Theorem bdsepnfALT 13087

Description: Alternate proof of bdsepnf 13086, not using bdsepnft 13085. (Contributed by BJ, 5-Oct-2019.) (Proof modification is discouraged.) (New usage is discouraged.)

Hypotheses

Ref	Expression
bdsepnf.nf	⊢ Ⅎ𝑏𝜑
bdsepnf.1	⊢ BOUNDED 𝜑

Assertion

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

Distinct variable group:   𝑎,𝑏,𝑥

Allowed substitution hints:   𝜑(𝑥,𝑎,𝑏)

Proof of Theorem bdsepnfALT

Dummy variable 𝑦 is distinct from all other variables.

Step	Hyp	Ref	Expression
1		bdsepnf.1	. . 3 ⊢ BOUNDED 𝜑
2	1	bdsep2 13084	. 2 ⊢ ∃𝑦∀𝑥(𝑥 ∈ 𝑦 ↔ (𝑥 ∈ 𝑎 ∧ 𝜑))
3		nfv 1508	. . . . 5 ⊢ Ⅎ𝑏 𝑥 ∈ 𝑦
4		nfv 1508	. . . . . 6 ⊢ Ⅎ𝑏 𝑥 ∈ 𝑎
5		bdsepnf.nf	. . . . . 6 ⊢ Ⅎ𝑏𝜑
6	4, 5	nfan 1544	. . . . 5 ⊢ Ⅎ𝑏(𝑥 ∈ 𝑎 ∧ 𝜑)
7	3, 6	nfbi 1568	. . . 4 ⊢ Ⅎ𝑏(𝑥 ∈ 𝑦 ↔ (𝑥 ∈ 𝑎 ∧ 𝜑))
8	7	nfal 1555	. . 3 ⊢ Ⅎ𝑏∀𝑥(𝑥 ∈ 𝑦 ↔ (𝑥 ∈ 𝑎 ∧ 𝜑))
9		nfv 1508	. . 3 ⊢ Ⅎ𝑦∀𝑥(𝑥 ∈ 𝑏 ↔ (𝑥 ∈ 𝑎 ∧ 𝜑))
10		elequ2 1691	. . . . 5 ⊢ (𝑦 = 𝑏 → (𝑥 ∈ 𝑦 ↔ 𝑥 ∈ 𝑏))
11	10	bibi1d 232	. . . 4 ⊢ (𝑦 = 𝑏 → ((𝑥 ∈ 𝑦 ↔ (𝑥 ∈ 𝑎 ∧ 𝜑)) ↔ (𝑥 ∈ 𝑏 ↔ (𝑥 ∈ 𝑎 ∧ 𝜑))))
12	11	albidv 1796	. . 3 ⊢ (𝑦 = 𝑏 → (∀𝑥(𝑥 ∈ 𝑦 ↔ (𝑥 ∈ 𝑎 ∧ 𝜑)) ↔ ∀𝑥(𝑥 ∈ 𝑏 ↔ (𝑥 ∈ 𝑎 ∧ 𝜑))))
13	8, 9, 12	cbvex 1729	. 2 ⊢ (∃𝑦∀𝑥(𝑥 ∈ 𝑦 ↔ (𝑥 ∈ 𝑎 ∧ 𝜑)) ↔ ∃𝑏∀𝑥(𝑥 ∈ 𝑏 ↔ (𝑥 ∈ 𝑎 ∧ 𝜑)))
14	2, 13	mpbi 144	1 ⊢ ∃𝑏∀𝑥(𝑥 ∈ 𝑏 ↔ (𝑥 ∈ 𝑎 ∧ 𝜑))

Syntax hints:   ∧ wa 103   ↔ wb 104  ∀wal 1329  Ⅎwnf 1436  ∃wex 1468  BOUNDED wbd 13010

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 105  ax-ia2 106  ax-ia3 107  ax-5 1423  ax-7 1424  ax-gen 1425  ax-ie1 1469  ax-ie2 1470  ax-8 1482  ax-4 1487  ax-14 1492  ax-17 1506  ax-i9 1510  ax-ial 1514  ax-i5r 1515  ax-ext 2121  ax-bdsep 13082

This theorem depends on definitions:  df-bi 116  df-tru 1334  df-nf 1437  df-cleq 2132  df-clel 2135

This theorem is referenced by: (None)