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Mirrors > Home > ILE Home > Th. List > reldisj | GIF version |
Description: Two ways of saying that two classes are disjoint, using the complement of 𝐵 relative to a universe 𝐶. (Contributed by NM, 15-Feb-2007.) (Proof shortened by Andrew Salmon, 26-Jun-2011.) |
Ref | Expression |
---|---|
reldisj | ⊢ (𝐴 ⊆ 𝐶 → ((𝐴 ∩ 𝐵) = ∅ ↔ 𝐴 ⊆ (𝐶 ∖ 𝐵))) |
Step | Hyp | Ref | Expression |
---|---|---|---|
1 | dfss2 3127 | . . . 4 ⊢ (𝐴 ⊆ 𝐶 ↔ ∀𝑥(𝑥 ∈ 𝐴 → 𝑥 ∈ 𝐶)) | |
2 | pm5.44 915 | . . . . . 6 ⊢ ((𝑥 ∈ 𝐴 → 𝑥 ∈ 𝐶) → ((𝑥 ∈ 𝐴 → ¬ 𝑥 ∈ 𝐵) ↔ (𝑥 ∈ 𝐴 → (𝑥 ∈ 𝐶 ∧ ¬ 𝑥 ∈ 𝐵)))) | |
3 | eldif 3121 | . . . . . . 7 ⊢ (𝑥 ∈ (𝐶 ∖ 𝐵) ↔ (𝑥 ∈ 𝐶 ∧ ¬ 𝑥 ∈ 𝐵)) | |
4 | 3 | imbi2i 225 | . . . . . 6 ⊢ ((𝑥 ∈ 𝐴 → 𝑥 ∈ (𝐶 ∖ 𝐵)) ↔ (𝑥 ∈ 𝐴 → (𝑥 ∈ 𝐶 ∧ ¬ 𝑥 ∈ 𝐵))) |
5 | 2, 4 | bitr4di 197 | . . . . 5 ⊢ ((𝑥 ∈ 𝐴 → 𝑥 ∈ 𝐶) → ((𝑥 ∈ 𝐴 → ¬ 𝑥 ∈ 𝐵) ↔ (𝑥 ∈ 𝐴 → 𝑥 ∈ (𝐶 ∖ 𝐵)))) |
6 | 5 | sps 1524 | . . . 4 ⊢ (∀𝑥(𝑥 ∈ 𝐴 → 𝑥 ∈ 𝐶) → ((𝑥 ∈ 𝐴 → ¬ 𝑥 ∈ 𝐵) ↔ (𝑥 ∈ 𝐴 → 𝑥 ∈ (𝐶 ∖ 𝐵)))) |
7 | 1, 6 | sylbi 120 | . . 3 ⊢ (𝐴 ⊆ 𝐶 → ((𝑥 ∈ 𝐴 → ¬ 𝑥 ∈ 𝐵) ↔ (𝑥 ∈ 𝐴 → 𝑥 ∈ (𝐶 ∖ 𝐵)))) |
8 | 7 | albidv 1811 | . 2 ⊢ (𝐴 ⊆ 𝐶 → (∀𝑥(𝑥 ∈ 𝐴 → ¬ 𝑥 ∈ 𝐵) ↔ ∀𝑥(𝑥 ∈ 𝐴 → 𝑥 ∈ (𝐶 ∖ 𝐵)))) |
9 | disj1 3455 | . 2 ⊢ ((𝐴 ∩ 𝐵) = ∅ ↔ ∀𝑥(𝑥 ∈ 𝐴 → ¬ 𝑥 ∈ 𝐵)) | |
10 | dfss2 3127 | . 2 ⊢ (𝐴 ⊆ (𝐶 ∖ 𝐵) ↔ ∀𝑥(𝑥 ∈ 𝐴 → 𝑥 ∈ (𝐶 ∖ 𝐵))) | |
11 | 8, 9, 10 | 3bitr4g 222 | 1 ⊢ (𝐴 ⊆ 𝐶 → ((𝐴 ∩ 𝐵) = ∅ ↔ 𝐴 ⊆ (𝐶 ∖ 𝐵))) |
Colors of variables: wff set class |
Syntax hints: ¬ wn 3 → wi 4 ∧ wa 103 ↔ wb 104 ∀wal 1340 = wceq 1342 ∈ wcel 2135 ∖ cdif 3109 ∩ cin 3111 ⊆ wss 3112 ∅c0 3405 |
This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-ia1 105 ax-ia2 106 ax-ia3 107 ax-in1 604 ax-in2 605 ax-io 699 ax-5 1434 ax-7 1435 ax-gen 1436 ax-ie1 1480 ax-ie2 1481 ax-8 1491 ax-10 1492 ax-11 1493 ax-i12 1494 ax-bndl 1496 ax-4 1497 ax-17 1513 ax-i9 1517 ax-ial 1521 ax-i5r 1522 ax-ext 2146 |
This theorem depends on definitions: df-bi 116 df-tru 1345 df-nf 1448 df-sb 1750 df-clab 2151 df-cleq 2157 df-clel 2160 df-nfc 2295 df-ral 2447 df-v 2724 df-dif 3114 df-in 3118 df-ss 3125 df-nul 3406 |
This theorem is referenced by: disj2 3460 ssdifsn 3699 structcnvcnv 12364 |
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