Proof 01. ⊢ u ∈ {x | P x} ↔ set u ∧ P u, comp. 02. ⊢ set u ∧ P u → u ∈ {x | P x}, bij_elimr 1. 03. 3 ⊢ set u, hypo. 04. 4 ⊢ P u, hypo. 05. 3, 4 ⊢ set u ∧ P u, hypo. 06. 3, 4 ⊢ u ∈ {x | P x}, subj_elim 2 5. comp_intro. ⊢ set u → P u → u ∈ {x | P x}, subj_intro_ii 6.
Dependencies
The given proof depends on one axiom:
comp.