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Mirrors > Home > ILE Home > Th. List > vss | 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 3201 | . . 3 ⊢ 𝐴 ⊆ V | |
2 | 1 | biantrur 303 | . 2 ⊢ (V ⊆ 𝐴 ↔ (𝐴 ⊆ V ∧ V ⊆ 𝐴)) |
3 | eqss 3194 | . 2 ⊢ (𝐴 = V ↔ (𝐴 ⊆ V ∧ V ⊆ 𝐴)) | |
4 | 2, 3 | bitr4i 187 | 1 ⊢ (V ⊆ 𝐴 ↔ 𝐴 = V) |
Colors of variables: wff set class |
Syntax hints: ∧ wa 104 ↔ wb 105 = wceq 1364 Vcvv 2760 ⊆ wss 3153 |
This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-ia1 106 ax-ia2 107 ax-ia3 108 ax-5 1458 ax-7 1459 ax-gen 1460 ax-ie1 1504 ax-ie2 1505 ax-8 1515 ax-11 1517 ax-4 1521 ax-17 1537 ax-i9 1541 ax-ial 1545 ax-i5r 1546 ax-ext 2175 |
This theorem depends on definitions: df-bi 117 df-nf 1472 df-sb 1774 df-clab 2180 df-cleq 2186 df-clel 2189 df-v 2762 df-in 3159 df-ss 3166 |
This theorem is referenced by: vdif0im 3512 |
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