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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 2918 | . . . 4 ⊢ (A ∩ B) = {x ∣ (x ∈ A ∧ x ∈ B)} | |
2 | 1 | eqeq1i 2044 | . . 3 ⊢ ((A ∩ B) = ∅ ↔ {x ∣ (x ∈ A ∧ x ∈ B)} = ∅) |
3 | abeq1 2144 | . . 3 ⊢ ({x ∣ (x ∈ A ∧ x ∈ B)} = ∅ ↔ ∀x((x ∈ A ∧ x ∈ B) ↔ x ∈ ∅)) | |
4 | imnan 623 | . . . . 5 ⊢ ((x ∈ A → ¬ x ∈ B) ↔ ¬ (x ∈ A ∧ x ∈ B)) | |
5 | noel 3222 | . . . . . 6 ⊢ ¬ x ∈ ∅ | |
6 | 5 | nbn 614 | . . . . 5 ⊢ (¬ (x ∈ A ∧ x ∈ B) ↔ ((x ∈ A ∧ x ∈ B) ↔ x ∈ ∅)) |
7 | 4, 6 | bitr2i 174 | . . . 4 ⊢ (((x ∈ A ∧ x ∈ B) ↔ x ∈ ∅) ↔ (x ∈ A → ¬ x ∈ B)) |
8 | 7 | albii 1356 | . . 3 ⊢ (∀x((x ∈ A ∧ x ∈ B) ↔ x ∈ ∅) ↔ ∀x(x ∈ A → ¬ x ∈ B)) |
9 | 2, 3, 8 | 3bitri 195 | . 2 ⊢ ((A ∩ B) = ∅ ↔ ∀x(x ∈ A → ¬ x ∈ B)) |
10 | df-ral 2305 | . 2 ⊢ (∀x ∈ A ¬ x ∈ B ↔ ∀x(x ∈ A → ¬ x ∈ B)) | |
11 | 9, 10 | bitr4i 176 | 1 ⊢ ((A ∩ B) = ∅ ↔ ∀x ∈ A ¬ x ∈ B) |
Colors of variables: wff set class |
Syntax hints: ¬ wn 3 → wi 4 ∧ wa 97 ↔ wb 98 ∀wal 1240 = wceq 1242 ∈ wcel 1390 {cab 2023 ∀wral 2300 ∩ cin 2910 ∅c0 3218 |
This theorem was proved from axioms: ax-1 5 ax-2 6 ax-mp 7 ax-ia1 99 ax-ia2 100 ax-ia3 101 ax-in1 544 ax-in2 545 ax-io 629 ax-5 1333 ax-7 1334 ax-gen 1335 ax-ie1 1379 ax-ie2 1380 ax-8 1392 ax-10 1393 ax-11 1394 ax-i12 1395 ax-bndl 1396 ax-4 1397 ax-17 1416 ax-i9 1420 ax-ial 1424 ax-i5r 1425 ax-ext 2019 |
This theorem depends on definitions: df-bi 110 df-tru 1245 df-nf 1347 df-sb 1643 df-clab 2024 df-cleq 2030 df-clel 2033 df-nfc 2164 df-ral 2305 df-v 2553 df-dif 2914 df-in 2918 df-nul 3219 |
This theorem is referenced by: disjr 3263 disj1 3264 disjne 3267 renfdisj 6876 |
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