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| Mirrors > Home > ILE Home > Th. List > disj3 | GIF version | ||
| Description: Two ways of saying that two classes are disjoint. (Contributed by NM, 19-May-1998.) |
| Ref | Expression |
|---|---|
| disj3 | ⊢ ((𝐴 ∩ 𝐵) = ∅ ↔ 𝐴 = (𝐴 ∖ 𝐵)) |
| Step | Hyp | Ref | Expression |
|---|---|---|---|
| 1 | pm4.71 389 | . . . 4 ⊢ ((𝑥 ∈ 𝐴 → ¬ 𝑥 ∈ 𝐵) ↔ (𝑥 ∈ 𝐴 ↔ (𝑥 ∈ 𝐴 ∧ ¬ 𝑥 ∈ 𝐵))) | |
| 2 | eldif 3183 | . . . . 5 ⊢ (𝑥 ∈ (𝐴 ∖ 𝐵) ↔ (𝑥 ∈ 𝐴 ∧ ¬ 𝑥 ∈ 𝐵)) | |
| 3 | 2 | bibi2i 227 | . . . 4 ⊢ ((𝑥 ∈ 𝐴 ↔ 𝑥 ∈ (𝐴 ∖ 𝐵)) ↔ (𝑥 ∈ 𝐴 ↔ (𝑥 ∈ 𝐴 ∧ ¬ 𝑥 ∈ 𝐵))) |
| 4 | 1, 3 | bitr4i 187 | . . 3 ⊢ ((𝑥 ∈ 𝐴 → ¬ 𝑥 ∈ 𝐵) ↔ (𝑥 ∈ 𝐴 ↔ 𝑥 ∈ (𝐴 ∖ 𝐵))) |
| 5 | 4 | albii 1494 | . 2 ⊢ (∀𝑥(𝑥 ∈ 𝐴 → ¬ 𝑥 ∈ 𝐵) ↔ ∀𝑥(𝑥 ∈ 𝐴 ↔ 𝑥 ∈ (𝐴 ∖ 𝐵))) |
| 6 | disj1 3519 | . 2 ⊢ ((𝐴 ∩ 𝐵) = ∅ ↔ ∀𝑥(𝑥 ∈ 𝐴 → ¬ 𝑥 ∈ 𝐵)) | |
| 7 | dfcleq 2201 | . 2 ⊢ (𝐴 = (𝐴 ∖ 𝐵) ↔ ∀𝑥(𝑥 ∈ 𝐴 ↔ 𝑥 ∈ (𝐴 ∖ 𝐵))) | |
| 8 | 5, 6, 7 | 3bitr4i 212 | 1 ⊢ ((𝐴 ∩ 𝐵) = ∅ ↔ 𝐴 = (𝐴 ∖ 𝐵)) |
| Colors of variables: wff set class |
| Syntax hints: ¬ wn 3 → wi 4 ∧ wa 104 ↔ wb 105 ∀wal 1371 = wceq 1373 ∈ wcel 2178 ∖ cdif 3171 ∩ cin 3173 ∅c0 3468 |
| 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 615 ax-in2 616 ax-io 711 ax-5 1471 ax-7 1472 ax-gen 1473 ax-ie1 1517 ax-ie2 1518 ax-8 1528 ax-10 1529 ax-11 1530 ax-i12 1531 ax-bndl 1533 ax-4 1534 ax-17 1550 ax-i9 1554 ax-ial 1558 ax-i5r 1559 ax-ext 2189 |
| This theorem depends on definitions: df-bi 117 df-tru 1376 df-nf 1485 df-sb 1787 df-clab 2194 df-cleq 2200 df-clel 2203 df-nfc 2339 df-ral 2491 df-v 2778 df-dif 3176 df-in 3180 df-nul 3469 |
| This theorem is referenced by: disjel 3523 uneqdifeqim 3554 difprsn1 3783 diftpsn3 3785 orddif 4613 phpm 6988 |
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