[ add ] Nat lemmas with _∸_, _⊔_ and _⊓_
#2924
Conversation
MatthewDaggitt
left a comment
MatthewDaggitt
left a comment
There was a problem hiding this comment.
Thanks for the PR!
- Naming and placement looks good to me.
- We tend to avoid rewriting in favour of using equational reasoning as its a) less brittle and b) clearer to the user what is going on. Could you switch these to use equational reasoning?
| ------------------------------------------------------------------------ | ||
| -- Properties of _∸_ and _⊓_ and _⊔_ | ||
|
|
||
| m∸n≤m⊔n : ∀ m n → m ∸ n ≤ m ⊔ n |
There was a problem hiding this comment.
This property doesn't feel natural to me, and the definition is short enough I don't really think it needs a name
There was a problem hiding this comment.
It looks kind of similar to the m⊔n≤m+n that is already here.
Some stuff that I've needed. I have no idea if those belong here in stdlib. I don't know if I've put those in the right place, or named them properly. Hence the draft and lack of changlog entry.
|
Monus |
| ∣m-n∣≤m⊔n (suc m) zero = ≤-refl | ||
| ∣m-n∣≤m⊔n (suc m) (suc n) = m≤n⇒m≤1+n (∣m-n∣≤m⊔n m n) | ||
|
|
||
| ∣m-n∣≡m⊔n∸m⊓n : ∀ m n → ∣ m - n ∣ ≡ m ⊔ n ∸ m ⊓ n |
There was a problem hiding this comment.
The structure of this proof recalls that of ∣m-n∣≡[m∸n]∨[n∸m] (L1863).
Is there a refactoring which allows you to prove each result from some other, more general lemma?
Separately, you might like to consider refactoring this proof to not use with, but (more) simply via [ branch1 , branch2 ]′ $ ≤-total m n for suitable subproofs branch1, branch2, should you ever need to reason about this operation. cf. 'with (sometimes) considered harmful' #2123
4 participants
Some stuff that I've needed. I have no idea if those belong here in stdlib. I don't know if I've put those in the right place, or named them properly. Hence the draft and lack of changlog entry.