![]() |
Metamath Proof Explorer |
< Previous
Next >
Nearby theorems |
|
Mirrors > Home > MPE Home > Th. List > vss | Structured version Visualization version GIF version |
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.) |
Ref | Expression |
---|---|
vss | ⊢ (V ⊆ 𝐴 ↔ 𝐴 = V) |
Step | Hyp | Ref | Expression |
---|---|---|---|
1 | ssv 4002 | . . 3 ⊢ 𝐴 ⊆ V | |
2 | 1 | biantrur 530 | . 2 ⊢ (V ⊆ 𝐴 ↔ (𝐴 ⊆ V ∧ V ⊆ 𝐴)) |
3 | eqss 3993 | . 2 ⊢ (𝐴 = V ↔ (𝐴 ⊆ V ∧ V ⊆ 𝐴)) | |
4 | 2, 3 | bitr4i 278 | 1 ⊢ (V ⊆ 𝐴 ↔ 𝐴 = V) |
Colors of variables: wff setvar class |
Syntax hints: ↔ wb 205 ∧ wa 395 = wceq 1534 Vcvv 3469 ⊆ wss 3944 |
This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-3 8 ax-gen 1790 ax-4 1804 ax-5 1906 ax-6 1964 ax-7 2004 ax-8 2101 ax-9 2109 ax-ext 2698 |
This theorem depends on definitions: df-bi 206 df-an 396 df-tru 1537 df-ex 1775 df-sb 2061 df-clab 2705 df-cleq 2719 df-clel 2805 df-v 3471 df-in 3951 df-ss 3961 |
This theorem is referenced by: vdif0 4464 |
Copyright terms: Public domain | W3C validator |