let X be set ; for R being Relation st R is co-well_founded & R is irreflexive & X is finite & R is finite holds
ex n being Nat st (iter (R,n)) .: X = {}
let R be Relation; ( R is co-well_founded & R is irreflexive & X is finite & R is finite implies ex n being Nat st (iter (R,n)) .: X = {} )
assume that
A1:
( R is co-well_founded & R is irreflexive & X is finite )
and
A2:
R is finite
and
A3:
for n being Nat holds (iter (R,n)) .: X <> {}
; contradiction
defpred S1[ object ] means for n being Nat holds Im ((iter (R,n)),$1) <> {} ;
consider x0 being set such that
A4:
( x0 in X & S1[x0] )
by A1, A3, Th19;
defpred S2[ object , object , object ] means ( S1[$2] implies ( $3 in Im (R,$2) & S1[$3] ) );
A5:
for n being Nat
for x being set ex y being set st S2[n,x,y]
proof
let n be
Nat;
for x being set ex y being set st S2[n,x,y]let x be
set ;
ex y being set st S2[n,x,y]
per cases
( not S1[x] or S1[x] )
;
suppose A6:
not
S1[
x]
;
ex y being set st S2[n,x,y]take
0
;
S2[n,x, 0 ]thus
S2[
n,
x,
0 ]
by A6;
verum end; suppose A7:
S1[
x]
;
ex y being set st S2[n,x,y]end; end;
end;
consider f being Function such that
A10:
( dom f = NAT & f . 0 = x0 & ( for n being Nat holds S2[n,f . n,f . (n + 1)] ) )
from RECDEF_1:sch 1(A5);
defpred S3[ Nat] means S1[f . $1];
A11:
S3[ 0 ]
by A4, A10;
A12:
for k being Nat st S3[k] holds
S3[k + 1]
by A10;
A13:
for k being Nat holds S3[k]
from NAT_1:sch 2(A11, A12);
A14:
rng f c= field R
then consider z being object such that
A16:
( z in rng f & ( for x being object st x in rng f & z <> x holds
[z,x] nin R ) )
by A1, A10, RELAT_1:42;
consider y being object such that
A17:
( y in NAT & z = f . y )
by A10, A16, FUNCT_1:def 3;
reconsider y = y as Element of NAT by A17;
( S2[y,z,f . (y + 1)] & S3[y] & y + 1 in NAT )
by A10, A13, A17, ORDINAL1:def 12;
then A18:
( [z,(f . (y + 1))] in R & f . (y + 1) in rng f )
by A10, A17, RELAT_1:169, FUNCT_1:def 3;
then
( z = f . (y + 1) & R is_irreflexive_in field R )
by A1, A16;
hence
contradiction
by A14, A18; verum