Theorem eqabcbw 2834

Description: Version of eqabcb 2900 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 2833	. 2 ⊢ (𝐴 = {𝑥 ∣ 𝜑} ↔ ∀𝑦(𝑦 ∈ 𝐴 ↔ 𝜓))
3		eqcom 2767	. 2 ⊢ ({𝑥 ∣ 𝜑} = 𝐴 ↔ 𝐴 = {𝑥 ∣ 𝜑})
4		bicom 225	. . 3 ⊢ ((𝜓 ↔ 𝑦 ∈ 𝐴) ↔ (𝑦 ∈ 𝐴 ↔ 𝜓))
5	4	albii 1851	. 2 ⊢ (∀𝑦(𝜓 ↔ 𝑦 ∈ 𝐴) ↔ ∀𝑦(𝑦 ∈ 𝐴 ↔ 𝜓))
6	2, 3, 5	3bitr4i 306	1 ⊢ ({𝑥 ∣ 𝜑} = 𝐴 ↔ ∀𝑦(𝜓 ↔ 𝑦 ∈ 𝐴))

This proof depends on syntax axioms:   → wi 4   ↔ wb 209  ∀wal 1567   = wceq 1569   ∈ wcel 2144  {cab 2738

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 1942  ax-6 1999  ax-7 2040  ax-9 2154  ax-ext 2732

This proof depends on definitions:  df-bi 210  df-an 402  df-ex 1812  df-sb 2099  df-clab 2739  df-cleq 2752

This theorem is used by:  ab0w  4327  ab0orv  4331  disj  4402  dm0rn0  5905  tz6.12-2  6873