Metamath Proof Explorer


Theorem divscld

Description: Surreal division closure law. (Contributed by Scott Fenton, 16-Mar-2025)

Ref Expression
Hypotheses divscld.1 ⊢ ( 𝜑 → 𝐴 ∈ No )
divscld.2 ⊢ ( 𝜑 → 𝐵 ∈ No )
divscld.3 ⊢ ( 𝜑 → 𝐵 ≠ 0s )
Assertion divscld ( 𝜑 → ( 𝐴 /su 𝐵 ) ∈ No )

Proof

Step Hyp Ref Expression
1 divscld.1 ⊢ ( 𝜑 → 𝐴 ∈ No )
2 divscld.2 ⊢ ( 𝜑 → 𝐵 ∈ No )
3 divscld.3 ⊢ ( 𝜑 → 𝐵 ≠ 0s )
4 divscl ⊢ ( ( 𝐴 ∈ No ∧ 𝐵 ∈ No ∧ 𝐵 ≠ 0s ) → ( 𝐴 /su 𝐵 ) ∈ No )
5 1 2 3 4 syl3anc ⊢ ( 𝜑 → ( 𝐴 /su 𝐵 ) ∈ No )