Theorem rabsneq 4666

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 4664	. . . . . 6 ⊢ (𝑥 ∈ {𝑁} ↔ 𝑥 = 𝑁)
2		eleq1a 2839	. . . . . . 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 630	. . . 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 2811	. 2 ⊢ (𝑁 ∈ 𝑉 → {𝑥 ∣ (𝑥 ∈ {𝑁} ∧ 𝜓)} = {𝑥 ∣ (𝑥 ∈ 𝑉 ∧ (𝑥 = 𝑁 ∧ 𝜓))})
9		df-rab 3444	. 2 ⊢ {𝑥 ∈ {𝑁} ∣ 𝜓} = {𝑥 ∣ (𝑥 ∈ {𝑁} ∧ 𝜓)}
10		df-rab 3444	. 2 ⊢ {𝑥 ∈ 𝑉 ∣ (𝑥 = 𝑁 ∧ 𝜓)} = {𝑥 ∣ (𝑥 ∈ 𝑉 ∧ (𝑥 = 𝑁 ∧ 𝜓))}
11	8, 9, 10	3eqtr4g 2805	1 ⊢ (𝑁 ∈ 𝑉 → {𝑥 ∈ {𝑁} ∣ 𝜓} = {𝑥 ∈ 𝑉 ∣ (𝑥 = 𝑁 ∧ 𝜓)})

Syntax hints:   → wi 4   ∧ wa 395   = wceq 1537   ∈ wcel 2108  {cab 2717  {crab 3443  {csn 4648

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1793  ax-4 1807  ax-5 1909  ax-6 1967  ax-7 2007  ax-8 2110  ax-9 2118  ax-ext 2711

This theorem depends on definitions:  df-bi 207  df-an 396  df-tru 1540  df-ex 1778  df-sb 2065  df-clab 2718  df-cleq 2732  df-clel 2819  df-rab 3444  df-v 3490  df-sn 4649

This theorem is referenced by:  dfsclnbgr6  47730