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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 3177 | . . 3 ⊢ 𝐴 ⊆ V | |
2 | 1 | biantrur 303 | . 2 ⊢ (V ⊆ 𝐴 ↔ (𝐴 ⊆ V ∧ V ⊆ 𝐴)) |
3 | eqss 3170 | . 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 1353 Vcvv 2737 ⊆ wss 3129 |
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 1447 ax-7 1448 ax-gen 1449 ax-ie1 1493 ax-ie2 1494 ax-8 1504 ax-11 1506 ax-4 1510 ax-17 1526 ax-i9 1530 ax-ial 1534 ax-i5r 1535 ax-ext 2159 |
This theorem depends on definitions: df-bi 117 df-nf 1461 df-sb 1763 df-clab 2164 df-cleq 2170 df-clel 2173 df-v 2739 df-in 3135 df-ss 3142 |
This theorem is referenced by: vdif0im 3488 |
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