Metamath Proof Explorer


Theorem 3olt9o

Description: Ordinal 3 is less than ordinal 9. (Contributed by BTernaryTau, 17-Oct-2026)

Ref Expression
Assertion 3olt9o 3o ∈ 9o

Proof

Step Hyp Ref Expression
1 3on ⊢ 3o ∈ On
2 sucidg ⊢ ( 3o ∈ On → 3o ∈ suc 3o )
3 1 2 ax-mp ⊢ 3o ∈ suc 3o
4 df-4o ⊢ 4o = suc 3o
5 3 4 eleqtrri ⊢ 3o ∈ 4o
6 4olt9o ⊢ 4o ∈ 9o
7 9on ⊢ 9o ∈ On
8 ontr1 ⊢ ( 9o ∈ On → ( ( 3o ∈ 4o ∧ 4o ∈ 9o ) → 3o ∈ 9o ) )
9 7 8 ax-mp ⊢ ( ( 3o ∈ 4o ∧ 4o ∈ 9o ) → 3o ∈ 9o )
10 5 6 9 mp2an ⊢ 3o ∈ 9o