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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 | pm4.71 393 | . . . 4 ⊢ ((𝑥 ∈ 𝐴 → ¬ 𝑥 ∈ 𝐵) ↔ (𝑥 ∈ 𝐴 ↔ (𝑥 ∈ 𝐴 ∧ ¬ 𝑥 ∈ 𝐵))) | |
| 2 | eldif 3229 | . . . . 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 1523 | . 2 ⊢ (∀𝑥(𝑥 ∈ 𝐴 → ¬ 𝑥 ∈ 𝐵) ↔ ∀𝑥(𝑥 ∈ 𝐴 ↔ 𝑥 ∈ (𝐴 ∖ 𝐵))) |
| 6 | disj1 3574 | . 2 ⊢ ((𝐴 ∩ 𝐵) = ∅ ↔ ∀𝑥(𝑥 ∈ 𝐴 → ¬ 𝑥 ∈ 𝐵)) | |
| 7 | dfcleq 2232 | . 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 1400 = wceq 1402 ∈ wcel 2209 ∖ cdif 3217 ∩ cin 3219 ∅c0 3520 |
| 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 623 ax-in2 624 ax-io 721 ax-5 1500 ax-7 1501 ax-gen 1502 ax-ie1 1546 ax-ie2 1547 ax-8 1557 ax-10 1558 ax-11 1559 ax-i12 1560 ax-bndl 1562 ax-4 1563 ax-17 1579 ax-i9 1583 ax-ial 1587 ax-i5r 1588 ax-ext 2220 |
| This theorem depends on definitions: df-bi 117 df-tru 1405 df-nf 1514 df-sb 1816 df-clab 2225 df-cleq 2231 df-clel 2234 df-nfc 2381 df-ral 2533 df-v 2823 df-dif 3222 df-in 3226 df-nul 3521 |
| This theorem is referenced by: disjel 3578 uneqdifeqim 3610 difprsn1 3849 diftpsn3 3851 orddif 4689 phpm 7157 ballotfilemfp1 13209 |
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