| 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 3990 | . . 3 ⊢ 𝐴 ⊆ V | |
| 2 | 1 | biantrur 530 | . 2 ⊢ (V ⊆ 𝐴 ↔ (𝐴 ⊆ V ∧ V ⊆ 𝐴)) |
| 3 | eqss 3981 | . 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 206 ∧ wa 395 = wceq 1539 Vcvv 3464 ⊆ wss 3933 |
| This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-3 8 ax-gen 1794 ax-4 1808 ax-5 1909 ax-6 1966 ax-7 2006 ax-8 2109 ax-9 2117 ax-ext 2706 |
| This theorem depends on definitions: df-bi 207 df-an 396 df-tru 1542 df-ex 1779 df-sb 2064 df-clab 2713 df-cleq 2726 df-clel 2808 df-v 3466 df-ss 3950 |
| This theorem is referenced by: vdif0 4451 |
| Copyright terms: Public domain | W3C validator |