YES

We show the termination of the TRS R:

  a__from(X) -> cons(mark(X),from(s(X)))
  a__sel(|0|(),cons(X,Y)) -> mark(X)
  a__sel(s(X),cons(Y,Z)) -> a__sel(mark(X),mark(Z))
  mark(from(X)) -> a__from(mark(X))
  mark(sel(X1,X2)) -> a__sel(mark(X1),mark(X2))
  mark(cons(X1,X2)) -> cons(mark(X1),X2)
  mark(s(X)) -> s(mark(X))
  mark(|0|()) -> |0|()
  a__from(X) -> from(X)
  a__sel(X1,X2) -> sel(X1,X2)

-- SCC decomposition.

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

p1: a__from#(X) -> mark#(X)
p2: a__sel#(|0|(),cons(X,Y)) -> mark#(X)
p3: a__sel#(s(X),cons(Y,Z)) -> a__sel#(mark(X),mark(Z))
p4: a__sel#(s(X),cons(Y,Z)) -> mark#(X)
p5: a__sel#(s(X),cons(Y,Z)) -> mark#(Z)
p6: mark#(from(X)) -> a__from#(mark(X))
p7: mark#(from(X)) -> mark#(X)
p8: mark#(sel(X1,X2)) -> a__sel#(mark(X1),mark(X2))
p9: mark#(sel(X1,X2)) -> mark#(X1)
p10: mark#(sel(X1,X2)) -> mark#(X2)
p11: mark#(cons(X1,X2)) -> mark#(X1)
p12: mark#(s(X)) -> mark#(X)

and R consists of:

r1: a__from(X) -> cons(mark(X),from(s(X)))
r2: a__sel(|0|(),cons(X,Y)) -> mark(X)
r3: a__sel(s(X),cons(Y,Z)) -> a__sel(mark(X),mark(Z))
r4: mark(from(X)) -> a__from(mark(X))
r5: mark(sel(X1,X2)) -> a__sel(mark(X1),mark(X2))
r6: mark(cons(X1,X2)) -> cons(mark(X1),X2)
r7: mark(s(X)) -> s(mark(X))
r8: mark(|0|()) -> |0|()
r9: a__from(X) -> from(X)
r10: a__sel(X1,X2) -> sel(X1,X2)

The estimated dependency graph contains the following SCCs:

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


-- Reduction pair.

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

p1: a__from#(X) -> mark#(X)
p2: mark#(s(X)) -> mark#(X)
p3: mark#(cons(X1,X2)) -> mark#(X1)
p4: mark#(sel(X1,X2)) -> mark#(X2)
p5: mark#(sel(X1,X2)) -> mark#(X1)
p6: mark#(sel(X1,X2)) -> a__sel#(mark(X1),mark(X2))
p7: a__sel#(s(X),cons(Y,Z)) -> mark#(Z)
p8: mark#(from(X)) -> mark#(X)
p9: mark#(from(X)) -> a__from#(mark(X))
p10: a__sel#(s(X),cons(Y,Z)) -> mark#(X)
p11: a__sel#(s(X),cons(Y,Z)) -> a__sel#(mark(X),mark(Z))
p12: a__sel#(|0|(),cons(X,Y)) -> mark#(X)

and R consists of:

r1: a__from(X) -> cons(mark(X),from(s(X)))
r2: a__sel(|0|(),cons(X,Y)) -> mark(X)
r3: a__sel(s(X),cons(Y,Z)) -> a__sel(mark(X),mark(Z))
r4: mark(from(X)) -> a__from(mark(X))
r5: mark(sel(X1,X2)) -> a__sel(mark(X1),mark(X2))
r6: mark(cons(X1,X2)) -> cons(mark(X1),X2)
r7: mark(s(X)) -> s(mark(X))
r8: mark(|0|()) -> |0|()
r9: a__from(X) -> from(X)
r10: a__sel(X1,X2) -> sel(X1,X2)

