Theorem eqabcbw 2805

Description: Version of eqabcb 2872 using implicit substitution, which requires fewer axioms. (Contributed by TM, 24-Jan-2026.)

Hypothesis

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

Assertion

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

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

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

Proof of Theorem eqabcbw

Step	Hyp	Ref	Expression
1		eqabbw.1	. . 3 ⊢ (𝑥 = 𝑦 → (𝜑 ↔ 𝜓))
2	1	eqabbw 2804	. 2 ⊢ (𝐴 = {𝑥 ∣ 𝜑} ↔ ∀𝑦(𝑦 ∈ 𝐴 ↔ 𝜓))
3		eqcom 2738	. 2 ⊢ ({𝑥 ∣ 𝜑} = 𝐴 ↔ 𝐴 = {𝑥 ∣ 𝜑})
4		bicom 222	. . 3 ⊢ ((𝜓 ↔ 𝑦 ∈ 𝐴) ↔ (𝑦 ∈ 𝐴 ↔ 𝜓))
5	4	albii 1820	. 2 ⊢ (∀𝑦(𝜓 ↔ 𝑦 ∈ 𝐴) ↔ ∀𝑦(𝑦 ∈ 𝐴 ↔ 𝜓))
6	2, 3, 5	3bitr4i 303	1 ⊢ ({𝑥 ∣ 𝜑} = 𝐴 ↔ ∀𝑦(𝜓 ↔ 𝑦 ∈ 𝐴))

Syntax hints:   → wi 4   ↔ wb 206  ∀wal 1539   = wceq 1541   ∈ wcel 2111  {cab 2709

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1796  ax-4 1810  ax-5 1911  ax-6 1968  ax-7 2009  ax-9 2121  ax-ext 2703

This theorem depends on definitions:  df-bi 207  df-an 396  df-ex 1781  df-sb 2068  df-clab 2710  df-cleq 2723

This theorem is referenced by:  disj  4400  dm0rn0  5864  tz6.12-2  6809