1. Not required to prove everything which is true, 2. Not limited to one set of axioms.
The nice thing about formalism is that you can use a computer program to check whether a proof is sound or not. Isn’t that nice? It has applications in e.g. computer science and engineering, where you can prove that a program does not crash, or prove that a traffic light never shows green in conflicting directions, or prove that a real-time kernel always provides enough computational time for your rocket computer to steer.
Working math, there are more and more complex proofs that cannot reasonably be checked by humans. Having a computer check them is a nice alternative.
Others hope that the next AI will come up with a sketch of a proof itself.
GPT-4 sucks at lean because lean had it's syntax changed too often, and GPT-4 shows limited reasoning capabilities
However it did come up with the following proof for my little order of pairs
instance LT : LT (ℕ × ℕ) where
lt f g := f.1 < g.1 ∨ (f.1 = g.1 ∧ f.2 < g.2)
instance : DecidableRel (LT.lt : ℕ × ℕ → ℕ × ℕ → Prop) :=
fun (a,b) (c,d) =>
if h₁ : a < c then isTrue (Or.inl h₁)
else if h₂ : a = c then
if h₃ : b < d then
isTrue (Or.inr ⟨h₂, h₃⟩)
-- the rest is even more silly since we have to turn everything around
else isFalse (λ (h : a < c ∨ (a = c ∧ b < d)) =>
Or.elim h
(λ hlt : a < c => absurd hlt h₁)
(λ heq : a = c ∧ b < d =>
absurd (And.right heq) h₃))
else isFalse (λ (h : a < c ∨ (a = c ∧ b < d)) =>
Or.elim h
(λ hlt : a < c => absurd hlt h₁)
(λ heq : a = c ∧ b < d =>
absurd (And.left heq) h₂)
)
This may lead back to the original question because Bool ≠ Prop and the formula for LT should ideally lead directly to decidability without all this redundancy?