Theorem vss 4395

Description: Only the universal class has the universal class as a subclass. (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 3991	. . 3 ⊢ 𝐴 ⊆ V
2	1	biantrur 533	. 2 ⊢ (V ⊆ 𝐴 ↔ (𝐴 ⊆ V ∧ V ⊆ 𝐴))
3		eqss 3982	. 2 ⊢ (𝐴 = V ↔ (𝐴 ⊆ V ∧ V ⊆ 𝐴))
4	2, 3	bitr4i 280	1 ⊢ (V ⊆ 𝐴 ↔ 𝐴 = V)

Syntax hints:   ↔ wb 208   ∧ wa 398   = wceq 1533  Vcvv 3495   ⊆ wss 3936

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1792  ax-4 1806  ax-5 1907  ax-6 1966  ax-7 2011  ax-8 2112  ax-9 2120  ax-10 2141  ax-11 2156  ax-12 2172  ax-ext 2793

This theorem depends on definitions:  df-bi 209  df-an 399  df-or 844  df-tru 1536  df-ex 1777  df-nf 1781  df-sb 2066  df-clab 2800  df-cleq 2814  df-clel 2893  df-v 3497  df-in 3943  df-ss 3952

This theorem is referenced by:  vdif0  4418