let Y be set ; for f being Function st rng f c= Y & ( for g, h being Function st dom g = Y & dom h = Y & g * f = h * f holds
g = h ) holds
Y = rng f
let f be Function; ( rng f c= Y & ( for g, h being Function st dom g = Y & dom h = Y & g * f = h * f holds
g = h ) implies Y = rng f )
assume that
A1:
rng f c= Y
and
A2:
for g, h being Function st dom g = Y & dom h = Y & g * f = h * f holds
g = h
; Y = rng f
Y c= rng f
proof
deffunc H1(
object )
-> set =
{} ;
let y be
object ;
TARSKI:def 3 ( not y in Y or y in rng f )
assume that A3:
y in Y
and A4:
not
y in rng f
;
contradiction
defpred S1[
object ,
object ]
means ( ( $1
= y implies $2
= {{}} ) & ( $1
<> y implies $2
= {} ) );
A5:
for
x being
object st
x in Y holds
ex
y1 being
object st
S1[
x,
y1]
proof
let x be
object ;
( x in Y implies ex y1 being object st S1[x,y1] )
assume
x in Y
;
ex y1 being object st S1[x,y1]
(
x = y implies ex
y1 being
object st
S1[
x,
y1] )
;
hence
ex
y1 being
object st
S1[
x,
y1]
;
verum
end;
A6:
for
x,
y1,
y2 being
object st
x in Y &
S1[
x,
y1] &
S1[
x,
y2] holds
y1 = y2
;
consider h being
Function such that A7:
dom h = Y
and A8:
for
x being
object st
x in Y holds
S1[
x,
h . x]
from FUNCT_1:sch 2(A6, A5);
A9:
dom (h * f) = dom f
by A1, A7, RELAT_1:27;
consider g being
Function such that A10:
dom g = Y
and A11:
for
x being
object st
x in Y holds
g . x = H1(
x)
from FUNCT_1:sch 3();
A12:
dom (g * f) = dom f
by A1, A10, RELAT_1:27;
for
x being
object st
x in dom f holds
(g * f) . x = (h * f) . x
then A15:
g = h
by A2, A10, A7, A12, A9, Th2;
g . y = {}
by A3, A11;
hence
contradiction
by A3, A8, A15;
verum
end;
hence
Y = rng f
by A1; verum