Metamath Proof Explorer


Theorem divscl

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

Ref Expression
Assertion divscl ( ( 𝐴 ∈ No ∧ 𝐵 ∈ No ∧ 𝐵 ≠ 0s ) → ( 𝐴 /su 𝐵 ) ∈ No )

Proof

Step Hyp Ref Expression
1 recsex ⊢ ( ( 𝐵 ∈ No ∧ 𝐵 ≠ 0s ) → ∃ 𝑥 ∈ No ( 𝐵 ·s 𝑥 ) = 1s )
2 1 3adant1 ⊢ ( ( 𝐴 ∈ No ∧ 𝐵 ∈ No ∧ 𝐵 ≠ 0s ) → ∃ 𝑥 ∈ No ( 𝐵 ·s 𝑥 ) = 1s )
3 divsclw ⊢ ( ( ( 𝐴 ∈ No ∧ 𝐵 ∈ No ∧ 𝐵 ≠ 0s ) ∧ ∃ 𝑥 ∈ No ( 𝐵 ·s 𝑥 ) = 1s ) → ( 𝐴 /su 𝐵 ) ∈ No )
4 2 3 mpdan ⊢ ( ( 𝐴 ∈ No ∧ 𝐵 ∈ No ∧ 𝐵 ≠ 0s ) → ( 𝐴 /su 𝐵 ) ∈ No )