Proof of Merge Law for Deployment

Question

I read an article by Jeremy Gibbons on origami programming , and I was stuck in exercise 3.7, which asks the reader to prove the merge law for a list unfolds:

unfoldL p f g . h = unfoldL p' f' g'

if

p . h = p'
f . h = f'
g . h = h . g'

The function unfoldLdeployed for lists is defined as follows:

unfoldL :: (b -> Bool) -> (b -> a) -> (b -> b) -> b -> List a
unfoldL p f g b = if p b then Nil else Cons (f b) (unfoldL p f g (g b))

Here is my current attempt at proof:

(unfoldL p f g . h) b
=   { composition }
unfoldL p f g (h b)
=   { unfoldL }
if p (h b) then Nil else Cons (f (h b)) (unfoldL p f g (g (h b)))
=   { composition }
if (p . h) b then Nil else Cons ((f . h) b) (unfoldL p f g ((g . h) b))
=   { assumptions }
if p' b then Nil else Cons (f' b) (unfoldL p f g ((h . g') b))
=   { composition }
if p' b then Nil else Cons (f' b) ((unfoldL p f g . h) (g' b))
=   { ??? }
if p' b then Nil else Cons (f' b) (unfoldL p' f' g' (g' b))
=   { unFoldL }
unFoldL p' f' g'

I am not sure how to justify the step noted ???. Perhaps I should use some kind of induction to apply functions on b? A related question: what are some good resources that explain and motivate various methods of induction, such as structural induction?

Answer 1

, , , .

, - , , , , . . , unfoldL . , "" - , . , .

, ( , ), . , , , , Nil, Cons , , , ( ).

, coinduction, b ( b s):

(unfoldL p f g . h) b ~~ unfoldL p' f' g' b

(unfoldL p f g . h) b
= { your reasoning }
if p' b then Nil else Cons (f' b) ((unfoldL p f g . h) (g' b))

p' b, p' b True,

if p' b then Nil else Cons (f' b) ((unfoldL p f g . h) (g' b))
= { p' b is True }
Nil
~~ { reflexivity }
Nil
= { p' b is True }
if p' b then Nil else Cons (f' b) (unfoldL p' f' g' (g' b))
= { unfoldL }
unfoldL p' f' g'

; p' b False,

if p' b then Nil else Cons (f' b) ((unfoldL p f g . h) (g' b))
= { p' b is False }
Cons (f' b) ((unfoldL p f g . h) (g' b))
*** ~~ { bisimilarity Cons rule, coinductive hypothesis } ***
Cons (f' b) (unfoldL p' f' g' (g' b))
= { p' b is False }
if p' b then Nil else Cons (f' b) (unfoldL p' f' g' (g' b))
= { unfoldL }

, ***, . -, ~~ =. , Cons. - , , .

Answer 2

( ), -; coinduction, , @luqui . 6 , . .

---

unfoldL:

h = unfoldL p f g
<=>
∀ b . h b = if p b then Nil else Cons (f b) (h (g b))

:

unfoldL p f g . h = unfoldL p' f' g'
<=> { universal property }
∀ b . (unfoldL p f g . h) b = if p' b then Nil else Cons (f' b) ((unfoldL p f g . h) (g' b))
<=> { composition }
∀ b . unfoldL p f g (h b) = if p' b then Nil else Cons (f' b) (unfoldL p f g (h (g' b)))
<=> { unfoldL }
∀ b . if p (h b) then Nil else Cons (p (h b)) (unfoldL p f g (g (h b))) = if p' b then Nil else Cons (f' b) (unfoldL p f g (h (g' b)))
<=> { composition }
∀ b . if (p . h) b then Nil else Cons (p . h) b (unfoldL p f g ((g . h) b)) = if p' b then Nil else Cons (f' b) (unfoldL p f g ((h . g') b))
<=  { assumptions }
(p . h = p') ^ (f . h = f') ^ (g . h = h . g')