Metamath Proof Explorer

Theorem 3olt9o

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

Ref Expression
Assertion 3olt9o 
|- 3o e. 9o

Proof

Step Hyp Ref Expression
1 3on 
 |-  3o e. On
2 sucidg 
 |-  ( 3o e. On -> 3o e. suc 3o )
3 1 2 ax-mp 
 |-  3o e. suc 3o
4 df-4o 
 |-  4o = suc 3o
5 3 4 eleqtrri 
 |-  3o e. 4o
6 4olt9o 
 |-  4o e. 9o
7 9on 
 |-  9o e. On
8 ontr1 
 |-  ( 9o e. On -> ( ( 3o e. 4o /\ 4o e. 9o ) -> 3o e. 9o ) )
9 7 8 ax-mp 
 |-  ( ( 3o e. 4o /\ 4o e. 9o ) -> 3o e. 9o )
10 5 6 9 mp2an 
 |-  3o e. 9o