Metamath Proof Explorer


Theorem 7olt9o

Description: Ordinal 7 is less than ordinal 9. (Contributed by BTernaryTau, 17-Oct-2026)

Ref Expression
Assertion 7olt9o 7o ∈ 9o

Proof

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