Theorem rabbidvaOLD 3476

Description: Obsolete proof of rabbidva 3475 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 3181	. 2 ⊢ (𝜑 → ∀𝑥 ∈ 𝐴 (𝜓 ↔ 𝜒))
3		rabbi 3382	. 2 ⊢ (∀𝑥 ∈ 𝐴 (𝜓 ↔ 𝜒) ↔ {𝑥 ∈ 𝐴 ∣ 𝜓} = {𝑥 ∈ 𝐴 ∣ 𝜒})
4	2, 3	sylib 220	1 ⊢ (𝜑 → {𝑥 ∈ 𝐴 ∣ 𝜓} = {𝑥 ∈ 𝐴 ∣ 𝜒})

Syntax hints:   → wi 4   ↔ wb 208   ∧ wa 398   = wceq 1536   ∈ wcel 2113  ∀wral 3137  {crab 3141

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1795  ax-4 1809  ax-5 1910  ax-6 1969  ax-7 2014  ax-9 2123  ax-10 2144  ax-11 2160  ax-12 2176  ax-ext 2792

This theorem depends on definitions:  df-bi 209  df-an 399  df-or 844  df-tru 1539  df-ex 1780  df-nf 1784  df-sb 2069  df-clab 2799  df-cleq 2813  df-ral 3142  df-rab 3146

This theorem is referenced by: (None)