Theorem rabsneq 4590

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 4587	. . . . . 6 ⊢ (𝑥 ∈ {𝑁} ↔ 𝑥 = 𝑁)
2		eleq1a 2826	. . . . . . 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 2797	. 2 ⊢ (𝑁 ∈ 𝑉 → {𝑥 ∣ (𝑥 ∈ {𝑁} ∧ 𝜓)} = {𝑥 ∣ (𝑥 ∈ 𝑉 ∧ (𝑥 = 𝑁 ∧ 𝜓))})
9		df-rab 3396	. 2 ⊢ {𝑥 ∈ {𝑁} ∣ 𝜓} = {𝑥 ∣ (𝑥 ∈ {𝑁} ∧ 𝜓)}
10		df-rab 3396	. 2 ⊢ {𝑥 ∈ 𝑉 ∣ (𝑥 = 𝑁 ∧ 𝜓)} = {𝑥 ∣ (𝑥 ∈ 𝑉 ∧ (𝑥 = 𝑁 ∧ 𝜓))}
11	8, 9, 10	3eqtr4g 2791	1 ⊢ (𝑁 ∈ 𝑉 → {𝑥 ∈ {𝑁} ∣ 𝜓} = {𝑥 ∈ 𝑉 ∣ (𝑥 = 𝑁 ∧ 𝜓)})

Syntax hints:   → wi 4   ∧ wa 395   = wceq 1541   ∈ wcel 2111  {cab 2709  {crab 3395  {csn 4571

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1796  ax-4 1810  ax-5 1911  ax-6 1968  ax-7 2009  ax-8 2113  ax-9 2121  ax-ext 2703

This theorem depends on definitions:  df-bi 207  df-an 396  df-tru 1544  df-ex 1781  df-sb 2068  df-clab 2710  df-cleq 2723  df-clel 2806  df-rab 3396  df-v 3438  df-sn 4572

This theorem is referenced by:  dfsclnbgr6  47889