Metamath Proof Explorer


Theorem har2o

Description: The least cardinal greater than 2 is 3. (Contributed by RP, 5-Nov-2023)

Ref Expression
Assertion har2o ( har ‘ 2o ) = 3o

Proof

Step Hyp Ref Expression
1 2onn 2o ∈ ω
2 harsucnn ( 2o ∈ ω → ( har ‘ 2o ) = suc 2o )
3 1 2 ax-mp ( har ‘ 2o ) = suc 2o
4 df-3o 3o = suc 2o
5 3 4 eqtr4i ( har ‘ 2o ) = 3o