Metamath Proof Explorer

Theorem 1olt9o

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

Ref Expression
Assertion 1olt9o 
|- 1o e. 9o

Proof

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