Theorem sbceqal 3807

Description: Class version of one implication of equvelv 2064. (Contributed by Andrew Salmon, 28-Jun-2011.) (Proof shortened by SN, 26-Oct-2024.)

Assertion

Ref	Expression
sbceqal	⊢ (𝐴 ∈ 𝑉 → (∀𝑥(𝑥 = 𝐴 → 𝑥 = 𝐵) → 𝐴 = 𝐵))

Distinct variable groups:   𝑥,𝐵   𝑥,𝐴

Allowed substitution hint:   𝑉(𝑥)

Proof of Theorem sbceqal

Step	Hyp	Ref	Expression
1		eqeq1 2769	. . . 4 ⊢ (𝑥 = 𝐴 → (𝑥 = 𝐴 ↔ 𝐴 = 𝐴))
2		eqeq1 2769	. . . 4 ⊢ (𝑥 = 𝐴 → (𝑥 = 𝐵 ↔ 𝐴 = 𝐵))
3	1, 2	imbi12d 347	. . 3 ⊢ (𝑥 = 𝐴 → ((𝑥 = 𝐴 → 𝑥 = 𝐵) ↔ (𝐴 = 𝐴 → 𝐴 = 𝐵)))
4		eqid 2765	. . . 4 ⊢ 𝐴 = 𝐴
5	4	a1bi 365	. . 3 ⊢ (𝐴 = 𝐵 ↔ (𝐴 = 𝐴 → 𝐴 = 𝐵))
6	3, 5	bitr4di 292	. 2 ⊢ (𝑥 = 𝐴 → ((𝑥 = 𝐴 → 𝑥 = 𝐵) ↔ 𝐴 = 𝐵))
7	6	spcgv 3557	1 ⊢ (𝐴 ∈ 𝑉 → (∀𝑥(𝑥 = 𝐴 → 𝑥 = 𝐵) → 𝐴 = 𝐵))

This proof depends on syntax axioms:   → wi 4  ∀wal 1568   = wceq 1570   ∈ wcel 2146

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-tru 1573  df-ex 1813  df-sb 2100  df-clab 2744  df-cleq 2757  df-clel 2840  df-v 3459

This theorem is used by:  sbeqalb  3808