Metamath Proof Explorer


Theorem vss

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

Ref Expression
Assertion vss ⊢ V ⊆ A ↔ A = V

Proof

Step Hyp Ref Expression
1 ssv ⊢ A ⊆ V
2 1 biantrur ⊢ V ⊆ A ↔ A ⊆ V ∧ V ⊆ A
3 eqss ⊢ A = V ↔ A ⊆ V ∧ V ⊆ A
4 2 3 bitr4i ⊢ V ⊆ A ↔ A = V