Metamath Proof Explorer


Theorem onnoxpg

Description: Every ordinal maps to a surreal number. (Contributed by RP, 21-Sep-2023)

Ref Expression
Assertion onnoxpg ⊢ A ∈ On ∧ B ∈ 1 𝑜 2 𝑜 → A × B ∈ No

Proof

Step Hyp Ref Expression
1 fconst6g ⊢ B ∈ 1 𝑜 2 𝑜 → A × B : A ⟶ 1 𝑜 2 𝑜
2 1 adantl ⊢ A ∈ On ∧ B ∈ 1 𝑜 2 𝑜 → A × B : A ⟶ 1 𝑜 2 𝑜
3 simp3 ⊢ A ∈ On ∧ B ∈ 1 𝑜 2 𝑜 ∧ A × B : A ⟶ 1 𝑜 2 𝑜 → A × B : A ⟶ 1 𝑜 2 𝑜
4 3 ffund ⊢ A ∈ On ∧ B ∈ 1 𝑜 2 𝑜 ∧ A × B : A ⟶ 1 𝑜 2 𝑜 → Fun ⁡ A × B
5 simp2 ⊢ A ∈ On ∧ B ∈ 1 𝑜 2 𝑜 ∧ A × B : A ⟶ 1 𝑜 2 𝑜 → B ∈ 1 𝑜 2 𝑜
6 snnzg ⊢ B ∈ 1 𝑜 2 𝑜 → B ≠ ∅
7 dmxp ⊢ B ≠ ∅ → dom ⁡ A × B = A
8 7 eqcomd ⊢ B ≠ ∅ → A = dom ⁡ A × B
9 5 6 8 3syl ⊢ A ∈ On ∧ B ∈ 1 𝑜 2 𝑜 ∧ A × B : A ⟶ 1 𝑜 2 𝑜 → A = dom ⁡ A × B
10 simp1 ⊢ A ∈ On ∧ B ∈ 1 𝑜 2 𝑜 ∧ A × B : A ⟶ 1 𝑜 2 𝑜 → A ∈ On
11 9 10 eqeltrrd ⊢ A ∈ On ∧ B ∈ 1 𝑜 2 𝑜 ∧ A × B : A ⟶ 1 𝑜 2 𝑜 → dom ⁡ A × B ∈ On
12 3 frnd ⊢ A ∈ On ∧ B ∈ 1 𝑜 2 𝑜 ∧ A × B : A ⟶ 1 𝑜 2 𝑜 → ran ⁡ A × B ⊆ 1 𝑜 2 𝑜
13 elno2 ⊢ A × B ∈ No ↔ Fun ⁡ A × B ∧ dom ⁡ A × B ∈ On ∧ ran ⁡ A × B ⊆ 1 𝑜 2 𝑜
14 4 11 12 13 syl3anbrc ⊢ A ∈ On ∧ B ∈ 1 𝑜 2 𝑜 ∧ A × B : A ⟶ 1 𝑜 2 𝑜 → A × B ∈ No
15 2 14 mpd3an3 ⊢ A ∈ On ∧ B ∈ 1 𝑜 2 𝑜 → A × B ∈ No