Theorem eqab 2872

Description: One direction of eqabb 2873 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 2870	. 2 ⊢ 𝐴 = {𝑥 ∣ 𝑥 ∈ 𝐴}
2		abbi 2800	. 2 ⊢ (∀𝑥(𝑥 ∈ 𝐴 ↔ 𝜑) → {𝑥 ∣ 𝑥 ∈ 𝐴} = {𝑥 ∣ 𝜑})
3	1, 2	eqtrid 2784	1 ⊢ (∀𝑥(𝑥 ∈ 𝐴 ↔ 𝜑) → 𝐴 = {𝑥 ∣ 𝜑})

Syntax hints:   → wi 4   ↔ wb 205  ∀wal 1539   = wceq 1541   ∈ wcel 2106  {cab 2709

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1797  ax-4 1811  ax-5 1913  ax-6 1971  ax-7 2011  ax-8 2108  ax-9 2116  ax-ext 2703

This theorem depends on definitions:  df-bi 206  df-an 397  df-tru 1544  df-ex 1782  df-sb 2068  df-clab 2710  df-cleq 2724  df-clel 2810

This theorem is referenced by: (None)