let a, b, c be ordinal number ; ( 0 in a & 1 in b & c = b -exponent a implies a div^ (exp b,c) in b )
assume A0:
( 0 in a & 1 in b & c = b -exponent a )
; a div^ (exp b,c) in b
set n = a div^ (exp b,c);
exp b,c <> 0
by A0, ORDINAL4:22;
then consider d being ordinal number such that
A2:
( a = ((a div^ (exp b,c)) *^ (exp b,c)) +^ d & d in exp b,c )
by ORDINAL3:def 7;
assume
not a div^ (exp b,c) in b
; contradiction
then
b *^ (exp b,c) c= (a div^ (exp b,c)) *^ (exp b,c)
by ORDINAL2:58, ORDINAL1:26;
then
( exp b,(succ c) c= (a div^ (exp b,c)) *^ (exp b,c) & (a div^ (exp b,c)) *^ (exp b,c) c= a )
by A2, ORDINAL2:61, ORDINAL3:27;
then
exp b,(succ c) c= a
by XBOOLE_1:1;
then
( succ c c= c & c in succ c )
by A0, EXP, ORDINAL1:10;
then
c in c
;
hence
contradiction
; verum