Theorem rabsneq 4643

Description: Equality of class abstractions restricted to a singleton. (Contributed by AV, 17-May-2025.)

Assertion

Ref	Expression
rabsneq	⊢ (𝑁 ∈ 𝑉 → {𝑥 ∈ {𝑁} ∣ 𝜓} = {𝑥 ∈ 𝑉 ∣ (𝑥 = 𝑁 ∧ 𝜓)})

Distinct variable groups:   𝑥,𝑁   𝑥,𝑉

Allowed substitution hint:   𝜓(𝑥)

Proof of Theorem rabsneq

Step	Hyp	Ref	Expression
1		velsn 4641	. . . . . 6 ⊢ (𝑥 ∈ {𝑁} ↔ 𝑥 = 𝑁)
2		eleq1a 2835	. . . . . . 7 ⊢ (𝑁 ∈ 𝑉 → (𝑥 = 𝑁 → 𝑥 ∈ 𝑉))
3	2	pm4.71rd 562	. . . . . 6 ⊢ (𝑁 ∈ 𝑉 → (𝑥 = 𝑁 ↔ (𝑥 ∈ 𝑉 ∧ 𝑥 = 𝑁)))
4	1, 3	bitrid 283	. . . . 5 ⊢ (𝑁 ∈ 𝑉 → (𝑥 ∈ {𝑁} ↔ (𝑥 ∈ 𝑉 ∧ 𝑥 = 𝑁)))
5	4	anbi1d 631	. . . 4 ⊢ (𝑁 ∈ 𝑉 → ((𝑥 ∈ {𝑁} ∧ 𝜓) ↔ ((𝑥 ∈ 𝑉 ∧ 𝑥 = 𝑁) ∧ 𝜓)))
6		anass 468	. . . 4 ⊢ (((𝑥 ∈ 𝑉 ∧ 𝑥 = 𝑁) ∧ 𝜓) ↔ (𝑥 ∈ 𝑉 ∧ (𝑥 = 𝑁 ∧ 𝜓)))
7	5, 6	bitrdi 287	. . 3 ⊢ (𝑁 ∈ 𝑉 → ((𝑥 ∈ {𝑁} ∧ 𝜓) ↔ (𝑥 ∈ 𝑉 ∧ (𝑥 = 𝑁 ∧ 𝜓))))
8	7	abbidv 2807	. 2 ⊢ (𝑁 ∈ 𝑉 → {𝑥 ∣ (𝑥 ∈ {𝑁} ∧ 𝜓)} = {𝑥 ∣ (𝑥 ∈ 𝑉 ∧ (𝑥 = 𝑁 ∧ 𝜓))})
9		df-rab 3436	. 2 ⊢ {𝑥 ∈ {𝑁} ∣ 𝜓} = {𝑥 ∣ (𝑥 ∈ {𝑁} ∧ 𝜓)}
10		df-rab 3436	. 2 ⊢ {𝑥 ∈ 𝑉 ∣ (𝑥 = 𝑁 ∧ 𝜓)} = {𝑥 ∣ (𝑥 ∈ 𝑉 ∧ (𝑥 = 𝑁 ∧ 𝜓))}
11	8, 9, 10	3eqtr4g 2801	1 ⊢ (𝑁 ∈ 𝑉 → {𝑥 ∈ {𝑁} ∣ 𝜓} = {𝑥 ∈ 𝑉 ∣ (𝑥 = 𝑁 ∧ 𝜓)})

Syntax hints:   → wi 4   ∧ wa 395   = wceq 1539   ∈ wcel 2107  {cab 2713  {crab 3435  {csn 4625

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-8 2109  ax-9 2117  ax-ext 2707

This theorem depends on definitions:  df-bi 207  df-an 396  df-tru 1542  df-ex 1779  df-sb 2064  df-clab 2714  df-cleq 2728  df-clel 2815  df-rab 3436  df-v 3481  df-sn 4626

This theorem is referenced by:  dfsclnbgr6  47849