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 & ( X <> f . i or Y <> f . j ) & product (f +* (i,X)) = product (f +* (j,Y)) holds
( i = j & X = Y )
let X, Y be non empty set ; :: thesis: for i, j being object st i in dom f & j in dom f & ( X <> f . i or Y <> f . j ) & product (f +* (i,X)) = product (f +* (j,Y)) holds
( i = j & X = Y )
let i, j be object ; :: thesis: ( i in dom f & j in dom f & ( X <> f . i or Y <> f . j ) & product (f +* (i,X)) = product (f +* (j,Y)) implies ( i = j & X = Y ) )
assume that
A1: ( i in dom f & j in dom f ) and
A2: ( X <> f . i or Y <> f . j ) and
A3: product (f +* (i,X)) = product (f +* (j,Y)) ; :: thesis: ( i = j & X = Y )
( f +* (i,X) is non-empty & f +* (j,Y) is non-empty ) by A1, Th35;
then A4: f +* (i,X) = f +* (j,Y) by A3, PUA2MSS1:2;
thus A5: i = j :: thesis: X = Y
proof
assume A6: i <> j ; :: thesis: contradiction
A7: X = (f +* (i,X)) . i by A1, FUNCT_7:31
.= f . i by A4, A6, FUNCT_7:32 ;
Y = (f +* (j,Y)) . j by A1, FUNCT_7:31
.= f . j by A4, A6, FUNCT_7:32 ;
hence contradiction by A2, A7; :: thesis: verum
end;
thus X = (f +* (j,Y)) . i by A1, A4, FUNCT_7:31
.= Y by A1, A5, FUNCT_7:31 ; :: thesis: verum