Theorem vss 4365

Description: Only the universal class has the universal class as a subclass. Dual of ss0b 4358. (Contributed by NM, 17-Sep-2003.) (Proof shortened by Andrew Salmon, 26-Jun-2011.)

Assertion

Ref	Expression
vss	⊢ (V ⊆ 𝐴 ↔ 𝐴 = V)

Proof of Theorem vss

Step	Hyp	Ref	Expression
1		ssv 3962	. . 3 ⊢ 𝐴 ⊆ V
2	1	biantrur 540	. 2 ⊢ (V ⊆ 𝐴 ↔ (𝐴 ⊆ V ∧ V ⊆ 𝐴))
3		eqss 3953	. 2 ⊢ (𝐴 = V ↔ (𝐴 ⊆ V ∧ V ⊆ 𝐴))
4	2, 3	bitr4i 281	1 ⊢ (V ⊆ 𝐴 ↔ 𝐴 = V)

This proof depends on syntax axioms:   ↔ wb 209   ∧ wa 401   = wceq 1570  Vcvv 3457   ⊆ wss 3906

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  df-ss 3923

This theorem is used by:  vvin  4366  vdif0  4429  fineqvr1ombregs  35567