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Add ∣m+n∣n⇒∣m to Data.Nat.Divisiblity #2989

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Add ∣m+n∣n⇒∣m to Data.Nat.Divisiblity #2989

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ec525fd
Add ∣m+n∣n⇒∣m
Taneb Apr 28, 2026
7487981
fix typo in variable name
Taneb Apr 29, 2026
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1 change: 1 addition & 0 deletions CHANGELOG.md
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Expand Up @@ -375,6 +375,7 @@ Additions to existing modules

* In `Data.Nat.Divisibility`:
```agda
∣m+n∣n⇒∣m : d ∣ m + n → d ∣ n → d ∣ m
m∣n⇒m^o∣n^o : ∀ o → m ∣ n → m ^ o ∣ n ^ o
n≤o⇒m^n∣m^o : ∀ m → .(n ≤ o) → m ^ n ∣ m ^ o
```
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12 changes: 10 additions & 2 deletions src/Data/Nat/Divisibility.agda
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Expand Up @@ -190,13 +190,21 @@ n∣n = ∣-refl
∣m+n∣m⇒∣n : d ∣ m + n → d ∣ m → d ∣ n
∣m+n∣m⇒∣n {d} {m} {n} (divides p m+n≡p*d) (divides-refl q) =
divides (p ∸ q) $ begin-equality
n ≡⟨ m+n∸n≡m n m ⟨
n + m ∸ m ≡⟨ cong (_∸ m) (+-comm n m) ⟩
n ≡⟨ m+n∸m≡n m n ⟨
m + n ∸ m ≡⟨ cong (_∸ m) m+n≡p*d ⟩
p * d ∸ q * d ≡⟨ *-distribʳ-∸ d p q ⟨
(p ∸ q) * d ∎
where open ∣-Reasoning

∣m+n∣n⇒∣m : d ∣ m + n → d ∣ n → d ∣ m
∣m+n∣n⇒∣m {d} {m} {n} (divides p m*n≡p*d) (divides-refl q) =
divides (p ∸ q) $ begin-equality
m ≡⟨ m+n∸n≡m m n ⟨
m + n ∸ n ≡⟨ cong (_∸ q * d) m*n≡p*d ⟩
p * d ∸ q * d ≡⟨ *-distribʳ-∸ d p q ⟨
(p ∸ q) * d ∎
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@gallais gallais Apr 29, 2026

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Based on the other proof, I presume that this is a typo on the variable name.

Suggested change
∣m+n∣n⇒∣m {d} {m} {n} (divides p m*n≡p*d) (divides-refl q) =
divides (p ∸ q) $ begin-equality
m ≡⟨ m+n∸n≡m m n ⟨
m + n ∸ n ≡⟨ cong (_∸ q * d) m*n≡p*d ⟩
p * d ∸ q * d ≡⟨ *-distribʳ-∸ d p q ⟨
(p ∸ q) * d ∎
∣m+n∣n⇒∣m {d} {m} {n} (divides p m+n≡p*d) (divides-refl q) =
divides (p ∸ q) $ begin-equality
m ≡⟨ m+n∸n≡m m n ⟨
m + n ∸ n ≡⟨ cong (_∸ q * d) m+n≡p*d ⟩
p * d ∸ q * d ≡⟨ *-distribʳ-∸ d p q ⟨
(p ∸ q) * d ∎

Also, why prove it by going back to the core argument instead of invoking ∣m+n∣m⇒∣n and
the fact _+_ is commutative?

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@Taneb Taneb Apr 29, 2026

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Also, why prove it by going back to the core argument instead of invoking ∣m+n∣m⇒∣n and
the fact _+_ is commutative?

I felt like this new proof is the more natural variant (the other already uses +-comm in its proof) and I don't like using subst if I can help it

where open ∣-Reasoning

------------------------------------------------------------------------
-- Properties of _∣_ and _*_

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