Theorem vss 4366

Description: Only the universal class has the universal class as a subclass. Dual of ss0b 4359. (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 539	. 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)

Syntax hints:   ↔ wb 209   ∧ wa 400   = wceq 1570  Vcvv 3455   ⊆ wss 3906

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1825  ax-4 1839  ax-5 1940  ax-6 1997  ax-7 2038  ax-8 2145  ax-9 2153  ax-ext 2735

This theorem depends on definitions:  df-bi 210  df-an 401  df-tru 1573  df-ex 1810  df-sb 2097  df-clab 2742  df-cleq 2755  df-clel 2838  df-v 3457  df-ss 3923

This theorem is referenced by:  vvin  4367  vdif0  4430  fineqvr1ombregs  35532