let a, b be Ordinal; ( a <> b implies ord-type {a,b} = 2 )
assume
a <> b
; ord-type {a,b} = 2
then A1:
card {a,b} = 2
by CARD_2:57;
( a c= a \/ b & b c= a \/ b )
by XBOOLE_1:7;
then
( a in succ (a \/ b) & b in succ (a \/ b) )
by ORDINAL1:22;
then A2:
{a,b} c= succ (a \/ b)
by ZFMISC_1:32;
then
On {a,b} = {a,b}
by ORDINAL3:6;
hence
ord-type {a,b} = 2
by A1, A2, CARD_5:36; verum