Theorem eqsnd 4798

Description: Deduce that a set is a singleton. (Contributed by Thierry Arnoux, 10-May-2023.) (Proof shortened by SN, 3-Jul-2025.)

Hypotheses

Ref	Expression
eqsnd.1	⊢ ((𝜑 ∧ 𝑥 ∈ 𝐴) → 𝑥 = 𝐵)
eqsnd.2	⊢ (𝜑 → 𝐵 ∈ 𝐴)

Assertion

Ref	Expression
eqsnd	⊢ (𝜑 → 𝐴 = {𝐵})

Distinct variable groups:   𝑥,𝐴   𝑥,𝐵   𝜑,𝑥

Proof of Theorem eqsnd

Step	Hyp	Ref	Expression
1		eqsnd.1	. . 3 ⊢ ((𝜑 ∧ 𝑥 ∈ 𝐴) → 𝑥 = 𝐵)
2	1	ralrimiva 3159	. 2 ⊢ (𝜑 → ∀𝑥 ∈ 𝐴 𝑥 = 𝐵)
3		eqsnd.2	. . . 4 ⊢ (𝜑 → 𝐵 ∈ 𝐴)
4	3	ne0d 4295	. . 3 ⊢ (𝜑 → 𝐴 ≠ ∅)
5		eqsn 4797	. . 3 ⊢ (𝐴 ≠ ∅ → (𝐴 = {𝐵} ↔ ∀𝑥 ∈ 𝐴 𝑥 = 𝐵))
6	4, 5	syl 18	. 2 ⊢ (𝜑 → (𝐴 = {𝐵} ↔ ∀𝑥 ∈ 𝐴 𝑥 = 𝐵))
7	2, 6	mpbird 260	1 ⊢ (𝜑 → 𝐴 = {𝐵})

This proof depends on syntax axioms:   → wi 4   ↔ wb 209   ∧ wa 401   = wceq 1570   ∈ wcel 2146   ≠ wne 2960  ∀wral 3081  ∅c0 4286  {csn 4591

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1828  ax-4 1842  ax-5 1943  ax-6 2000  ax-7 2041  ax-8 2148  ax-9 2156  ax-ext 2737

This proof depends on definitions:  df-bi 210  df-an 402  df-or 862  df-tru 1573  df-fal 1583  df-ex 1813  df-sb 2100  df-clab 2744  df-cleq 2757  df-clel 2840  df-ne 2961  df-ral 3082  df-v 3459  df-dif 3909  df-ss 3923  df-nul 4287  df-sn 4592

This theorem is used by:  0ringidl  21389  tglineinsn  28946  dflring3  33810  dflring4  33811  lbsdiflsp0  34039  fiabv  43337  thinchom  50238