Theorem eqabbw 2834

Description: Version of eqabb 2900 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 2754	. 2 ⊢ (𝐴 = {𝑥 ∣ 𝜑} ↔ ∀𝑦(𝑦 ∈ 𝐴 ↔ 𝑦 ∈ {𝑥 ∣ 𝜑}))
2		df-clab 2740	. . . . 5 ⊢ (𝑦 ∈ {𝑥 ∣ 𝜑} ↔ [𝑦 / 𝑥]𝜑)
3		eqabbw.1	. . . . . 6 ⊢ (𝑥 = 𝑦 → (𝜑 ↔ 𝜓))
4	3	sbievw 2131	. . . . 5 ⊢ ([𝑦 / 𝑥]𝜑 ↔ 𝜓)
5	2, 4	bitri 278	. . . 4 ⊢ (𝑦 ∈ {𝑥 ∣ 𝜑} ↔ 𝜓)
6	5	bibi2i 340	. . 3 ⊢ ((𝑦 ∈ 𝐴 ↔ 𝑦 ∈ {𝑥 ∣ 𝜑}) ↔ (𝑦 ∈ 𝐴 ↔ 𝜓))
7	6	albii 1852	. 2 ⊢ (∀𝑦(𝑦 ∈ 𝐴 ↔ 𝑦 ∈ {𝑥 ∣ 𝜑}) ↔ ∀𝑦(𝑦 ∈ 𝐴 ↔ 𝜓))
8	1, 7	bitri 278	1 ⊢ (𝐴 = {𝑥 ∣ 𝜑} ↔ ∀𝑦(𝑦 ∈ 𝐴 ↔ 𝜓))

This proof depends on syntax axioms:   → wi 4   ↔ wb 209  ∀wal 1568   = wceq 1570  [wsb 2099   ∈ wcel 2145  {cab 2739

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1828  ax-4 1842  ax-5 1943  ax-6 2000  ax-7 2041  ax-9 2155  ax-ext 2733

This proof depends on definitions:  df-bi 210  df-an 402  df-ex 1813  df-sb 2100  df-clab 2740  df-cleq 2753

This theorem is used by:  eqabcbw  2835  ru  3738  vn0  4291  vn0OLD  4292  eq0  4297  vpwex  5339  fineqvpow  35783  bj-ru1  37856