Theorem rabbidvaOLD 3389

Description: Obsolete proof of rabbidva 3388 as of 4-Dec-2023. (Contributed by NM, 28-Nov-2003.) (New usage is discouraged.) (Proof modification is discouraged.)

Hypothesis

Ref	Expression
rabbidva.1	⊢ ((𝜑 ∧ 𝑥 ∈ 𝐴) → (𝜓 ↔ 𝜒))

Assertion

Ref	Expression
rabbidvaOLD	⊢ (𝜑 → {𝑥 ∈ 𝐴 ∣ 𝜓} = {𝑥 ∈ 𝐴 ∣ 𝜒})

Distinct variable group:   𝜑,𝑥

Allowed substitution hints:   𝜓(𝑥)   𝜒(𝑥)   𝐴(𝑥)

Proof of Theorem rabbidvaOLD

Step	Hyp	Ref	Expression
1		rabbidva.1	. . 3 ⊢ ((𝜑 ∧ 𝑥 ∈ 𝐴) → (𝜓 ↔ 𝜒))
2	1	ralrimiva 3105	. 2 ⊢ (𝜑 → ∀𝑥 ∈ 𝐴 (𝜓 ↔ 𝜒))
3		rabbi 3295	. 2 ⊢ (∀𝑥 ∈ 𝐴 (𝜓 ↔ 𝜒) ↔ {𝑥 ∈ 𝐴 ∣ 𝜓} = {𝑥 ∈ 𝐴 ∣ 𝜒})
4	2, 3	sylib 221	1 ⊢ (𝜑 → {𝑥 ∈ 𝐴 ∣ 𝜓} = {𝑥 ∈ 𝐴 ∣ 𝜒})

Syntax hints:   → wi 4   ↔ wb 209   ∧ wa 399   = wceq 1543   ∈ wcel 2110  ∀wral 3061  {crab 3065

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1803  ax-4 1817  ax-5 1918  ax-6 1976  ax-7 2016  ax-9 2120  ax-10 2141  ax-11 2158  ax-12 2175  ax-ext 2708

This theorem depends on definitions:  df-bi 210  df-an 400  df-or 848  df-tru 1546  df-ex 1788  df-nf 1792  df-sb 2071  df-clab 2715  df-cleq 2729  df-ral 3066  df-rab 3070

This theorem is referenced by: (None)