Metamath Proof Explorer


Theorem mulscomd

Description: Surreal multiplication is commutative. Part of theorem 7 of Conway p. 19. (Contributed by Scott Fenton, 6-Mar-2025)

Ref Expression
Hypotheses mulscomd.1 ⊢ φ → A ∈ No
mulscomd.2 ⊢ φ → B ∈ No
Assertion mulscomd ⊢ φ → A ⋅ s B = B ⋅ s A

Proof

Step Hyp Ref Expression
1 mulscomd.1 ⊢ φ → A ∈ No
2 mulscomd.2 ⊢ φ → B ∈ No
3 mulscom ⊢ A ∈ No ∧ B ∈ No → A ⋅ s B = B ⋅ s A
4 1 2 3 syl2anc ⊢ φ → A ⋅ s B = B ⋅ s A