Metamath Proof Explorer


Theorem 1olt9o

Description: Ordinal 1 is less than ordinal 9. (Contributed by BTernaryTau, 17-Oct-2026)

Ref Expression
Assertion 1olt9o 1o ∈ 9o

Proof

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