YES

We show the termination of the TRS R:

  a__from(X) -> cons(mark(X),from(s(X)))
  a__length(nil()) -> |0|()
  a__length(cons(X,Y)) -> s(a__length1(Y))
  a__length1(X) -> a__length(X)
  mark(from(X)) -> a__from(mark(X))
  mark(length(X)) -> a__length(X)
  mark(length1(X)) -> a__length1(X)
  mark(cons(X1,X2)) -> cons(mark(X1),X2)
  mark(s(X)) -> s(mark(X))
  mark(nil()) -> nil()
  mark(|0|()) -> |0|()
  a__from(X) -> from(X)
  a__length(X) -> length(X)
  a__length1(X) -> length1(X)

-- SCC decomposition.

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

p1: a__from#(X) -> mark#(X)
p2: a__length#(cons(X,Y)) -> a__length1#(Y)
p3: a__length1#(X) -> a__length#(X)
p4: mark#(from(X)) -> a__from#(mark(X))
p5: mark#(from(X)) -> mark#(X)
p6: mark#(length(X)) -> a__length#(X)
p7: mark#(length1(X)) -> a__length1#(X)
p8: mark#(cons(X1,X2)) -> mark#(X1)
p9: mark#(s(X)) -> mark#(X)

and R consists of:

r1: a__from(X) -> cons(mark(X),from(s(X)))
r2: a__length(nil()) -> |0|()
r3: a__length(cons(X,Y)) -> s(a__length1(Y))
r4: a__length1(X) -> a__length(X)
r5: mark(from(X)) -> a__from(mark(X))
r6: mark(length(X)) -> a__length(X)
r7: mark(length1(X)) -> a__length1(X)
r8: mark(cons(X1,X2)) -> cons(mark(X1),X2)
r9: mark(s(X)) -> s(mark(X))
r10: mark(nil()) -> nil()
r11: mark(|0|()) -> |0|()
r12: a__from(X) -> from(X)
r13: a__length(X) -> length(X)
r14: a__length1(X) -> length1(X)

The estimated dependency graph contains the following SCCs:

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


-- 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#(from(X)) -> mark#(X)
p5: mark#(from(X)) -> a__from#(mark(X))

and R consists of:

r1: a__from(X) -> cons(mark(X),from(s(X)))
r2: a__length(nil()) -> |0|()
r3: a__length(cons(X,Y)) -> s(a__length1(Y))
r4: a__length1(X) -> a__length(X)
r5: mark(from(X)) -> a__from(mark(X))
r6: mark(length(X)) -> a__length(X)
r7: mark(length1(X)) -> a__length1(X)
r8: mark(cons(X1,X2)) -> cons(mark(X1),X2)
r9: mark(s(X)) -> s(mark(X))
r10: mark(nil()) -> nil()
r11: mark(|0|()) -> |0|()
r12: a__from(X) -> from(X)
r13: a__length(X) -> length(X)
r14: a__length1(X) -> length1(X)

The set of usable rules consists of

  r1, r2, r3, r4, r5, r6, r7, r8, r9, r10, r11, r12, r13, r14

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{11, x1 + 5}
          mark#_A(x1) = x1 + 5
          s_A(x1) = max{2, x1}
          cons_A(x1,x2) = max{14, x1 + 4, x2}
          from_A(x1) = max{13, x1 + 12}
          mark_A(x1) = x1 + 6
          a__from_A(x1) = max{14, x1 + 12}
          a__length_A(x1) = max{5, x1}
          nil_A = 7
          |0|_A = 6
          a__length1_A(x1) = max{5, x1}
          length_A(x1) = max{5, x1}
          length1_A(x1) = max{1, x1}
    
      precedence:
      
        a__from# = mark = nil = |0| > a__from > cons > from = a__length1 > s = a__length > mark# = length = length1
      
      partial status:
    
        pi(a__from#) = [1]
        pi(mark#) = [1]
        pi(s) = [1]
        pi(cons) = [1, 2]
        pi(from) = [1]
        pi(mark) = [1]
        pi(a__from) = [1]
        pi(a__length) = [1]
        pi(nil) = []
        pi(|0|) = []
        pi(a__length1) = [1]
        pi(length) = [1]
        pi(length1) = [1]
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          a__from#_A(x1) = x1 + 26
          mark#_A(x1) = max{25, x1 + 22}
          s_A(x1) = x1 + 21
          cons_A(x1,x2) = max{x1 + 10, x2 + 8}
          from_A(x1) = x1 + 5
          mark_A(x1) = x1
          a__from_A(x1) = x1 + 6
          a__length_A(x1) = max{10, x1 + 9}
          nil_A = 10
          |0|_A = 11
          a__length1_A(x1) = max{9, x1 + 8}
          length_A(x1) = x1 + 1
          length1_A(x1) = x1 + 1
    
      precedence:
      
        a__from# > mark# = s = cons = from > mark > a__from = a__length = nil = |0| = a__length1 = length = length1
      
      partial status:
    
        pi(a__from#) = []
        pi(mark#) = [1]
        pi(s) = [1]
        pi(cons) = []
        pi(from) = [1]
        pi(mark) = [1]
        pi(a__from) = []
        pi(a__length) = [1]
        pi(nil) = []
        pi(|0|) = []
        pi(a__length1) = [1]
        pi(length) = [1]
        pi(length1) = [1]
    

The next rules are strictly ordered:

  p1, p2, p3, p4, p5

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__length#(cons(X,Y)) -> a__length1#(Y)
p2: a__length1#(X) -> a__length#(X)

and R consists of:

r1: a__from(X) -> cons(mark(X),from(s(X)))
r2: a__length(nil()) -> |0|()
r3: a__length(cons(X,Y)) -> s(a__length1(Y))
r4: a__length1(X) -> a__length(X)
r5: mark(from(X)) -> a__from(mark(X))
r6: mark(length(X)) -> a__length(X)
r7: mark(length1(X)) -> a__length1(X)
r8: mark(cons(X1,X2)) -> cons(mark(X1),X2)
r9: mark(s(X)) -> s(mark(X))
r10: mark(nil()) -> nil()
r11: mark(|0|()) -> |0|()
r12: a__from(X) -> from(X)
r13: a__length(X) -> length(X)
r14: a__length1(X) -> length1(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:
    
        max/plus interpretations on natural numbers:
        
          a__length#_A(x1) = max{5, x1 + 3}
          cons_A(x1,x2) = max{x1 - 1, x2 + 3}
          a__length1#_A(x1) = max{6, x1 + 4}
    
      precedence:
      
        a__length# = cons = a__length1#
      
      partial status:
    
        pi(a__length#) = [1]
        pi(cons) = [2]
        pi(a__length1#) = [1]
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          a__length#_A(x1) = x1 + 3
          cons_A(x1,x2) = x2 + 1
          a__length1#_A(x1) = x1 + 4
    
      precedence:
      
        a__length# = cons = a__length1#
      
      partial status:
    
        pi(a__length#) = [1]
        pi(cons) = [2]
        pi(a__length1#) = [1]
    

The next rules are strictly ordered:

  p1, p2

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