Theorem eqabbw 2810

Description: Version of eqabb 2876 using implicit substitution, which requires fewer axioms. (Contributed by GG and AV, 18-Sep-2024.)

Hypothesis

Ref	Expression
eqabbw.1	⊢ (𝑥 = 𝑦 → (𝜑 ↔ 𝜓))

Assertion

Ref	Expression
eqabbw	⊢ (𝐴 = {𝑥 ∣ 𝜑} ↔ ∀𝑦(𝑦 ∈ 𝐴 ↔ 𝜓))

Distinct variable groups:   𝑥,𝑦   𝑦,𝐴   𝜑,𝑦   𝜓,𝑥

Allowed substitution hints:   𝜑(𝑥)   𝜓(𝑦)   𝐴(𝑥)

Proof of Theorem eqabbw

Step	Hyp	Ref	Expression
1		dfcleq 2730	. 2 ⊢ (𝐴 = {𝑥 ∣ 𝜑} ↔ ∀𝑦(𝑦 ∈ 𝐴 ↔ 𝑦 ∈ {𝑥 ∣ 𝜑}))
2		df-clab 2716	. . . . 5 ⊢ (𝑦 ∈ {𝑥 ∣ 𝜑} ↔ [𝑦 / 𝑥]𝜑)
3		eqabbw.1	. . . . . 6 ⊢ (𝑥 = 𝑦 → (𝜑 ↔ 𝜓))
4	3	sbievw 2099	. . . . 5 ⊢ ([𝑦 / 𝑥]𝜑 ↔ 𝜓)
5	2, 4	bitri 275	. . . 4 ⊢ (𝑦 ∈ {𝑥 ∣ 𝜑} ↔ 𝜓)
6	5	bibi2i 337	. . 3 ⊢ ((𝑦 ∈ 𝐴 ↔ 𝑦 ∈ {𝑥 ∣ 𝜑}) ↔ (𝑦 ∈ 𝐴 ↔ 𝜓))
7	6	albii 1821	. 2 ⊢ (∀𝑦(𝑦 ∈ 𝐴 ↔ 𝑦 ∈ {𝑥 ∣ 𝜑}) ↔ ∀𝑦(𝑦 ∈ 𝐴 ↔ 𝜓))
8	1, 7	bitri 275	1 ⊢ (𝐴 = {𝑥 ∣ 𝜑} ↔ ∀𝑦(𝑦 ∈ 𝐴 ↔ 𝜓))

Syntax hints:   → wi 4   ↔ wb 206  ∀wal 1540   = wceq 1542  [wsb 2068   ∈ wcel 2114  {cab 2715

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1797  ax-4 1811  ax-5 1912  ax-6 1969  ax-7 2010  ax-9 2124  ax-ext 2709

This theorem depends on definitions:  df-bi 207  df-an 396  df-ex 1782  df-sb 2069  df-clab 2716  df-cleq 2729

This theorem is referenced by:  eqabcbw  2811  ru  3740  vn0  4299  eq0  4304  vpwex  5324  fineqvpow  35293  bj-ru1  37191