Metamath Proof Explorer


Theorem en2other2

Description: Taking the other element twice in a pair gets back to the original element. (Contributed by Stefan O'Rear, 22-Aug-2015)

Ref Expression
Assertion en2other2 ⊢ X ∈ P ∧ P ≈ 2 𝑜 → ⋃ P ∖ ⋃ P ∖ X = X

Proof

Step Hyp Ref Expression
1 en2eleq ⊢ X ∈ P ∧ P ≈ 2 𝑜 → P = X ⋃ P ∖ X
2 prcom ⊢ X ⋃ P ∖ X = ⋃ P ∖ X X
3 1 2 eqtrdi ⊢ X ∈ P ∧ P ≈ 2 𝑜 → P = ⋃ P ∖ X X
4 3 difeq1d ⊢ X ∈ P ∧ P ≈ 2 𝑜 → P ∖ ⋃ P ∖ X = ⋃ P ∖ X X ∖ ⋃ P ∖ X
5 difprsnss ⊢ ⋃ P ∖ X X ∖ ⋃ P ∖ X ⊆ X
6 4 5 eqsstrdi ⊢ X ∈ P ∧ P ≈ 2 𝑜 → P ∖ ⋃ P ∖ X ⊆ X
7 simpl ⊢ X ∈ P ∧ P ≈ 2 𝑜 → X ∈ P
8 1onn ⊢ 1 𝑜 ∈ ω
9 simpr ⊢ X ∈ P ∧ P ≈ 2 𝑜 → P ≈ 2 𝑜
10 df-2o ⊢ 2 𝑜 = suc ⁡ 1 𝑜
11 9 10 breqtrdi ⊢ X ∈ P ∧ P ≈ 2 𝑜 → P ≈ suc ⁡ 1 𝑜
12 dif1ennn ⊢ 1 𝑜 ∈ ω ∧ P ≈ suc ⁡ 1 𝑜 ∧ X ∈ P → P ∖ X ≈ 1 𝑜
13 8 11 7 12 mp3an2i ⊢ X ∈ P ∧ P ≈ 2 𝑜 → P ∖ X ≈ 1 𝑜
14 en1uniel ⊢ P ∖ X ≈ 1 𝑜 → ⋃ P ∖ X ∈ P ∖ X
15 eldifsni ⊢ ⋃ P ∖ X ∈ P ∖ X → ⋃ P ∖ X ≠ X
16 13 14 15 3syl ⊢ X ∈ P ∧ P ≈ 2 𝑜 → ⋃ P ∖ X ≠ X
17 16 necomd ⊢ X ∈ P ∧ P ≈ 2 𝑜 → X ≠ ⋃ P ∖ X
18 eldifsn ⊢ X ∈ P ∖ ⋃ P ∖ X ↔ X ∈ P ∧ X ≠ ⋃ P ∖ X
19 7 17 18 sylanbrc ⊢ X ∈ P ∧ P ≈ 2 𝑜 → X ∈ P ∖ ⋃ P ∖ X
20 19 snssd ⊢ X ∈ P ∧ P ≈ 2 𝑜 → X ⊆ P ∖ ⋃ P ∖ X
21 6 20 eqssd ⊢ X ∈ P ∧ P ≈ 2 𝑜 → P ∖ ⋃ P ∖ X = X
22 21 unieqd ⊢ X ∈ P ∧ P ≈ 2 𝑜 → ⋃ P ∖ ⋃ P ∖ X = ⋃ X
23 unisng ⊢ X ∈ P → ⋃ X = X
24 23 adantr ⊢ X ∈ P ∧ P ≈ 2 𝑜 → ⋃ X = X
25 22 24 eqtrd ⊢ X ∈ P ∧ P ≈ 2 𝑜 → ⋃ P ∖ ⋃ P ∖ X = X