let f be non-empty Function; :: thesis: for X, Y being non empty set
for i, j being object st i in dom f & j in dom f & ( not X c= f . i or not f . j c= Y ) & product (f +* (i,X)) c= product (f +* (j,Y)) holds
( i = j & X c= Y )

let X, Y be non empty set ; :: thesis: for i, j being object st i in dom f & j in dom f & ( not X c= f . i or not f . j c= Y ) & product (f +* (i,X)) c= product (f +* (j,Y)) holds
( i = j & X c= Y )

let i, j be object ; :: thesis: ( i in dom f & j in dom f & ( not X c= f . i or not f . j c= Y ) & product (f +* (i,X)) c= product (f +* (j,Y)) implies ( i = j & X c= Y ) )
assume that
A1: ( i in dom f & j in dom f ) and
A2: ( not X c= f . i or not f . j c= Y ) and
A3: product (f +* (i,X)) c= product (f +* (j,Y)) ; :: thesis: ( i = j & X c= Y )
a4: ( f +* (i,X) is non-empty & f +* (j,Y) is non-empty ) by ;
thus A5: i = j :: thesis: X c= Y
proof
assume A6: i <> j ; :: thesis: contradiction
A7: ( i in dom (f +* (i,X)) & j in dom (f +* (i,X)) ) by ;
(f +* (i,X)) . i c= (f +* (j,Y)) . i by ;
then a8: X c= (f +* (j,Y)) . i by ;
(f +* (i,X)) . j c= (f +* (j,Y)) . j by ;
then (f +* (i,X)) . j c= Y by ;
hence contradiction by A2, a8, A6, FUNCT_7:32; :: thesis: verum
end;
let x be object ; :: according to TARSKI:def 3 :: thesis: ( not x in X or x in Y )
assume A9: x in X ; :: thesis: x in Y
set g = the Element of product f;
A10: the Element of product f +* (i,x) in product (f +* (i,X)) by A1, A9, Th37;
i in dom (f +* (j,Y)) by ;
then ( the Element of product f +* (i,x)) . i in (f +* (j,Y)) . i by ;
then A11: ( the Element of product f +* (i,x)) . i in Y by ;
i in dom the Element of product f by ;
hence x in Y by ; :: thesis: verum