Metamath Proof Explorer

Theorem 6olt9o

Description: Ordinal 6 is less than ordinal 9. (Contributed by BTernaryTau, 17-Oct-2026)

Ref Expression
Assertion 6olt9o 
|- 6o e. 9o

Proof

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