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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 3019 | . . 3 ⊢ 𝐴 ⊆ V | |
2 | 1 | biantrur 297 | . 2 ⊢ (V ⊆ 𝐴 ↔ (𝐴 ⊆ V ∧ V ⊆ 𝐴)) |
3 | eqss 3014 | . 2 ⊢ (𝐴 = V ↔ (𝐴 ⊆ V ∧ V ⊆ 𝐴)) | |
4 | 2, 3 | bitr4i 185 | 1 ⊢ (V ⊆ 𝐴 ↔ 𝐴 = V) |
Colors of variables: wff set class |
Syntax hints: ∧ wa 102 ↔ wb 103 = wceq 1284 Vcvv 2601 ⊆ wss 2973 |
This theorem was proved from axioms: ax-1 5 ax-2 6 ax-mp 7 ax-ia1 104 ax-ia2 105 ax-ia3 106 ax-5 1376 ax-7 1377 ax-gen 1378 ax-ie1 1422 ax-ie2 1423 ax-8 1435 ax-11 1437 ax-4 1440 ax-17 1459 ax-i9 1463 ax-ial 1467 ax-i5r 1468 ax-ext 2063 |
This theorem depends on definitions: df-bi 115 df-nf 1390 df-sb 1686 df-clab 2068 df-cleq 2074 df-clel 2077 df-v 2603 df-in 2979 df-ss 2986 |
This theorem is referenced by: vdif0im 3309 |
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