Theorem eqab 2899

Description: One direction of eqabb 2900 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 2897	. 2 ⊢ 𝐴 = {𝑥 ∣ 𝑥 ∈ 𝐴}
2		abbi 2826	. 2 ⊢ (∀𝑥(𝑥 ∈ 𝐴 ↔ 𝜑) → {𝑥 ∣ 𝑥 ∈ 𝐴} = {𝑥 ∣ 𝜑})
3	1, 2	eqtrid 2808	1 ⊢ (∀𝑥(𝑥 ∈ 𝐴 ↔ 𝜑) → 𝐴 = {𝑥 ∣ 𝜑})

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-8 2143  ax-9 2151  ax-ext 2733

This theorem depends on definitions:  df-bi 209  df-an 400  df-tru 1562  df-ex 1799  df-sb 2090  df-clab 2740  df-cleq 2753  df-clel 2836

This theorem is referenced by:  rabid2im  3445