Theorem bdsep2 9274

Description: Version of ax-bdsep 9273 with one DV condition removed and without initial universal quantifier. (Contributed by BJ, 5-Oct-2019.)

Hypothesis

Ref	Expression
bdsep2.1	⊢ BOUNDED φ

Assertion

Ref	Expression
bdsep2	⊢ ∃𝑏∀x(x ∈ 𝑏 ↔ (x ∈ 𝑎 ∧ φ))

Distinct variable groups:   𝑎,𝑏,x   φ,𝑏

Allowed substitution hints:   φ(x,𝑎)

Proof of Theorem bdsep2

Dummy variable y is distinct from all other variables.

Step	Hyp	Ref	Expression
1		eleq2 2098	. . . . . 6 ⊢ (y = 𝑎 → (x ∈ y ↔ x ∈ 𝑎))
2	1	anbi1d 438	. . . . 5 ⊢ (y = 𝑎 → ((x ∈ y ∧ φ) ↔ (x ∈ 𝑎 ∧ φ)))
3	2	bibi2d 221	. . . 4 ⊢ (y = 𝑎 → ((x ∈ 𝑏 ↔ (x ∈ y ∧ φ)) ↔ (x ∈ 𝑏 ↔ (x ∈ 𝑎 ∧ φ))))
4	3	albidv 1702	. . 3 ⊢ (y = 𝑎 → (∀x(x ∈ 𝑏 ↔ (x ∈ y ∧ φ)) ↔ ∀x(x ∈ 𝑏 ↔ (x ∈ 𝑎 ∧ φ))))
5	4	exbidv 1703	. 2 ⊢ (y = 𝑎 → (∃𝑏∀x(x ∈ 𝑏 ↔ (x ∈ y ∧ φ)) ↔ ∃𝑏∀x(x ∈ 𝑏 ↔ (x ∈ 𝑎 ∧ φ))))
6		bdsep2.1	. . . 4 ⊢ BOUNDED φ
7	6	ax-bdsep 9273	. . 3 ⊢ ∀y∃𝑏∀x(x ∈ 𝑏 ↔ (x ∈ y ∧ φ))
8	7	spi 1426	. 2 ⊢ ∃𝑏∀x(x ∈ 𝑏 ↔ (x ∈ y ∧ φ))
9	5, 8	chvarv 1809	1 ⊢ ∃𝑏∀x(x ∈ 𝑏 ↔ (x ∈ 𝑎 ∧ φ))

Syntax hints:   ∧ wa 97   ↔ wb 98  ∀wal 1240  ∃wex 1378  BOUNDED wbd 9201

This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 99  ax-ia2 100  ax-ia3 101  ax-5 1333  ax-gen 1335  ax-ie1 1379  ax-ie2 1380  ax-4 1397  ax-17 1416  ax-i9 1420  ax-ial 1424  ax-ext 2019  ax-bdsep 9273

This theorem depends on definitions:  df-bi 110  df-nf 1347  df-cleq 2030  df-clel 2033

This theorem is referenced by:  bdsepnft  9275  bdsepnfALT  9277  bdzfauscl  9278  bdbm1.3ii  9279  bj-nalset  9280