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Mirrors > Home > ILE Home > Th. List > disj | GIF version |
Description: Two ways of saying that two classes are disjoint (have no members in common). (Contributed by NM, 17-Feb-2004.) |
Ref | Expression |
---|---|
disj | ⊢ ((𝐴 ∩ 𝐵) = ∅ ↔ ∀𝑥 ∈ 𝐴 ¬ 𝑥 ∈ 𝐵) |
Step | Hyp | Ref | Expression |
---|---|---|---|
1 | df-in 3005 | . . . 4 ⊢ (𝐴 ∩ 𝐵) = {𝑥 ∣ (𝑥 ∈ 𝐴 ∧ 𝑥 ∈ 𝐵)} | |
2 | 1 | eqeq1i 2095 | . . 3 ⊢ ((𝐴 ∩ 𝐵) = ∅ ↔ {𝑥 ∣ (𝑥 ∈ 𝐴 ∧ 𝑥 ∈ 𝐵)} = ∅) |
3 | abeq1 2197 | . . 3 ⊢ ({𝑥 ∣ (𝑥 ∈ 𝐴 ∧ 𝑥 ∈ 𝐵)} = ∅ ↔ ∀𝑥((𝑥 ∈ 𝐴 ∧ 𝑥 ∈ 𝐵) ↔ 𝑥 ∈ ∅)) | |
4 | imnan 659 | . . . . 5 ⊢ ((𝑥 ∈ 𝐴 → ¬ 𝑥 ∈ 𝐵) ↔ ¬ (𝑥 ∈ 𝐴 ∧ 𝑥 ∈ 𝐵)) | |
5 | noel 3290 | . . . . . 6 ⊢ ¬ 𝑥 ∈ ∅ | |
6 | 5 | nbn 650 | . . . . 5 ⊢ (¬ (𝑥 ∈ 𝐴 ∧ 𝑥 ∈ 𝐵) ↔ ((𝑥 ∈ 𝐴 ∧ 𝑥 ∈ 𝐵) ↔ 𝑥 ∈ ∅)) |
7 | 4, 6 | bitr2i 183 | . . . 4 ⊢ (((𝑥 ∈ 𝐴 ∧ 𝑥 ∈ 𝐵) ↔ 𝑥 ∈ ∅) ↔ (𝑥 ∈ 𝐴 → ¬ 𝑥 ∈ 𝐵)) |
8 | 7 | albii 1404 | . . 3 ⊢ (∀𝑥((𝑥 ∈ 𝐴 ∧ 𝑥 ∈ 𝐵) ↔ 𝑥 ∈ ∅) ↔ ∀𝑥(𝑥 ∈ 𝐴 → ¬ 𝑥 ∈ 𝐵)) |
9 | 2, 3, 8 | 3bitri 204 | . 2 ⊢ ((𝐴 ∩ 𝐵) = ∅ ↔ ∀𝑥(𝑥 ∈ 𝐴 → ¬ 𝑥 ∈ 𝐵)) |
10 | df-ral 2364 | . 2 ⊢ (∀𝑥 ∈ 𝐴 ¬ 𝑥 ∈ 𝐵 ↔ ∀𝑥(𝑥 ∈ 𝐴 → ¬ 𝑥 ∈ 𝐵)) | |
11 | 9, 10 | bitr4i 185 | 1 ⊢ ((𝐴 ∩ 𝐵) = ∅ ↔ ∀𝑥 ∈ 𝐴 ¬ 𝑥 ∈ 𝐵) |
Colors of variables: wff set class |
Syntax hints: ¬ wn 3 → wi 4 ∧ wa 102 ↔ wb 103 ∀wal 1287 = wceq 1289 ∈ wcel 1438 {cab 2074 ∀wral 2359 ∩ cin 2998 ∅c0 3286 |
This theorem was proved from axioms: ax-1 5 ax-2 6 ax-mp 7 ax-ia1 104 ax-ia2 105 ax-ia3 106 ax-in1 579 ax-in2 580 ax-io 665 ax-5 1381 ax-7 1382 ax-gen 1383 ax-ie1 1427 ax-ie2 1428 ax-8 1440 ax-10 1441 ax-11 1442 ax-i12 1443 ax-bndl 1444 ax-4 1445 ax-17 1464 ax-i9 1468 ax-ial 1472 ax-i5r 1473 ax-ext 2070 |
This theorem depends on definitions: df-bi 115 df-tru 1292 df-nf 1395 df-sb 1693 df-clab 2075 df-cleq 2081 df-clel 2084 df-nfc 2217 df-ral 2364 df-v 2621 df-dif 3001 df-in 3005 df-nul 3287 |
This theorem is referenced by: disjr 3332 disj1 3333 disjne 3336 f0rn0 5205 renfdisj 7546 fvinim0ffz 9652 fxnn0nninf 9844 exmidsbthrlem 11912 |
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