Proof
01. 1 ⊢ x ∈ A ∪ B, hypo.
02. 1 ⊢ x ∈ {x | x ∈ A ∨ x ∈ B}, eq_subst union_eq 1, P u ↔ x ∈ u.
03. 1 ⊢ x ∈ A ∨ x ∈ B, comp_elim 2.
union_elim. ⊢ x ∈ A ∪ B → x ∈ A ∨ x ∈ B, subj_intro 3.
Dependencies
The given proof depends on two axioms:
comp, eq_subst.