Metamath Proof Explorer


Theorem zexpscl

Description: Closure law for surreal integer exponentiation. (Contributed by Scott Fenton, 11-Dec-2025)

Ref Expression
Assertion zexpscl ( ( 𝐴 ∈ ℤs ∧ 𝑁 ∈ ℕ0s ) → ( 𝐴 ↑s 𝑁 ) ∈ ℤs )

Proof

Step Hyp Ref Expression
1 zssno ⊢ ℤs ⊆ No
2 simpl ⊢ ( ( 𝑥 ∈ ℤs ∧ 𝑦 ∈ ℤs ) → 𝑥 ∈ ℤs )
3 simpr ⊢ ( ( 𝑥 ∈ ℤs ∧ 𝑦 ∈ ℤs ) → 𝑦 ∈ ℤs )
4 2 3 zmulscld ⊢ ( ( 𝑥 ∈ ℤs ∧ 𝑦 ∈ ℤs ) → ( 𝑥 ·s 𝑦 ) ∈ ℤs )
5 1zs ⊢ 1s ∈ ℤs
6 1 4 5 expscllem ⊢ ( ( 𝐴 ∈ ℤs ∧ 𝑁 ∈ ℕ0s ) → ( 𝐴 ↑s 𝑁 ) ∈ ℤs )