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Mirrors > Home > ILE Home > Th. List > difdif | GIF version |
Description: Double class difference. Exercise 11 of [TakeutiZaring] p. 22. (Contributed by NM, 17-May-1998.) |
Ref | Expression |
---|---|
difdif | ⊢ (𝐴 ∖ (𝐵 ∖ 𝐴)) = 𝐴 |
Step | Hyp | Ref | Expression |
---|---|---|---|
1 | simpl 108 | . . 3 ⊢ ((𝑥 ∈ 𝐴 ∧ ¬ 𝑥 ∈ (𝐵 ∖ 𝐴)) → 𝑥 ∈ 𝐴) | |
2 | pm4.45im 332 | . . . 4 ⊢ (𝑥 ∈ 𝐴 ↔ (𝑥 ∈ 𝐴 ∧ (𝑥 ∈ 𝐵 → 𝑥 ∈ 𝐴))) | |
3 | imanim 662 | . . . . . 6 ⊢ ((𝑥 ∈ 𝐵 → 𝑥 ∈ 𝐴) → ¬ (𝑥 ∈ 𝐵 ∧ ¬ 𝑥 ∈ 𝐴)) | |
4 | eldif 3050 | . . . . . 6 ⊢ (𝑥 ∈ (𝐵 ∖ 𝐴) ↔ (𝑥 ∈ 𝐵 ∧ ¬ 𝑥 ∈ 𝐴)) | |
5 | 3, 4 | sylnibr 651 | . . . . 5 ⊢ ((𝑥 ∈ 𝐵 → 𝑥 ∈ 𝐴) → ¬ 𝑥 ∈ (𝐵 ∖ 𝐴)) |
6 | 5 | anim2i 339 | . . . 4 ⊢ ((𝑥 ∈ 𝐴 ∧ (𝑥 ∈ 𝐵 → 𝑥 ∈ 𝐴)) → (𝑥 ∈ 𝐴 ∧ ¬ 𝑥 ∈ (𝐵 ∖ 𝐴))) |
7 | 2, 6 | sylbi 120 | . . 3 ⊢ (𝑥 ∈ 𝐴 → (𝑥 ∈ 𝐴 ∧ ¬ 𝑥 ∈ (𝐵 ∖ 𝐴))) |
8 | 1, 7 | impbii 125 | . 2 ⊢ ((𝑥 ∈ 𝐴 ∧ ¬ 𝑥 ∈ (𝐵 ∖ 𝐴)) ↔ 𝑥 ∈ 𝐴) |
9 | 8 | difeqri 3166 | 1 ⊢ (𝐴 ∖ (𝐵 ∖ 𝐴)) = 𝐴 |
Colors of variables: wff set class |
Syntax hints: ¬ wn 3 → wi 4 ∧ wa 103 = wceq 1316 ∈ wcel 1465 ∖ cdif 3038 |
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 588 ax-in2 589 ax-io 683 ax-5 1408 ax-7 1409 ax-gen 1410 ax-ie1 1454 ax-ie2 1455 ax-8 1467 ax-10 1468 ax-11 1469 ax-i12 1470 ax-bndl 1471 ax-4 1472 ax-17 1491 ax-i9 1495 ax-ial 1499 ax-i5r 1500 ax-ext 2099 |
This theorem depends on definitions: df-bi 116 df-tru 1319 df-nf 1422 df-sb 1721 df-clab 2104 df-cleq 2110 df-clel 2113 df-nfc 2247 df-v 2662 df-dif 3043 |
This theorem is referenced by: dif0 3403 |
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