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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 2990 | . . . 4 ⊢ (𝐴 ∩ 𝐵) = {𝑥 ∣ (𝑥 ∈ 𝐴 ∧ 𝑥 ∈ 𝐵)} | |
2 | 1 | eqeq1i 2090 | . . 3 ⊢ ((𝐴 ∩ 𝐵) = ∅ ↔ {𝑥 ∣ (𝑥 ∈ 𝐴 ∧ 𝑥 ∈ 𝐵)} = ∅) |
3 | abeq1 2192 | . . 3 ⊢ ({𝑥 ∣ (𝑥 ∈ 𝐴 ∧ 𝑥 ∈ 𝐵)} = ∅ ↔ ∀𝑥((𝑥 ∈ 𝐴 ∧ 𝑥 ∈ 𝐵) ↔ 𝑥 ∈ ∅)) | |
4 | imnan 657 | . . . . 5 ⊢ ((𝑥 ∈ 𝐴 → ¬ 𝑥 ∈ 𝐵) ↔ ¬ (𝑥 ∈ 𝐴 ∧ 𝑥 ∈ 𝐵)) | |
5 | noel 3273 | . . . . . 6 ⊢ ¬ 𝑥 ∈ ∅ | |
6 | 5 | nbn 648 | . . . . 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 1400 | . . 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 2358 | . 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 1283 = wceq 1285 ∈ wcel 1434 {cab 2069 ∀wral 2353 ∩ cin 2983 ∅c0 3269 |
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 577 ax-in2 578 ax-io 663 ax-5 1377 ax-7 1378 ax-gen 1379 ax-ie1 1423 ax-ie2 1424 ax-8 1436 ax-10 1437 ax-11 1438 ax-i12 1439 ax-bndl 1440 ax-4 1441 ax-17 1460 ax-i9 1464 ax-ial 1468 ax-i5r 1469 ax-ext 2065 |
This theorem depends on definitions: df-bi 115 df-tru 1288 df-nf 1391 df-sb 1688 df-clab 2070 df-cleq 2076 df-clel 2079 df-nfc 2212 df-ral 2358 df-v 2614 df-dif 2986 df-in 2990 df-nul 3270 |
This theorem is referenced by: disjr 3314 disj1 3315 disjne 3318 f0rn0 5153 renfdisj 7449 fvinim0ffz 9541 fxnn0nninf 9733 |
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