Structural induction proof for reverse(push(as, bs)) = push(reverse(bs), reverse(as))

Question

I need to prove:

reverse(push(as,bs)) = push(reverse(bs), reverse(as))

where:

def push[T](as: List[T], bs: List[T]): 
  List [T] = as match {
      case Nil => bs
      case x::xs => x::push(xs, bs)
  }
def reverse[T](ls: List[T]): 
  List[T] = ls match {
     case Nil => Nil
     case x::xs => push(reverse(xs), x::Nil)
  }

I am already stuck in the base case, since i cant figure out a lemma. The only thing I have this far is:

reverse(push(Nil,bs)) = reverse(bs)

and I'm stuck here since it felt eternity.

Does anyone have an idea for a lemma?

score 1 · Answer 1 · answered May 04 '22 at 23:45

I think what would help a lot is proving first that push(push(a, b), c) = push(a, push(b, c)).

Once you have that result, you can make an induction on as only in the equality reverse(push(as,bs)) = push(reverse(bs), reverse(as)).

I will add some details if necessary.

Structural induction proof for reverse(push(as, bs)) = push(reverse(bs), reverse(as))

1 Answers1