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| Mirrors > Home > ILE Home > Th. List > disjr | GIF version | ||
| Description: Two ways of saying that two classes are disjoint. (Contributed by Jeff Madsen, 19-Jun-2011.) |
| Ref | Expression |
|---|---|
| disjr | ⊢ ((𝐴 ∩ 𝐵) = ∅ ↔ ∀𝑥 ∈ 𝐵 ¬ 𝑥 ∈ 𝐴) |
| Step | Hyp | Ref | Expression |
|---|---|---|---|
| 1 | incom 3396 | . . 3 ⊢ (𝐴 ∩ 𝐵) = (𝐵 ∩ 𝐴) | |
| 2 | 1 | eqeq1i 2237 | . 2 ⊢ ((𝐴 ∩ 𝐵) = ∅ ↔ (𝐵 ∩ 𝐴) = ∅) |
| 3 | disj 3540 | . 2 ⊢ ((𝐵 ∩ 𝐴) = ∅ ↔ ∀𝑥 ∈ 𝐵 ¬ 𝑥 ∈ 𝐴) | |
| 4 | 2, 3 | bitri 184 | 1 ⊢ ((𝐴 ∩ 𝐵) = ∅ ↔ ∀𝑥 ∈ 𝐵 ¬ 𝑥 ∈ 𝐴) |
| Colors of variables: wff set class |
| Syntax hints: ¬ wn 3 ↔ wb 105 = wceq 1395 ∈ wcel 2200 ∀wral 2508 ∩ cin 3196 ∅c0 3491 |
| 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-in1 617 ax-in2 618 ax-io 714 ax-5 1493 ax-7 1494 ax-gen 1495 ax-ie1 1539 ax-ie2 1540 ax-8 1550 ax-10 1551 ax-11 1552 ax-i12 1553 ax-bndl 1555 ax-4 1556 ax-17 1572 ax-i9 1576 ax-ial 1580 ax-i5r 1581 ax-ext 2211 |
| This theorem depends on definitions: df-bi 117 df-tru 1398 df-nf 1507 df-sb 1809 df-clab 2216 df-cleq 2222 df-clel 2225 df-nfc 2361 df-ral 2513 df-v 2801 df-dif 3199 df-in 3203 df-nul 3492 |
| This theorem is referenced by: (None) |
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