Metamath Proof Explorer


Theorem 4olt9o

Description: Ordinal 4 is less than ordinal 9. (Contributed by BTernaryTau, 17-Oct-2026)

Ref Expression
Assertion 4olt9o 4o ∈ 9o

Proof

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