Theorem eqsnd 29939

Description: Deduce that a set is a singleton. (Contributed by Thierry Arnoux, 10-May-2023.)

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	. . . 4 ⊢ ((𝜑 ∧ 𝑥 ∈ 𝐴) → 𝑥 = 𝐵)
2		simpr 479	. . . . 5 ⊢ ((𝜑 ∧ 𝑥 = 𝐵) → 𝑥 = 𝐵)
3		eqsnd.2	. . . . . 6 ⊢ (𝜑 → 𝐵 ∈ 𝐴)
4	3	adantr 474	. . . . 5 ⊢ ((𝜑 ∧ 𝑥 = 𝐵) → 𝐵 ∈ 𝐴)
5	2, 4	eqeltrd 2859	. . . 4 ⊢ ((𝜑 ∧ 𝑥 = 𝐵) → 𝑥 ∈ 𝐴)
6	1, 5	impbida 791	. . 3 ⊢ (𝜑 → (𝑥 ∈ 𝐴 ↔ 𝑥 = 𝐵))
7		vex 3401	. . . 4 ⊢ 𝑥 ∈ V
8	7	elsn 4413	. . 3 ⊢ (𝑥 ∈ {𝐵} ↔ 𝑥 = 𝐵)
9	6, 8	syl6bbr 281	. 2 ⊢ (𝜑 → (𝑥 ∈ 𝐴 ↔ 𝑥 ∈ {𝐵}))
10	9	eqrdv 2776	1 ⊢ (𝜑 → 𝐴 = {𝐵})

Syntax hints:   → wi 4   ∧ wa 386   = wceq 1601   ∈ wcel 2107  {csn 4398

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1839  ax-4 1853  ax-5 1953  ax-6 2021  ax-7 2055  ax-9 2116  ax-10 2135  ax-11 2150  ax-12 2163  ax-ext 2754

This theorem depends on definitions:  df-bi 199  df-an 387  df-or 837  df-tru 1605  df-ex 1824  df-nf 1828  df-sb 2012  df-clab 2764  df-cleq 2770  df-clel 2774  df-nfc 2921  df-v 3400  df-sn 4399

This theorem is referenced by:  lbsdiflsp0  30448