) x2 + (5,0)
            mark#_A(x1) = ((0,1),(0,1)) x1 + (11,6)
            |2nd|_A(x1) = ((0,1),(0,1)) x1 + (2,0)
            mark_A(x1) = ((0,1),(0,1)) x1 + (8,0)
            from_A(x1) = ((0,0),(0,1)) x1
            s_A(x1) = ((0,0),(0,1)) x1 + (7,0)
            a__2nd_A(x1) = ((0,1),(0,1)) x1 + (3,0)
            a__from_A(x1) = ((0,1),(0,1)) x1 + (6,0)
    
      precedence:
      
        a__2nd > mark > a__2nd# = mark# = |2nd| > a__from > cons > from = s
      
      partial status:
    
        pi(a__2nd#) = [1]
        pi(cons) = []
        pi(mark#) = []
        pi(|2nd|) = []
        pi(mark) = []
        pi(from) = []
        pi(s) = []
        pi(a__2nd) = []
        pi(a__from) = []
    
    2. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: standard order
          interpretations:
            a__2nd#_A(x1) = ((1,0),(1,1)) x1 + (2,1)
            cons_A(x1,x2) = (0,1)
            mark#_A(x1) = (1,2)
            |2nd|_A(x1) = (2,3)
            mark_A(x1) = (3,3)
            from_A(x1) = (3,3)
            s_A(x1) = (2,2)
            a__2nd_A(x1) = (3,3)
            a__from_A(x1) = (3,3)
    
      precedence:
      
        cons = |2nd| = mark = from = a__2nd = a__from > s > a__2nd# = mark#
      
      partial status:
    
        pi(a__2nd#) = [1]
        pi(cons) = []
        pi(mark#) = []
        pi(|2nd|) = []
        pi(mark) = []
        pi(from) = []
        pi(s) = []
        pi(a__2nd) = []
        pi(a__from) = []
    

The next rules are strictly ordered:

  p2

We remove them from the problem.

-- SCC decomposition.

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

p1: a__2nd#(cons(X,cons(Y,Z))) -> mark#(Y)
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)

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:

  {p2, p3, p4, p5}


-- Reduction pair.

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

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

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

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#(cons(X1,X2)) -> mark#(X1)

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}


-- Reduction pair.

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

p1: mark#(cons(X1,X2)) -> mark#(X1)

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

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

The next rules are strictly ordered:

  p1

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