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Mirrors > Home > MPE Home > Th. List > Mathboxes > disjdifr | Structured version Visualization version GIF version |
Description: A class and its relative complement are disjoint. (Contributed by Thierry Arnoux, 29-Nov-2023.) |
Ref | Expression |
---|---|
disjdifr | ⊢ ((𝐵 ∖ 𝐴) ∩ 𝐴) = ∅ |
Step | Hyp | Ref | Expression |
---|---|---|---|
1 | incom 4171 | . 2 ⊢ (𝐴 ∩ (𝐵 ∖ 𝐴)) = ((𝐵 ∖ 𝐴) ∩ 𝐴) | |
2 | disjdif 4414 | . 2 ⊢ (𝐴 ∩ (𝐵 ∖ 𝐴)) = ∅ | |
3 | 1, 2 | eqtr3i 2845 | 1 ⊢ ((𝐵 ∖ 𝐴) ∩ 𝐴) = ∅ |
Colors of variables: wff setvar class |
Syntax hints: = wceq 1536 ∖ cdif 3926 ∩ cin 3928 ∅c0 4284 |
This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-3 8 ax-gen 1795 ax-4 1809 ax-5 1910 ax-6 1969 ax-7 2014 ax-8 2115 ax-9 2123 ax-10 2144 ax-11 2160 ax-12 2176 ax-ext 2792 |
This theorem depends on definitions: df-bi 209 df-an 399 df-or 844 df-tru 1539 df-ex 1780 df-nf 1784 df-sb 2069 df-clab 2799 df-cleq 2813 df-clel 2892 df-nfc 2962 df-rab 3146 df-v 3493 df-dif 3932 df-in 3936 df-ss 3945 df-nul 4285 |
This theorem is referenced by: (None) |
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