Proof
01. 1 ⊢ x ∈ A ∩ B, hypo.
02. 1 ⊢ x ∈ {x | x ∈ A ∧ x ∈ B},
eq_subst intersection_eq 1, P X ↔ x ∈ X.
03. 1 ⊢ x ∈ A ∧ x ∈ B, comp_elim 2.
intersection_elim. ⊢ x ∈ A ∩ B → x ∈ A ∧ x ∈ B, subj_intro 3.
Dependencies
The given proof depends on two axioms:
comp, eq_subst.