Metamath Proof Explorer


Theorem expscl

Description: Closure law for surreal exponentiation. (Contributed by Scott Fenton, 7-Aug-2025)

Ref Expression
Assertion expscl ( ( 𝐴 ∈ No ∧ 𝑁 ∈ ℕ0s ) → ( 𝐴 ↑s 𝑁 ) ∈ No )

Proof

Step Hyp Ref Expression
1 ssid ⊢ No ⊆ No
2 mulscl ⊢ ( ( 𝑥 ∈ No ∧ 𝑦 ∈ No ) → ( 𝑥 ·s 𝑦 ) ∈ No )
3 1no ⊢ 1s ∈ No
4 1 2 3 expscllem ⊢ ( ( 𝐴 ∈ No ∧ 𝑁 ∈ ℕ0s ) → ( 𝐴 ↑s 𝑁 ) ∈ No )