Theorem eqab 2901

Description: One direction of eqabb 2902 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 2899	. 2 ⊢ 𝐴 = {𝑥 ∣ 𝑥 ∈ 𝐴}
2		abbi 2828	. 2 ⊢ (∀𝑥(𝑥 ∈ 𝐴 ↔ 𝜑) → {𝑥 ∣ 𝑥 ∈ 𝐴} = {𝑥 ∣ 𝜑})
3	1, 2	eqtrid 2810	1 ⊢ (∀𝑥(𝑥 ∈ 𝐴 ↔ 𝜑) → 𝐴 = {𝑥 ∣ 𝜑})

This proof depends on syntax axioms:   → wi 4   ↔ wb 209  ∀wal 1567   = wceq 1569   ∈ wcel 2146  {cab 2741

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 1944  ax-6 2001  ax-7 2042  ax-8 2148  ax-9 2156  ax-ext 2735

This proof depends on definitions:  df-bi 210  df-an 402  df-tru 1572  df-ex 1812  df-sb 2101  df-clab 2742  df-cleq 2755  df-clel 2838

This theorem is used by:  rabid2im  3446