let D be non empty set ; for X, Y, Z, S being set
for f1, f2 being Function of [:X,Y,Z,S:],D st ( for x, y, z, s being object st x in X & y in Y & z in Z & s in S holds
f1 . [x,y,z,s] = f2 . [x,y,z,s] ) holds
f1 = f2
let X, Y, Z, S be set ; for f1, f2 being Function of [:X,Y,Z,S:],D st ( for x, y, z, s being object st x in X & y in Y & z in Z & s in S holds
f1 . [x,y,z,s] = f2 . [x,y,z,s] ) holds
f1 = f2
let f1, f2 be Function of [:X,Y,Z,S:],D; ( ( for x, y, z, s being object st x in X & y in Y & z in Z & s in S holds
f1 . [x,y,z,s] = f2 . [x,y,z,s] ) implies f1 = f2 )
assume A1:
for x, y, z, s being object st x in X & y in Y & z in Z & s in S holds
f1 . [x,y,z,s] = f2 . [x,y,z,s]
; f1 = f2
for t being object st t in [:X,Y,Z,S:] holds
f1 . t = f2 . t
proof
let t be
object ;
( t in [:X,Y,Z,S:] implies f1 . t = f2 . t )
assume
t in [:X,Y,Z,S:]
;
f1 . t = f2 . t
then
ex
x,
y,
z,
s being
object st
(
x in X &
y in Y &
z in Z &
s in S &
t = [x,y,z,s] )
by MCART_1:79;
hence
f1 . t = f2 . t
by A1;
verum
end;
hence
f1 = f2
by FUNCT_2:12; verum