let b, a be ordinal number ; for n being natural number st 0 in n & n in b holds
b -exponent (n *^ (exp b,a)) = a
let n be natural number ; ( 0 in n & n in b implies b -exponent (n *^ (exp b,a)) = a )
assume Z0:
( 0 in n & n in b )
; b -exponent (n *^ (exp b,a)) = a
then A0:
succ 0 c= n
by ORDINAL1:33;
then Z1:
( 1 *^ (exp b,a) = exp b,a & 1 *^ (exp b,a) c= n *^ (exp b,a) )
by ORDINAL2:56, ORDINAL2:58;
Z5:
( exp b,a <> 0 & n <> 0 )
by Z0, ORDINAL4:22;
then
0 <> n *^ (exp b,a)
by ORDINAL3:34;
then Z2:
( 1 in b & 0 in n *^ (exp b,a) )
by A0, Z0, ORDINAL1:22, ORDINAL3:10;
now let c be
ordinal number ;
( exp b,c c= n *^ (exp b,a) implies not c c/= a )assume Z3:
(
exp b,
c c= n *^ (exp b,a) &
c c/= a )
;
contradictionthen
a in c
by ORDINAL1:26;
then
succ a c= c
by ORDINAL1:33;
then Z4:
exp b,
(succ a) c= exp b,
c
by Z0, ORDINAL4:27;
exp b,
(succ a) = b *^ (exp b,a)
by ORDINAL2:61;
then
b *^ (exp b,a) c= n *^ (exp b,a)
by Z3, Z4, XBOOLE_1:1;
then
b c= n
by Z5, ORDINAL3:38;
then
n in n
by Z0;
hence
contradiction
;
verum end;
hence
b -exponent (n *^ (exp b,a)) = a
by Z1, Z2, EXP; verum