Theorem eqabcbw 2835

Description: Version of eqabcb 2901 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 2834	. 2 ⊢ (𝐴 = {𝑥 ∣ 𝜑} ↔ ∀𝑦(𝑦 ∈ 𝐴 ↔ 𝜓))
3		eqcom 2768	. 2 ⊢ ({𝑥 ∣ 𝜑} = 𝐴 ↔ 𝐴 = {𝑥 ∣ 𝜑})
4		bicom 224	. . 3 ⊢ ((𝜓 ↔ 𝑦 ∈ 𝐴) ↔ (𝑦 ∈ 𝐴 ↔ 𝜓))
5	4	albii 1838	. 2 ⊢ (∀𝑦(𝜓 ↔ 𝑦 ∈ 𝐴) ↔ ∀𝑦(𝑦 ∈ 𝐴 ↔ 𝜓))
6	2, 3, 5	3bitr4i 305	1 ⊢ ({𝑥 ∣ 𝜑} = 𝐴 ↔ ∀𝑦(𝜓 ↔ 𝑦 ∈ 𝐴))

Syntax hints:   → wi 4   ↔ wb 208  ∀wal 1557   = wceq 1559   ∈ wcel 2141  {cab 2739

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1814  ax-4 1828  ax-5 1929  ax-6 1986  ax-7 2027  ax-9 2151  ax-ext 2733

This theorem depends on definitions:  df-bi 209  df-an 400  df-ex 1799  df-sb 2090  df-clab 2740  df-cleq 2753

This theorem is referenced by:  ab0w  4331  ab0orv  4335  disj  4403  dm0rn0  5898  tz6.12-2  6850