Assuming p(x) ⊢ q(x), apply the definition to the subobject ι : {x ∈ X | p(x)} ↪ X. Certainly the map p ◦ ι : {x ∈ X | p(x)} ↪ X → Ω will factor through 1. So q ◦ ι : {x ∈ X | p(x)} ↪ X → Ω will also factor through 1 and we can work from there.
Assuming p(x) ⊢ q(x), apply the definition to the subobject ι : {x ∈ X | p(x)} ↪ X. Certainly the map p ◦ ι : {x ∈ X | p(x)} ↪ X → Ω will factor through 1. So q ◦ ι : {x ∈ X | p(x)} ↪ X → Ω will also factor through 1 and we can work from there.
(yes, @gro-tsen.bsky.social, I know you're going to react with some combination of 🙄, 😫 and 😱 by emoji kitchen 😉)
(yes, @gro-tsen.bsky.social, I know you're going to react with some combination of 🙄, 😫 and 😱 by emoji kitchen 😉)
jean.abou-samra.fr/talks/2025-1...
jean.abou-samra.fr/talks/2025-1...