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Mirrors > Home > MPE Home > Th. List > disj2 | Structured version Visualization version GIF version |
Description: Two ways of saying that two classes are disjoint. (Contributed by NM, 17-May-1998.) |
Ref | Expression |
---|---|
disj2 | ⊢ ((𝐴 ∩ 𝐵) = ∅ ↔ 𝐴 ⊆ (V ∖ 𝐵)) |
Step | Hyp | Ref | Expression |
---|---|---|---|
1 | ssv 4020 | . 2 ⊢ 𝐴 ⊆ V | |
2 | reldisj 4459 | . 2 ⊢ (𝐴 ⊆ V → ((𝐴 ∩ 𝐵) = ∅ ↔ 𝐴 ⊆ (V ∖ 𝐵))) | |
3 | 1, 2 | ax-mp 5 | 1 ⊢ ((𝐴 ∩ 𝐵) = ∅ ↔ 𝐴 ⊆ (V ∖ 𝐵)) |
Colors of variables: wff setvar class |
Syntax hints: ↔ wb 206 = wceq 1537 Vcvv 3478 ∖ cdif 3960 ∩ cin 3962 ⊆ wss 3963 ∅c0 4339 |
This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-3 8 ax-gen 1792 ax-4 1806 ax-5 1908 ax-6 1965 ax-7 2005 ax-8 2108 ax-9 2116 ax-ext 2706 |
This theorem depends on definitions: df-bi 207 df-an 396 df-tru 1540 df-fal 1550 df-ex 1777 df-sb 2063 df-clab 2713 df-cleq 2727 df-clel 2814 df-ral 3060 df-v 3480 df-dif 3966 df-in 3970 df-ss 3980 df-nul 4340 |
This theorem is referenced by: ssindif0 4470 intirr 6141 setsres 17212 setscom 17214 f1omvdco3 19482 psgnunilem5 19527 opsrtoslem2 22098 clsconn 23454 cldsubg 24135 uniinn0 32571 imadifxp 32621 |
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