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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 102 | . . 3 ⊢ ((x ∈ A ∧ ¬ x ∈ (B ∖ A)) → x ∈ A) | |
2 | pm4.45im 317 | . . . 4 ⊢ (x ∈ A ↔ (x ∈ A ∧ (x ∈ B → x ∈ A))) | |
3 | imanim 784 | . . . . . 6 ⊢ ((x ∈ B → x ∈ A) → ¬ (x ∈ B ∧ ¬ x ∈ A)) | |
4 | eldif 2921 | . . . . . 6 ⊢ (x ∈ (B ∖ A) ↔ (x ∈ B ∧ ¬ x ∈ A)) | |
5 | 3, 4 | sylnibr 601 | . . . . 5 ⊢ ((x ∈ B → x ∈ A) → ¬ x ∈ (B ∖ A)) |
6 | 5 | anim2i 324 | . . . 4 ⊢ ((x ∈ A ∧ (x ∈ B → x ∈ A)) → (x ∈ A ∧ ¬ x ∈ (B ∖ A))) |
7 | 2, 6 | sylbi 114 | . . 3 ⊢ (x ∈ A → (x ∈ A ∧ ¬ x ∈ (B ∖ A))) |
8 | 1, 7 | impbii 117 | . 2 ⊢ ((x ∈ A ∧ ¬ x ∈ (B ∖ A)) ↔ x ∈ A) |
9 | 8 | difeqri 3058 | 1 ⊢ (A ∖ (B ∖ A)) = A |
Colors of variables: wff set class |
Syntax hints: ¬ wn 3 → wi 4 ∧ wa 97 = wceq 1242 ∈ wcel 1390 ∖ cdif 2908 |
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-v 2553 df-dif 2914 |
This theorem is referenced by: dif0 3288 |
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