Metamath Proof Explorer


Theorem bj-sepg

Description: Version of sepg which does not require df-clab , thanks to the use of bj-vtoclg (and ultimately, to the use of elissetv instead of elisset ). This axiom save is not very important, since this theorem uses df-cleq and df-clel . This theorem is a remnant of a previous state of set.mm where the axiom saving was larger. (Contributed by BJ, 2-Jul-2022) (Proof modification is discouraged.)

Ref Expression
Assertion bj-sepg ( 𝐴 ∈ 𝑉 → ∃ 𝑦 ∀ 𝑥 ( 𝑥 ∈ 𝑦 ↔ ( 𝑥 ∈ 𝐴 ∧ 𝜑 ) ) )

Proof

Step Hyp Ref Expression
1 eleq2 ⊢ ( 𝑧 = 𝐴 → ( 𝑥 ∈ 𝑧 ↔ 𝑥 ∈ 𝐴 ) )
2 1 anbi1d ⊢ ( 𝑧 = 𝐴 → ( ( 𝑥 ∈ 𝑧 ∧ 𝜑 ) ↔ ( 𝑥 ∈ 𝐴 ∧ 𝜑 ) ) )
3 2 bibi2d ⊢ ( 𝑧 = 𝐴 → ( ( 𝑥 ∈ 𝑦 ↔ ( 𝑥 ∈ 𝑧 ∧ 𝜑 ) ) ↔ ( 𝑥 ∈ 𝑦 ↔ ( 𝑥 ∈ 𝐴 ∧ 𝜑 ) ) ) )
4 3 biimpd ⊢ ( 𝑧 = 𝐴 → ( ( 𝑥 ∈ 𝑦 ↔ ( 𝑥 ∈ 𝑧 ∧ 𝜑 ) ) → ( 𝑥 ∈ 𝑦 ↔ ( 𝑥 ∈ 𝐴 ∧ 𝜑 ) ) ) )
5 4 alimdv ⊢ ( 𝑧 = 𝐴 → ( ∀ 𝑥 ( 𝑥 ∈ 𝑦 ↔ ( 𝑥 ∈ 𝑧 ∧ 𝜑 ) ) → ∀ 𝑥 ( 𝑥 ∈ 𝑦 ↔ ( 𝑥 ∈ 𝐴 ∧ 𝜑 ) ) ) )
6 5 eximdv ⊢ ( 𝑧 = 𝐴 → ( ∃ 𝑦 ∀ 𝑥 ( 𝑥 ∈ 𝑦 ↔ ( 𝑥 ∈ 𝑧 ∧ 𝜑 ) ) → ∃ 𝑦 ∀ 𝑥 ( 𝑥 ∈ 𝑦 ↔ ( 𝑥 ∈ 𝐴 ∧ 𝜑 ) ) ) )
7 ax-sep ⊢ ∃ 𝑦 ∀ 𝑥 ( 𝑥 ∈ 𝑦 ↔ ( 𝑥 ∈ 𝑧 ∧ 𝜑 ) )
8 6 7 bj-vtoclg ⊢ ( 𝐴 ∈ 𝑉 → ∃ 𝑦 ∀ 𝑥 ( 𝑥 ∈ 𝑦 ↔ ( 𝑥 ∈ 𝐴 ∧ 𝜑 ) ) )