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:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        a__2nd#_A(x1) = max{14, x1 + 9}
        cons_A(x1,x2) = max{24, x1, x2 - 10}
        mark#_A(x1) = max{13, x1 - 2}
        s_A(x1) = max{4, x1 + 3}
        from_A(x1) = x1 + 30
        a__from#_A(x1) = max{28, x1 + 27}
        mark_A(x1) = x1
        |2nd|_A(x1) = x1 + 16
        a__2nd_A(x1) = x1 + 16
        a__from_A(x1) = x1 + 30
  
    precedence:
    
      a__2nd# = cons = mark# = s = mark > a__from# > |2nd| = a__2nd > from = a__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:

  p5

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: a__from#(X) -> 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:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        a__2nd#_A(x1) = x1 + 9
        cons_A(x1,x2) = max{x1 + 11, x2 - 5}
        mark#_A(x1) = x1 + 5
        |2nd|_A(x1) = x1 + 4
        mark_A(x1) = x1
        from_A(x1) = x1 + 12
        s_A(x1) = x1 + 5
        a__2nd_A(x1) = x1 + 4
        a__from_A(x1) = x1 + 12
  
    precedence:
    
      a__2nd# = cons = mark# = |2nd| = mark = from = s = a__2nd = a__from
    
    partial status:
  
      pi(a__2nd#) = []
      pi(cons) = []
      pi(mark#) = [1]
      pi(|2nd|) = [1]
      pi(mark) = [1]
      pi(from) = []
      pi(s) = [1]
      pi(a__2nd) = [1]
      pi(a__from) = []

The next rules are strictly ordered:

  p4

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)) -> a__2nd#(mark(X))
p3: mark#(|2nd|(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:

  {p1, p2, p3, p4, p5}


-- 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#(|2nd|(X)) -> mark#(X)
p5: 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:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        a__2nd#_A(x1) = max{23, x1 + 22}
        cons_A(x1,x2) = max{x1 + 15, x2 - 8}
        mark#_A(x1) = x1 + 8
        s_A(x1) = x1 + 8
        |2nd|_A(x1) = x1 + 17
        mark_A(x1) = x1
        a__2nd_A(x1) = x1 + 17
        a__from_A(x1) = x1 + 15
        from_A(x1) = x1 + 15
  
    precedence:
    
      a__2nd# = cons = mark# = s = |2nd| = mark = a__2nd = a__from = from
    
    partial status:
  
      pi(a__2nd#) = [1]
      pi(cons) = []
      pi(mark#) = [1]
      pi(s) = [1]
      pi(|2nd|) = []
      pi(mark) = [1]
      pi(a__2nd) = [1]
      pi(a__from) = []
      pi(from) = []

The next rules are strictly ordered:

  p1

We remove them from the problem.

-- SCC decomposition.

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

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


-- Reduction pair.

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

p1: mark#(s(X)) -> mark#(X)
p2: mark#(|2nd|(X)) -> mark#(X)
p3: 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:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        mark#_A(x1) = max{4, x1 + 2}
        s_A(x1) = max{3, x1 + 1}
        |2nd|_A(x1) = x1 + 2
        cons_A(x1,x2) = max{x1, x2}
  
    precedence:
    
      s = |2nd| > mark# = cons
    
    partial status:
  
      pi(mark#) = []
      pi(s) = [1]
      pi(|2nd|) = [1]
      pi(cons) = [2]

The next rules are strictly ordered:

  p1

We remove them from the problem.

-- SCC decomposition.

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

p1: mark#(|2nd|(X)) -> mark#(X)
p2: 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, p2}


-- Reduction pair.

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

p1: mark#(|2nd|(X)) -> mark#(X)
p2: 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 monotone reduction pair:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        mark#_A(x1) = x1 + 1
        |2nd|_A(x1) = x1
        cons_A(x1,x2) = max{x1, x2}
  
    precedence:
    
      mark# = |2nd| = cons
    
    partial status:
  
      pi(mark#) = [1]
      pi(|2nd|) = [1]
      pi(cons) = [1, 2]

The next rules are strictly ordered:

  p1

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:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        mark#_A(x1) = x1 + 2
        cons_A(x1,x2) = max{x1 + 1, x2 + 1}
  
    precedence:
    
      mark# = cons
    
    partial status:
  
      pi(mark#) = []
      pi(cons) = [2]

The next rules are strictly ordered:

  p1

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