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__from#_A(x1) = max{2, x1 + 1}
          mark#_A(x1) = max{0, x1 - 2}
          s_A(x1) = x1
          cons_A(x1,x2) = max{x1 + 5, x2}
          sel_A(x1,x2) = max{10, x1 + 9, x2 + 5}
          a__sel#_A(x1,x2) = max{x1 + 2, x2 - 1}
          mark_A(x1) = x1
          from_A(x1) = max{6, x1 + 5}
          |0|_A = 1
          a__from_A(x1) = max{6, x1 + 5}
          a__sel_A(x1,x2) = max{10, x1 + 9, x2 + 5}
    
      precedence:
      
        a__from# = mark# = s = cons = sel = a__sel# = mark = from = |0| = a__from = a__sel
      
      partial status:
    
        pi(a__from#) = [1]
        pi(mark#) = []
        pi(s) = []
        pi(cons) = []
        pi(sel) = []
        pi(a__sel#) = [1]
        pi(mark) = []
        pi(from) = []
        pi(|0|) = []
        pi(a__from) = []
        pi(a__sel) = []
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          a__from#_A(x1) = 7
          mark#_A(x1) = 7
          s_A(x1) = 4
          cons_A(x1,x2) = 4
          sel_A(x1,x2) = 0
          a__sel#_A(x1,x2) = 4
          mark_A(x1) = 4
          from_A(x1) = 4
          |0|_A = 4
          a__from_A(x1) = 4
          a__sel_A(x1,x2) = 4
    
      precedence:
      
        mark = |0| = a__sel > a__from > from > cons = sel > a__from# = mark# > s = a__sel#
      
      partial status:
    
        pi(a__from#) = []
        pi(mark#) = []
        pi(s) = []
        pi(cons) = []
        pi(sel) = []
        pi(a__sel#) = []
        pi(mark) = []
        pi(from) = []
        pi(|0|) = []
        pi(a__from) = []
        pi(a__sel) = []
    

The next rules are strictly ordered:

  p7, p10, p12

We remove them from the problem.

-- SCC decomposition.

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

p1: a__from#(X) -> mark#(X)
p2: mark#(s(X)) -> mark#(X)
p3: mark#(cons(X1,X2)) -> mark#(X1)
p4: mark#(sel(X1,X2)) -> mark#(X2)
p5: mark#(sel(X1,X2)) -> mark#(X1)
p6: mark#(sel(X1,X2)) -> a__sel#(mark(X1),mark(X2))
p7: mark#(from(X)) -> mark#(X)
p8: mark#(from(X)) -> a__from#(mark(X))
p9: a__sel#(s(X),cons(Y,Z)) -> a__sel#(mark(X),mark(Z))

and R consists of:

r1: a__from(X) -> cons(mark(X),from(s(X)))
r2: a__sel(|0|(),cons(X,Y)) -> mark(X)
r3: a__sel(s(X),cons(Y,Z)) -> a__sel(mark(X),mark(Z))
r4: mark(from(X)) -> a__from(mark(X))
r5: mark(sel(X1,X2)) -> a__sel(mark(X1),mark(X2))
r6: mark(cons(X1,X2)) -> cons(mark(X1),X2)
r7: mark(s(X)) -> s(mark(X))
r8: mark(|0|()) -> |0|()
r9: a__from(X) -> from(X)
r10: a__sel(X1,X2) -> sel(X1,X2)

The estimated dependency graph contains the following SCCs:

  {p1, p2, p3, p4, p5, p7, p8}
  {p9}


-- Reduction pair.

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

p1: a__from#(X) -> mark#(X)
p2: mark#(from(X)) -> a__from#(mark(X))
p3: mark#(from(X)) -> mark#(X)
p4: mark#(sel(X1,X2)) -> mark#(X1)
p5: mark#(sel(X1,X2)) -> mark#(X2)
p6: mark#(cons(X1,X2)) -> mark#(X1)
p7: mark#(s(X)) -> mark#(X)

and R consists of:

r1: a__from(X) -> cons(mark(X),from(s(X)))
r2: a__sel(|0|(),cons(X,Y)) -> mark(X)
r3: a__sel(s(X),cons(Y,Z)) -> a__sel(mark(X),mark(Z))
r4: mark(from(X)) -> a__from(mark(X))
r5: mark(sel(X1,X2)) -> a__sel(mark(X1),mark(X2))
r6: mark(cons(X1,X2)) -> cons(mark(X1),X2)
r7: mark(s(X)) -> s(mark(X))
r8: mark(|0|()) -> |0|()
r9: a__from(X) -> from(X)
r10: a__sel(X1,X2) -> sel(X1,X2)

