Metamath Proof Explorer |
< Previous
Next >
Nearby theorems |
||
Mirrors > Home > MPE Home > Th. List > ddif | Structured version Visualization version GIF version |
Description: Double complement under universal class. Exercise 4.10(s) of [Mendelson] p. 231. (Contributed by NM, 8-Jan-2002.) |
Ref | Expression |
---|---|
ddif | ⊢ (V ∖ (V ∖ 𝐴)) = 𝐴 |
Step | Hyp | Ref | Expression |
---|---|---|---|
1 | vex 3176 | . . . . 5 ⊢ 𝑥 ∈ V | |
2 | eldif 3550 | . . . . 5 ⊢ (𝑥 ∈ (V ∖ 𝐴) ↔ (𝑥 ∈ V ∧ ¬ 𝑥 ∈ 𝐴)) | |
3 | 1, 2 | mpbiran 955 | . . . 4 ⊢ (𝑥 ∈ (V ∖ 𝐴) ↔ ¬ 𝑥 ∈ 𝐴) |
4 | 3 | con2bii 346 | . . 3 ⊢ (𝑥 ∈ 𝐴 ↔ ¬ 𝑥 ∈ (V ∖ 𝐴)) |
5 | 1 | biantrur 526 | . . 3 ⊢ (¬ 𝑥 ∈ (V ∖ 𝐴) ↔ (𝑥 ∈ V ∧ ¬ 𝑥 ∈ (V ∖ 𝐴))) |
6 | 4, 5 | bitr2i 264 | . 2 ⊢ ((𝑥 ∈ V ∧ ¬ 𝑥 ∈ (V ∖ 𝐴)) ↔ 𝑥 ∈ 𝐴) |
7 | 6 | difeqri 3692 | 1 ⊢ (V ∖ (V ∖ 𝐴)) = 𝐴 |
Colors of variables: wff setvar class |
Syntax hints: ¬ wn 3 ∧ wa 383 = wceq 1475 ∈ wcel 1977 Vcvv 3173 ∖ cdif 3537 |
This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-3 8 ax-gen 1713 ax-4 1728 ax-5 1827 ax-6 1875 ax-7 1922 ax-10 2006 ax-11 2021 ax-12 2034 ax-13 2234 ax-ext 2590 |
This theorem depends on definitions: df-bi 196 df-or 384 df-an 385 df-tru 1478 df-ex 1696 df-nf 1701 df-sb 1868 df-clab 2597 df-cleq 2603 df-clel 2606 df-nfc 2740 df-v 3175 df-dif 3543 |
This theorem is referenced by: complss 3713 dfun3 3824 dfin3 3825 invdif 3827 ssindif0 3983 difdifdir 4008 |
Copyright terms: Public domain | W3C validator |