Metamath Proof Explorer


Theorem 6onn

Description: The ordinal 6 is a natural number. (Contributed by BTernaryTau, 4-Sep-2026)

Ref Expression
Assertion 6onn 6o ∈ ω

Proof

Step Hyp Ref Expression
1 df-6o 6o = suc 5o
2 5onn 5o ∈ ω
3 peano2 ( 5o ∈ ω → suc 5o ∈ ω )
4 2 3 ax-mp suc 5o ∈ ω
5 1 4 eqeltri 6o ∈ ω