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__from#_A(x1) = x1 + 7
          mark#_A(x1) = x1 + 6
          from_A(x1) = x1 + 6
          mark_A(x1) = x1
          sel_A(x1,x2) = max{x1 + 4, x2 + 2}
          cons_A(x1,x2) = max{x1 + 6, x2}
          s_A(x1) = x1
          a__from_A(x1) = x1 + 6
          a__sel_A(x1,x2) = max{x1 + 4, x2 + 2}
          |0|_A = 1
    
      precedence:
      
        a__from# = mark# = from = mark = sel = cons = s = a__from = a__sel = |0|
      
      partial status:
    
        pi(a__from#) = []
        pi(mark#) = []
        pi(from) = []
        pi(mark) = []
        pi(sel) = []
        pi(cons) = []
        pi(s) = []
        pi(a__from) = []
        pi(a__sel) = []
        pi(|0|) = []
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          a__from#_A(x1) = max{56, x1 + 51}
          mark#_A(x1) = max{34, x1 + 28}
          from_A(x1) = x1 + 27
          mark_A(x1) = x1 + 12
          sel_A(x1,x2) = 11
          cons_A(x1,x2) = 1
          s_A(x1) = x1 + 13
          a__from_A(x1) = max{28, x1 + 27}
          a__sel_A(x1,x2) = 12
          |0|_A = 6
    
      precedence:
      
        a__from# = mark# > cons = a__sel > sel > mark = s = a__from > |0| > from
      
      partial status:
    
        pi(a__from#) = []
        pi(mark#) = []
        pi(from) = [1]
        pi(mark) = [1]
        pi(sel) = []
        pi(cons) = []
        pi(s) = [1]
        pi(a__from) = []
        pi(a__sel) = []
        pi(|0|) = []
    

The next rules are strictly ordered:

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

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

-- Reduction pair.

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

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

and R consists of:

r1: a__from(X) -> cons(mark(X),from(s(X)))
r2: a__sel(|0|(),cons(X,Y)) -> mark(X)
r3: a__sel(s(X),cons(Y,Z)) -> a__sel(mark(X),mark(Z))
r4: mark(from(X)) -> a__from(mark(X))
r5: mark(sel(X1,X2)) -> a__sel(mark(X1),mark(X2))
r6: mark(cons(X1,X2)) -> cons(mark(X1),X2)
r7: mark(s(X)) -> s(mark(X))
r8: mark(|0|()) -> |0|()
r9: a__from(X) -> from(X)
r10: a__sel(X1,X2) -> sel(X1,X2)

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__sel#_A(x1,x2) = max{x1 + 2, x2 + 2}
          s_A(x1) = x1
          cons_A(x1,x2) = max{4, x1 + 3, x2}
          mark_A(x1) = x1
          a__from_A(x1) = max{6, x1 + 5}
          from_A(x1) = max{6, x1 + 5}
          a__sel_A(x1,x2) = max{1, x2 - 1}
          |0|_A = 2
          sel_A(x1,x2) = max{1, x2 - 1}
    
      precedence:
      
        a__from = from > a__sel# = a__sel = sel > cons > s = mark = |0|
      
      partial status:
    
        pi(a__sel#) = [1]
        pi(s) = [1]
        pi(cons) = []
        pi(mark) = [1]
        pi(a__from) = []
        pi(from) = []
        pi(a__sel) = []
        pi(|0|) = []
        pi(sel) = []
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          a__sel#_A(x1,x2) = max{1, x1}
          s_A(x1) = x1 + 5
          cons_A(x1,x2) = 1
          mark_A(x1) = x1
          a__from_A(x1) = 2
          from_A(x1) = 2
          a__sel_A(x1,x2) = 2
          |0|_A = 1
          sel_A(x1,x2) = 1
    
      precedence:
      
        s = a__sel > a__sel# = cons = mark = a__from = |0| > from > sel
      
      partial status:
    
        pi(a__sel#) = [1]
        pi(s) = []
        pi(cons) = []
        pi(mark) = [1]
        pi(a__from) = []
        pi(from) = []
        pi(a__sel) = []
        pi(|0|) = []
        pi(sel) = []
    

The next rules are strictly ordered:

  p1

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