let x, X, y be set ; ( x in X & y in X implies X \ {x},X \ {y} are_equipotent )
assume that
A1:
x in X
and
A2:
y in X
; X \ {x},X \ {y} are_equipotent
defpred S1[ set , set ] means ( ( $1 = y & $2 = x ) or ( $1 <> y & $1 = $2 ) );
A3:
for a being set st a in X \ {x} holds
ex b being set st S1[a,b]
proof
let a be
set ;
( a in X \ {x} implies ex b being set st S1[a,b] )
assume
a in X \ {x}
;
ex b being set st S1[a,b]
(
a = y implies ex
b being
set st
S1[
a,
b] )
;
hence
ex
b being
set st
S1[
a,
b]
;
verum
end;
A4:
for a, b1, b2 being set st a in X \ {x} & S1[a,b1] & S1[a,b2] holds
b1 = b2
;
consider f being Function such that
A5:
( dom f = X \ {x} & ( for a being set st a in X \ {x} holds
S1[a,f . a] ) )
from FUNCT_1:sch 2(A4, A3);
take
f
; WELLORD2:def 4 ( f is one-to-one & proj1 f = X \ {x} & proj2 f = X \ {y} )
thus
f is one-to-one
( proj1 f = X \ {x} & proj2 f = X \ {y} )
thus
dom f = X \ {x}
by A5; proj2 f = X \ {y}
thus
rng f c= X \ {y}
XBOOLE_0:def 10 X \ {y} c= proj2 f
let z be set ; TARSKI:def 3 ( not z in X \ {y} or z in proj2 f )
assume A14:
z in X \ {y}
; z in proj2 f
then
not z in {y}
by XBOOLE_0:def 5;
then A15:
z <> y
by TARSKI:def 1;
hence
z in proj2 f
by A16; verum