let R be Order of X; R is well_founded
let Y be set ; WELLORD1:def 2 ( not Y c= field R or Y = {} or ex b1 being object st
( b1 in Y & R -Seg b1 misses Y ) )
assume A1:
( Y c= field R & Y <> {} )
; ex b1 being object st
( b1 in Y & R -Seg b1 misses Y )
defpred S1[ set ] means ( X <> {} implies ex a being set st
( a in X & R -Seg a misses X ) );
A2:
Y is finite
by A1;
A3:
S1[ {} ]
;
A4:
for x, B being set st x in Y & B c= Y & S1[B] holds
S1[B \/ {x}]
proof
let x,
B be
set ;
( x in Y & B c= Y & S1[B] implies S1[B \/ {x}] )
assume that A5:
(
x in Y &
B c= Y &
S1[
B] )
and
B \/ {x} <> {}
;
ex a being set st
( a in B \/ {x} & R -Seg a misses B \/ {x} )
per cases
( B = {} or B <> {} )
;
suppose
B <> {}
;
ex a being set st
( a in B \/ {x} & R -Seg a misses B \/ {x} )then consider a being
set such that A7:
(
a in B &
R -Seg a misses B )
by A5;
per cases
( x nin R -Seg a or x in R -Seg a )
;
suppose
x in R -Seg a
;
ex a being set st
( a in B \/ {x} & R -Seg a misses B \/ {x} )then A9:
(
x <> a &
[x,a] in R )
by WELLORD1:1;
take b =
x;
( b in B \/ {x} & R -Seg b misses B \/ {x} )
b in {x}
by TARSKI:def 1;
hence
b in B \/ {x}
by XBOOLE_0:def 3;
R -Seg b misses B \/ {x}assume
R -Seg b meets B \/ {x}
;
contradictionthen consider c being
object such that A10:
(
c in R -Seg b &
c in B \/ {x} )
by XBOOLE_0:3;
reconsider cc =
c,
xx =
x,
aa =
a as
set by TARSKI:1;
A11:
(
c <> b &
[c,b] in R )
by A10, WELLORD1:1;
then
(
cc,
xx in R &
x,
a in R )
by A9;
then A12:
(
cc,
aa in R &
c <> a )
by A11, Def2, Def1;
then
(
[c,a] in R & (
c in B or
c in {x} ) )
by A10, XBOOLE_0:def 3;
then
(
c in R -Seg a &
c in B )
by A11, A12, TARSKI:def 1, WELLORD1:1;
hence
contradiction
by A7, XBOOLE_0:3;
verum end; end; end; end;
end;
S1[Y]
from FINSET_1:sch 2(A2, A3, A4);
hence
ex a being object st
( a in Y & R -Seg a misses Y )
by A1; verum