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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 3319 | . . 3 ⊢ (𝐴 ∩ 𝐵) = (𝐵 ∩ 𝐴) | |
2 | 1 | eqeq1i 2178 | . 2 ⊢ ((𝐴 ∩ 𝐵) = ∅ ↔ (𝐵 ∩ 𝐴) = ∅) |
3 | disj 3463 | . 2 ⊢ ((𝐵 ∩ 𝐴) = ∅ ↔ ∀𝑥 ∈ 𝐵 ¬ 𝑥 ∈ 𝐴) | |
4 | 2, 3 | bitri 183 | 1 ⊢ ((𝐴 ∩ 𝐵) = ∅ ↔ ∀𝑥 ∈ 𝐵 ¬ 𝑥 ∈ 𝐴) |
Colors of variables: wff set class |
Syntax hints: ¬ wn 3 ↔ wb 104 = wceq 1348 ∈ wcel 2141 ∀wral 2448 ∩ cin 3120 ∅c0 3414 |
This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-ia1 105 ax-ia2 106 ax-ia3 107 ax-in1 609 ax-in2 610 ax-io 704 ax-5 1440 ax-7 1441 ax-gen 1442 ax-ie1 1486 ax-ie2 1487 ax-8 1497 ax-10 1498 ax-11 1499 ax-i12 1500 ax-bndl 1502 ax-4 1503 ax-17 1519 ax-i9 1523 ax-ial 1527 ax-i5r 1528 ax-ext 2152 |
This theorem depends on definitions: df-bi 116 df-tru 1351 df-nf 1454 df-sb 1756 df-clab 2157 df-cleq 2163 df-clel 2166 df-nfc 2301 df-ral 2453 df-v 2732 df-dif 3123 df-in 3127 df-nul 3415 |
This theorem is referenced by: (None) |
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