Metamath Proof Explorer


Theorem subscl

Description: Closure law for surreal subtraction. (Contributed by Scott Fenton, 3-Feb-2025)

Ref Expression
Assertion subscl ⊢ A ∈ No ∧ B ∈ No → A - s B ∈ No

Proof

Step Hyp Ref Expression
1 subsval ⊢ A ∈ No ∧ B ∈ No → A - s B = A + s + s ⁡ B
2 negscl ⊢ B ∈ No → + s ⁡ B ∈ No
3 addscl ⊢ A ∈ No ∧ + s ⁡ B ∈ No → A + s + s ⁡ B ∈ No
4 2 3 sylan2 ⊢ A ∈ No ∧ B ∈ No → A + s + s ⁡ B ∈ No
5 1 4 eqeltrd ⊢ A ∈ No ∧ B ∈ No → A - s B ∈ No