Theorem eqabbw 2836

Description: Version of eqabb 2902 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 2756	. 2 ⊢ (𝐴 = {𝑥 ∣ 𝜑} ↔ ∀𝑦(𝑦 ∈ 𝐴 ↔ 𝑦 ∈ {𝑥 ∣ 𝜑}))
2		df-clab 2742	. . . . 5 ⊢ (𝑦 ∈ {𝑥 ∣ 𝜑} ↔ [𝑦 / 𝑥]𝜑)
3		eqabbw.1	. . . . . 6 ⊢ (𝑥 = 𝑦 → (𝜑 ↔ 𝜓))
4	3	sbievw 2132	. . . . 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 1851	. 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 1567   = wceq 1569  [wsb 2100   ∈ wcel 2146  {cab 2741

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1827  ax-4 1841  ax-5 1944  ax-6 2001  ax-7 2042  ax-9 2156  ax-ext 2735

This proof depends on definitions:  df-bi 210  df-an 402  df-ex 1812  df-sb 2101  df-clab 2742  df-cleq 2755

This theorem is used by:  eqabcbw  2837  ru  3740  vn0  4293  vn0OLD  4294  eq0  4299  vpwex  5342  fineqvpow  35854  bj-ru1  37983