YES

We show the termination of the TRS R:

  a__eq(|0|(),|0|()) -> true()
  a__eq(s(X),s(Y)) -> a__eq(X,Y)
  a__eq(X,Y) -> false()
  a__inf(X) -> cons(X,inf(s(X)))
  a__take(|0|(),X) -> nil()
  a__take(s(X),cons(Y,L)) -> cons(Y,take(X,L))
  a__length(nil()) -> |0|()
  a__length(cons(X,L)) -> s(length(L))
  mark(eq(X1,X2)) -> a__eq(X1,X2)
  mark(inf(X)) -> a__inf(mark(X))
  mark(take(X1,X2)) -> a__take(mark(X1),mark(X2))
  mark(length(X)) -> a__length(mark(X))
  mark(|0|()) -> |0|()
  mark(true()) -> true()
  mark(s(X)) -> s(X)
  mark(false()) -> false()
  mark(cons(X1,X2)) -> cons(X1,X2)
  mark(nil()) -> nil()
  a__eq(X1,X2) -> eq(X1,X2)
  a__inf(X) -> inf(X)
  a__take(X1,X2) -> take(X1,X2)
  a__length(X) -> length(X)

-- SCC decomposition.

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

p1: a__eq#(s(X),s(Y)) -> a__eq#(X,Y)
p2: mark#(eq(X1,X2)) -> a__eq#(X1,X2)
p3: mark#(inf(X)) -> a__inf#(mark(X))
p4: mark#(inf(X)) -> mark#(X)
p5: mark#(take(X1,X2)) -> a__take#(mark(X1),mark(X2))
p6: mark#(take(X1,X2)) -> mark#(X1)
p7: mark#(take(X1,X2)) -> mark#(X2)
p8: mark#(length(X)) -> a__length#(mark(X))
p9: mark#(length(X)) -> mark#(X)

and R consists of:

r1: a__eq(|0|(),|0|()) -> true()
r2: a__eq(s(X),s(Y)) -> a__eq(X,Y)
r3: a__eq(X,Y) -> false()
r4: a__inf(X) -> cons(X,inf(s(X)))
r5: a__take(|0|(),X) -> nil()
r6: a__take(s(X),cons(Y,L)) -> cons(Y,take(X,L))
r7: a__length(nil()) -> |0|()
r8: a__length(cons(X,L)) -> s(length(L))
r9: mark(eq(X1,X2)) -> a__eq(X1,X2)
r10: mark(inf(X)) -> a__inf(mark(X))
r11: mark(take(X1,X2)) -> a__take(mark(X1),mark(X2))
r12: mark(length(X)) -> a__length(mark(X))
r13: mark(|0|()) -> |0|()
r14: mark(true()) -> true()
r15: mark(s(X)) -> s(X)
r16: mark(false()) -> false()
r17: mark(cons(X1,X2)) -> cons(X1,X2)
r18: mark(nil()) -> nil()
r19: a__eq(X1,X2) -> eq(X1,X2)
r20: a__inf(X) -> inf(X)
r21: a__take(X1,X2) -> take(X1,X2)
r22: a__length(X) -> length(X)

The estimated dependency graph contains the following SCCs:

  {p4, p6, p7, p9}
  {p1}


-- Reduction pair.

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

p1: mark#(length(X)) -> mark#(X)
p2: mark#(take(X1,X2)) -> mark#(X2)
p3: mark#(take(X1,X2)) -> mark#(X1)
p4: mark#(inf(X)) -> mark#(X)

and R consists of:

r1: a__eq(|0|(),|0|()) -> true()
r2: a__eq(s(X),s(Y)) -> a__eq(X,Y)
r3: a__eq(X,Y) -> false()
r4: a__inf(X) -> cons(X,inf(s(X)))
r5: a__take(|0|(),X) -> nil()
r6: a__take(s(X),cons(Y,L)) -> cons(Y,take(X,L))
r7: a__length(nil()) -> |0|()
r8: a__length(cons(X,L)) -> s(length(L))
r9: mark(eq(X1,X2)) -> a__eq(X1,X2)
r10: mark(inf(X)) -> a__inf(mark(X))
r11: mark(take(X1,X2)) -> a__take(mark(X1),mark(X2))
r12: mark(length(X)) -> a__length(mark(X))
r13: mark(|0|()) -> |0|()
r14: mark(true()) -> true()
r15: mark(s(X)) -> s(X)
r16: mark(false()) -> false()
r17: mark(cons(X1,X2)) -> cons(X1,X2)
r18: mark(nil()) -> nil()
r19: a__eq(X1,X2) -> eq(X1,X2)
r20: a__inf(X) -> inf(X)
r21: a__take(X1,X2) -> take(X1,X2)
r22: a__length(X) -> length(X)

The set of usable rules consists of

  (no rules)

Take the monotone reduction pair:

  lexicographic combination of reduction pairs:
  
    1. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        mark#_A(x1) = ((1,1),(1,0)) x1
        length_A(x1) = ((1,1),(1,1)) x1 + (1,1)
        take_A(x1,x2) = ((1,1),(1,1)) x1 + ((1,1),(1,1)) x2 + (1,1)
        inf_A(x1) = ((1,1),(1,1)) x1 + (1,1)
    
    2. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        mark#_A(x1) = ((1,0),(1,1)) x1
        length_A(x1) = ((1,0),(1,1)) x1 + (1,1)
        take_A(x1,x2) = ((1,1),(1,1)) x1 + ((1,1),(0,1)) x2 + (1,1)
        inf_A(x1) = ((1,1),(1,1)) x1 + (1,1)
    

The next rules are strictly ordered:

  p1, p2, p3, p4
  r1, r2, r3, r4, r5, r6, r7, r8, r9, r10, r11, r12, r13, r14, r15, r16, r17, r18, r19, r20, r21, r22

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__eq#(s(X),s(Y)) -> a__eq#(X,Y)

and R consists of:

r1: a__eq(|0|(),|0|()) -> true()
r2: a__eq(s(X),s(Y)) -> a__eq(X,Y)
r3: a__eq(X,Y) -> false()
r4: a__inf(X) -> cons(X,inf(s(X)))
r5: a__take(|0|(),X) -> nil()
r6: a__take(s(X),cons(Y,L)) -> cons(Y,take(X,L))
r7: a__length(nil()) -> |0|()
r8: a__length(cons(X,L)) -> s(length(L))
r9: mark(eq(X1,X2)) -> a__eq(X1,X2)
r10: mark(inf(X)) -> a__inf(mark(X))
r11: mark(take(X1,X2)) -> a__take(mark(X1),mark(X2))
r12: mark(length(X)) -> a__length(mark(X))
r13: mark(|0|()) -> |0|()
r14: mark(true()) -> true()
r15: mark(s(X)) -> s(X)
r16: mark(false()) -> false()
r17: mark(cons(X1,X2)) -> cons(X1,X2)
r18: mark(nil()) -> nil()
r19: a__eq(X1,X2) -> eq(X1,X2)
r20: a__inf(X) -> inf(X)
r21: a__take(X1,X2) -> take(X1,X2)
r22: a__length(X) -> length(X)

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        a__eq#_A(x1,x2) = ((1,1),(0,0)) x1 + ((1,1),(1,1)) x2
        s_A(x1) = ((1,1),(1,1)) x1 + (1,1)
    
    2. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        a__eq#_A(x1,x2) = ((0,1),(1,0)) x1 + ((1,1),(0,1)) x2
        s_A(x1) = ((0,1),(1,1)) x1 + (1,1)
    

The next rules are strictly ordered:

  p1

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