Theorem bdsep2 7112

Description: Version of ax-bdsep 7111 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 2083	. . . . . 6 ⊢ (y = 𝑎 → (x ∈ y ↔ x ∈ 𝑎))
2	1	anbi1d 441	. . . . 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 1687	. . 3 ⊢ (y = 𝑎 → (∀x(x ∈ 𝑏 ↔ (x ∈ y ∧ φ)) ↔ ∀x(x ∈ 𝑏 ↔ (x ∈ 𝑎 ∧ φ))))
5	4	exbidv 1688	. 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 7111	. . 3 ⊢ ∀y∃𝑏∀x(x ∈ 𝑏 ↔ (x ∈ y ∧ φ))
8	7	spi 1411	. 2 ⊢ ∃𝑏∀x(x ∈ 𝑏 ↔ (x ∈ y ∧ φ))
9	5, 8	chvarv 1794	1 ⊢ ∃𝑏∀x(x ∈ 𝑏 ↔ (x ∈ 𝑎 ∧ φ))

Syntax hints:   ∧ wa 97   ↔ wb 98  ∀wal 1226  ∃wex 1362  BOUNDED wbd 7039

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 1316  ax-gen 1318  ax-ie1 1363  ax-ie2 1364  ax-4 1381  ax-17 1400  ax-i9 1404  ax-ial 1409  ax-ext 2004  ax-bdsep 7111

This theorem depends on definitions:  df-bi 110  df-nf 1330  df-cleq 2015  df-clel 2018

This theorem is referenced by:  bdsepnft  7113  bdsepnfALT  7115  bdzfauscl  7116  bdbm1.3ii  7117  bj-nalset  7118