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. matrix interpretations:
    
      carrier: N^1
      order: standard order
      interpretations:
        a__from#_A(x1) = x1 + 1
        mark#_A(x1) = x1
        s_A(x1) = x1
        cons_A(x1,x2) = x1 + 1
        from_A(x1) = x1 + 4
        mark_A(x1) = x1 + 2
        a__from_A(x1) = x1 + 4
        a__length_A(x1) = 2
        nil_A() = 1
        |0|_A() = 1
        a__length1_A(x1) = 2
        length_A(x1) = 1
        length1_A(x1) = 1
    
    2. matrix interpretations:
    
      carrier: N^1
      order: standard order
      interpretations:
        a__from#_A(x1) = 0
        mark#_A(x1) = 1
        s_A(x1) = 0
        cons_A(x1,x2) = x1
        from_A(x1) = 0
        mark_A(x1) = 3
        a__from_A(x1) = x1
        a__length_A(x1) = 1
        nil_A() = 4
        |0|_A() = 2
        a__length1_A(x1) = 4
        length_A(x1) = 2
        length1_A(x1) = 1
    

The next rules are strictly ordered:

  p1, p3, p4, p5

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)

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}


-- Reduction pair.

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

p1: 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 set of usable rules consists of

  (no rules)

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. matrix interpretations:
    
      carrier: N^1
      order: standard order
      interpretations:
        mark#_A(x1) = x1
        s_A(x1) = x1 + 1
    
    2. matrix interpretations:
    
      carrier: N^1
      order: standard order
      interpretations:
        mark#_A(x1) = 0
        s_A(x1) = x1 + 1
    

The next rules are strictly ordered:

  p1

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. matrix interpretations:
    
      carrier: N^1
      order: standard order
      interpretations:
        a__length#_A(x1) = x1
        cons_A(x1,x2) = x1 + x2 + 2
        a__length1#_A(x1) = x1 + 1
    
    2. matrix interpretations:
    
      carrier: N^1
      order: standard order
      interpretations:
        a__length#_A(x1) = x1
        cons_A(x1,x2) = x1
        a__length1#_A(x1) = 1
    

The next rules are strictly ordered:

  p1, p2

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