Theorem rabbiiaOLD 3425

Description: Obsolete version of rabbiia 3424 as of 12-Jan-2025. (Contributed by NM, 22-May-1999.) (Proof modification is discouraged.) (New usage is discouraged.)

Hypothesis

Ref	Expression
rabbiia.1	⊢ (𝑥 ∈ 𝐴 → (𝜑 ↔ 𝜓))

Assertion

Ref	Expression
rabbiiaOLD	⊢ {𝑥 ∈ 𝐴 ∣ 𝜑} = {𝑥 ∈ 𝐴 ∣ 𝜓}

Proof of Theorem rabbiiaOLD

Step	Hyp	Ref	Expression
1		rabbiia.1	. . . 4 ⊢ (𝑥 ∈ 𝐴 → (𝜑 ↔ 𝜓))
2	1	pm5.32i 574	. . 3 ⊢ ((𝑥 ∈ 𝐴 ∧ 𝜑) ↔ (𝑥 ∈ 𝐴 ∧ 𝜓))
3	2	abbii 2801	. 2 ⊢ {𝑥 ∣ (𝑥 ∈ 𝐴 ∧ 𝜑)} = {𝑥 ∣ (𝑥 ∈ 𝐴 ∧ 𝜓)}
4		df-rab 3421	. 2 ⊢ {𝑥 ∈ 𝐴 ∣ 𝜑} = {𝑥 ∣ (𝑥 ∈ 𝐴 ∧ 𝜑)}
5		df-rab 3421	. 2 ⊢ {𝑥 ∈ 𝐴 ∣ 𝜓} = {𝑥 ∣ (𝑥 ∈ 𝐴 ∧ 𝜓)}
6	3, 4, 5	3eqtr4i 2767	1 ⊢ {𝑥 ∈ 𝐴 ∣ 𝜑} = {𝑥 ∈ 𝐴 ∣ 𝜓}

Syntax hints:   → wi 4   ↔ wb 206   ∧ wa 395   = wceq 1539   ∈ wcel 2107  {cab 2712  {crab 3420

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1794  ax-4 1808  ax-5 1909  ax-6 1966  ax-7 2006  ax-9 2117  ax-ext 2706

This theorem depends on definitions:  df-bi 207  df-an 396  df-ex 1779  df-sb 2064  df-clab 2713  df-cleq 2726  df-rab 3421

This theorem is referenced by: (None)