Metamath Proof Explorer


Theorem ord1eln01

Description: An ordinal that is not 0 or 1 contains 1. (Contributed by BTernaryTau, 1-Dec-2024)

Ref Expression
Assertion ord1eln01 ⊢ Ord ⁡ A → 1 𝑜 ∈ A ↔ A ≠ ∅ ∧ A ≠ 1 𝑜

Proof

Step Hyp Ref Expression
1 ne0i ⊢ 1 𝑜 ∈ A → A ≠ ∅
2 1on ⊢ 1 𝑜 ∈ On
3 2 onirri ⊢ ¬ 1 𝑜 ∈ 1 𝑜
4 eleq2 ⊢ A = 1 𝑜 → 1 𝑜 ∈ A ↔ 1 𝑜 ∈ 1 𝑜
5 3 4 mtbiri ⊢ A = 1 𝑜 → ¬ 1 𝑜 ∈ A
6 5 necon2ai ⊢ 1 𝑜 ∈ A → A ≠ 1 𝑜
7 1 6 jca ⊢ 1 𝑜 ∈ A → A ≠ ∅ ∧ A ≠ 1 𝑜
8 el1o ⊢ A ∈ 1 𝑜 ↔ A = ∅
9 8 biimpi ⊢ A ∈ 1 𝑜 → A = ∅
10 9 necon3ai ⊢ A ≠ ∅ → ¬ A ∈ 1 𝑜
11 nesym ⊢ A ≠ 1 𝑜 ↔ ¬ 1 𝑜 = A
12 11 biimpi ⊢ A ≠ 1 𝑜 → ¬ 1 𝑜 = A
13 10 12 anim12ci ⊢ A ≠ ∅ ∧ A ≠ 1 𝑜 → ¬ 1 𝑜 = A ∧ ¬ A ∈ 1 𝑜
14 pm4.56 ⊢ ¬ 1 𝑜 = A ∧ ¬ A ∈ 1 𝑜 ↔ ¬ 1 𝑜 = A ∨ A ∈ 1 𝑜
15 13 14 sylib ⊢ A ≠ ∅ ∧ A ≠ 1 𝑜 → ¬ 1 𝑜 = A ∨ A ∈ 1 𝑜
16 2 onordi ⊢ Ord ⁡ 1 𝑜
17 ordtri2 ⊢ Ord ⁡ 1 𝑜 ∧ Ord ⁡ A → 1 𝑜 ∈ A ↔ ¬ 1 𝑜 = A ∨ A ∈ 1 𝑜
18 16 17 mpan ⊢ Ord ⁡ A → 1 𝑜 ∈ A ↔ ¬ 1 𝑜 = A ∨ A ∈ 1 𝑜
19 15 18 imbitrrid ⊢ Ord ⁡ A → A ≠ ∅ ∧ A ≠ 1 𝑜 → 1 𝑜 ∈ A
20 7 19 impbid2 ⊢ Ord ⁡ A → 1 𝑜 ∈ A ↔ A ≠ ∅ ∧ A ≠ 1 𝑜