Theorem eqab 2864

Description: One direction of eqabb 2865 is provable from fewer axioms. (Contributed by Wolf Lammen, 13-Feb-2025.)

Assertion

Ref	Expression
eqab	⊢ (∀𝑥(𝑥 ∈ 𝐴 ↔ 𝜑) → 𝐴 = {𝑥 ∣ 𝜑})

Distinct variable group:   𝑥,𝐴

Allowed substitution hint:   𝜑(𝑥)

Proof of Theorem eqab

Step	Hyp	Ref	Expression
1		abid1 2862	. 2 ⊢ 𝐴 = {𝑥 ∣ 𝑥 ∈ 𝐴}
2		abbi 2792	. 2 ⊢ (∀𝑥(𝑥 ∈ 𝐴 ↔ 𝜑) → {𝑥 ∣ 𝑥 ∈ 𝐴} = {𝑥 ∣ 𝜑})
3	1, 2	eqtrid 2776	1 ⊢ (∀𝑥(𝑥 ∈ 𝐴 ↔ 𝜑) → 𝐴 = {𝑥 ∣ 𝜑})

Syntax hints:   → wi 4   ↔ wb 205  ∀wal 1531   = wceq 1533   ∈ wcel 2098  {cab 2701

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1789  ax-4 1803  ax-5 1905  ax-6 1963  ax-7 2003  ax-8 2100  ax-9 2108  ax-ext 2695

This theorem depends on definitions:  df-bi 206  df-an 396  df-tru 1536  df-ex 1774  df-sb 2060  df-clab 2702  df-cleq 2716  df-clel 2802

This theorem is referenced by: (None)