hereby ( ( not 1 in b or not 0 in a ) implies ex b1 being Ordinal st b1 = 0 )
assume A:
( 1
in b &
0 in a )
;
ex c being ordinal number st
( exp b,c c= a & ( for d being ordinal number st exp b,d c= a holds
not d c/= c ) )defpred S1[
Ordinal]
means a c= exp b,$1;
a c= exp b,
a
by A, ORDINAL4:32;
then A0:
ex
c being
ordinal number st
S1[
c]
;
consider c being
ordinal number such that A1:
(
S1[
c] & ( for
d being
ordinal number st
S1[
d] holds
c c= d ) )
from ORDINAL1:sch 1(A0);
0 in 1
by NAT_1:45;
then A2:
0 in b
by A, ORDINAL1:19;
per cases
( a = exp b,c or a c< exp b,c )
by A1, XBOOLE_0:def 8;
suppose
a c< exp b,
c
;
ex d being ordinal number st
( exp b,d c= a & ( for e being ordinal number st exp b,e c= a holds
not e c/= d ) )then C1:
a in exp b,
c
by ORDINAL1:21;
succ 0 c= a
by A, ORDINAL1:33;
then
( 1
in exp b,
c &
exp b,
0 = 1 )
by C1, ORDINAL1:22, ORDINAL2:60;
then C4:
c <> 0
;
then consider d being
ordinal number such that C3:
c = succ d
by ORDINAL1:42;
take d =
d;
( exp b,d c= a & ( for e being ordinal number st exp b,e c= a holds
not e c/= d ) )thus
exp b,
d c= a
for e being ordinal number st exp b,e c= a holds
not e c/= dlet e be
ordinal number ;
( exp b,e c= a implies not e c/= d )assume B2:
(
exp b,
e c= a &
e c/= d )
;
contradictionthen
d in e
by ORDINAL1:26;
then
c c= e
by C3, ORDINAL1:33;
then
exp b,
c c= exp b,
e
by A, ORDINAL4:27;
then
a in exp b,
e
by C1;
then
a in a
by B2;
hence
contradiction
;
verum end; end;
end;
thus
( ( not 1 in b or not 0 in a ) implies ex b1 being Ordinal st b1 = 0 )
; verum