Metamath Proof Explorer


Theorem 6onn

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

Ref Expression
Assertion 6onn Could not format assertion : No typesetting found for |- 6o e. _om with typecode |-

Proof

Step Hyp Ref Expression
1 df-6o Could not format 6o = suc 5o : No typesetting found for |- 6o = suc 5o with typecode |-
2 5onn Could not format 5o e. _om : No typesetting found for |- 5o e. _om with typecode |-
3 peano2 Could not format ( 5o e. _om -> suc 5o e. _om ) : No typesetting found for |- ( 5o e. _om -> suc 5o e. _om ) with typecode |-
4 2 3 ax-mp Could not format suc 5o e. _om : No typesetting found for |- suc 5o e. _om with typecode |-
5 1 4 eqeltri Could not format 6o e. _om : No typesetting found for |- 6o e. _om with typecode |-