Metamath Proof Explorer


Theorem omsson

Description: Omega is a subset of On . (Contributed by NM, 13-Jun-1994) (Proof shortened by Andrew Salmon, 27-Aug-2011)

Ref Expression
Assertion omsson ω ⊆ On

Proof

Step Hyp Ref Expression
1 dfom2 ω = { 𝑥 ∈ On ∣ suc 𝑥 ⊆ { 𝑦 ∈ On ∣ ¬ Lim 𝑦 } }
2 1 ssrab3 ω ⊆ On