let X, Y be set ; for f being Function of X,Y st X <> {} & Y <> {} holds
( f is one-to-one iff for Z being set
for g, h being Function of Z,X st f * g = f * h holds
g = h )
let f be Function of X,Y; ( X <> {} & Y <> {} implies ( f is one-to-one iff for Z being set
for g, h being Function of Z,X st f * g = f * h holds
g = h ) )
assume that
A1:
X <> {}
and
A2:
Y <> {}
; ( f is one-to-one iff for Z being set
for g, h being Function of Z,X st f * g = f * h holds
g = h )
A3:
dom f = X
by A2, Def1;
thus
( f is one-to-one implies for Z being set
for g, h being Function of Z,X st f * g = f * h holds
g = h )
( ( for Z being set
for g, h being Function of Z,X st f * g = f * h holds
g = h ) implies f is one-to-one )proof
assume A4:
f is
one-to-one
;
for Z being set
for g, h being Function of Z,X st f * g = f * h holds
g = h
let Z be
set ;
for g, h being Function of Z,X st f * g = f * h holds
g = hlet g,
h be
Function of
Z,
X;
( f * g = f * h implies g = h )
A5:
(
rng g c= X &
rng h c= X )
;
(
dom g = Z &
dom h = Z )
by A1, Def1;
hence
(
f * g = f * h implies
g = h )
by A3, A4, A5, FUNCT_1:27;
verum
end;
assume A6:
for Z being set
for g, h being Function of Z,X st f * g = f * h holds
g = h
; f is one-to-one
hence
f is one-to-one
by FUNCT_1:27; verum