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Mirrors > Home > ILE Home > Th. List > disjr | GIF version |
Description: Two ways of saying that two classes are disjoint. (Contributed by Jeff Madsen, 19-Jun-2011.) |
Ref | Expression |
---|---|
disjr | ⊢ ((𝐴 ∩ 𝐵) = ∅ ↔ ∀𝑥 ∈ 𝐵 ¬ 𝑥 ∈ 𝐴) |
Step | Hyp | Ref | Expression |
---|---|---|---|
1 | incom 3342 | . . 3 ⊢ (𝐴 ∩ 𝐵) = (𝐵 ∩ 𝐴) | |
2 | 1 | eqeq1i 2197 | . 2 ⊢ ((𝐴 ∩ 𝐵) = ∅ ↔ (𝐵 ∩ 𝐴) = ∅) |
3 | disj 3486 | . 2 ⊢ ((𝐵 ∩ 𝐴) = ∅ ↔ ∀𝑥 ∈ 𝐵 ¬ 𝑥 ∈ 𝐴) | |
4 | 2, 3 | bitri 184 | 1 ⊢ ((𝐴 ∩ 𝐵) = ∅ ↔ ∀𝑥 ∈ 𝐵 ¬ 𝑥 ∈ 𝐴) |
Colors of variables: wff set class |
Syntax hints: ¬ wn 3 ↔ wb 105 = wceq 1364 ∈ wcel 2160 ∀wral 2468 ∩ cin 3143 ∅c0 3437 |
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 710 ax-5 1458 ax-7 1459 ax-gen 1460 ax-ie1 1504 ax-ie2 1505 ax-8 1515 ax-10 1516 ax-11 1517 ax-i12 1518 ax-bndl 1520 ax-4 1521 ax-17 1537 ax-i9 1541 ax-ial 1545 ax-i5r 1546 ax-ext 2171 |
This theorem depends on definitions: df-bi 117 df-tru 1367 df-nf 1472 df-sb 1774 df-clab 2176 df-cleq 2182 df-clel 2185 df-nfc 2321 df-ral 2473 df-v 2754 df-dif 3146 df-in 3150 df-nul 3438 |
This theorem is referenced by: (None) |
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