Metamath Proof Explorer


Theorem cutsfo

Description: The surreal cut function is onto. (Contributed by Scott Fenton, 23-Aug-2024)

Ref Expression
Assertion cutsfo ⊢ | s : ≪ s ⟶ onto No

Proof

Step Hyp Ref Expression
1 cutsf ⊢ | s : ≪ s ⟶ No
2 lltr ⊢ L ⁡ x ≪ s R ⁡ x
3 df-br ⊢ L ⁡ x ≪ s R ⁡ x ↔ L ⁡ x R ⁡ x ∈ ≪ s
4 2 3 mpbi ⊢ L ⁡ x R ⁡ x ∈ ≪ s
5 lrcut ⊢ x ∈ No → L ⁡ x | s R ⁡ x = x
6 5 eqcomd ⊢ x ∈ No → x = L ⁡ x | s R ⁡ x
7 fveq2 ⊢ y = L ⁡ x R ⁡ x → | s ⁡ y = | s ⁡ L ⁡ x R ⁡ x
8 df-ov ⊢ L ⁡ x | s R ⁡ x = | s ⁡ L ⁡ x R ⁡ x
9 7 8 eqtr4di ⊢ y = L ⁡ x R ⁡ x → | s ⁡ y = L ⁡ x | s R ⁡ x
10 9 rspceeqv ⊢ L ⁡ x R ⁡ x ∈ ≪ s ∧ x = L ⁡ x | s R ⁡ x → ∃ y ∈ ≪ s x = | s ⁡ y
11 4 6 10 sylancr ⊢ x ∈ No → ∃ y ∈ ≪ s x = | s ⁡ y
12 11 rgen ⊢ ∀ x ∈ No ∃ y ∈ ≪ s x = | s ⁡ y
13 dffo3 ⊢ | s : ≪ s ⟶ onto No ↔ | s : ≪ s ⟶ No ∧ ∀ x ∈ No ∃ y ∈ ≪ s x = | s ⁡ y
14 1 12 13 mpbir2an ⊢ | s : ≪ s ⟶ onto No