let A be QC-alphabet ; for s, t being QC-symbol of A holds
( s < t iff not t <= s )
let s, t be QC-symbol of A; ( s < t iff not t <= s )
thus
( s < t implies not t <= s )
( not t <= s implies s < t )proof
assume
s < t
;
not t <= s
then A1:
(
s <= t & not
s = t )
by Def34;
assume
t <= s
;
contradiction
hence
contradiction
by A1, Th23;
verum
end;
assume
not t <= s
; s < t
then
( s <= t & not t = s )
by Th24;
hence
s < t
by Def34; verum