Theorem eqabbw 2801

Description: Version of eqabb 2865 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 2717	. 2 ⊢ (𝐴 = {𝑥 ∣ 𝜑} ↔ ∀𝑦(𝑦 ∈ 𝐴 ↔ 𝑦 ∈ {𝑥 ∣ 𝜑}))
2		df-clab 2702	. . . . 5 ⊢ (𝑦 ∈ {𝑥 ∣ 𝜑} ↔ [𝑦 / 𝑥]𝜑)
3		eqabbw.1	. . . . . 6 ⊢ (𝑥 = 𝑦 → (𝜑 ↔ 𝜓))
4	3	sbievw 2087	. . . . 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 1813	. 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 205  ∀wal 1531   = wceq 1533  [wsb 2059   ∈ wcel 2098  {cab 2701

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1789  ax-4 1803  ax-5 1905  ax-6 1963  ax-7 2003  ax-9 2108  ax-ext 2695

This theorem depends on definitions:  df-bi 206  df-an 396  df-ex 1774  df-sb 2060  df-clab 2702  df-cleq 2716

This theorem is referenced by:  vpwex  5365  fineqvpow  34585