Combining proofs of commutativity and associativity of addition - agda

Question

I am trying to proof the below lemma
infixr 5 _~_
_~_ = trans
lemma-+swap : ∀ a b c → a + (b + c) ≡ b + (a + c)
lemma-+swap zero b c = refl
lemma-+swap (suc a) b c = (+-assoc a b c) ~ (comm-+ a (b + c)) ~ (+-assoc b c a)
Note : I imported this file
open import Relation.Binary.PropositionalEquality as PropEq
using (_≡_; refl; sym; trans; cong; cong₂)
While on paper I tried in this way :
(a + b) + c equivalent a + (b + c) -- associativity
a + (b + c) equivalent to (b + c) + a -- commutativity
(b + c) + a equi to b + (c + a) -- associativity
I wrote this in goal but getting error. I have proof of associative and commutative property. Please help.

Answer 1

Transcribing the proof you already have on paper
A very nice way of writing proofs like this is to use the Relation.Binary.PropositionalEquality.≡-Reasoning module, because it enables you to write out your proof exactly how you would do it on paper:
open import Relation.Binary.PropositionalEquality as PropEq using (_≡_)
open import Data.Nat
open import Data.Nat.Properties.Simple using (+-assoc; +-comm)
lemma-+swap : ∀ a b c → a + (b + c) ≡ b + (a + c)
lemma-+swap a b c = begin
a + (b + c) ≡⟨ {!!} ⟩
(a + b) + c ≡⟨ {!!} ⟩
(b + a) + c ≡⟨ {!!} ⟩
b + (a + c) ∎
where
open PropEq.≡-Reasoning
Now all you need to fill in are the three holes corresponding to the three steps of the proof.
Using the semiring solver
The hassle-free way is to just let the semiring solver take care of your equality, since addition and multiplication of natural numbers form a semiring. However, I don't recommend you do this for now, since it seems you need way more experience with the fundamentals to ensure you won't fall into the trap of thinking the semiring solver is black-box magic that shouldn't be understood. So the below is meant more for other readers of this answer and not so much to OP:
open import Relation.Binary.PropositionalEquality as PropEq using (_≡_; refl)
open import Data.Nat
lemma-+swap : ∀ a b c → a + (b + c) ≡ b + (a + c)
lemma-+swap = solve 3 (λ a b c → a :+ (b :+ c) := b :+ (a :+ c)) refl
where
open import Data.Nat.Properties as Nat
open Nat.SemiringSolver

Answer 2

You are basically doing needless case splitting. Let's check the goal in both function clauses. In the first one, we get:
lemma-+swap zero b c = ?
-- goal: b + c ≡ b + c
which is satisfied by simple refl. The second one is however:
lemma-+swap (suc a) b c = ?
-- goal: suc (a + (b + c)) ≡ b + suc (a + c)
And notice that +-assoc a b c (from Data.Nat.Properties.Simple) has a type a + b + c ≡ a + (b + c) - with no suc in sight.
So you have two options: the prefered one is to avoid case splitting and use the properties directly. The other one is to use the properties with suc a instead of a.
Your implementation also most likely won't work even if you fix the above problem.
Here, I assume your properties have the same type as the ones from the module mentioned above (which is a fair assumption, given that for the other variations the subexpression +-assoc a b c ~ comm-+ a (b + c) ~ +-assoc b c a does not typecheck).
+-assoc a b c : a + b + c ≡ a + (b + c)
+-comm a (b + c) : a + (b + c) ≡ b + c + a
+-assoc b c a : b + c + a ≡ b + (c + a)
+-assoc a b c ~ comm-+ a (b + c) ~ +-assoc b c a : a + b + c ≡ b + (c + a) (*)
So, the subexpression has the type (*), but your goal is a + (b + c) ≡ b + (a + c).