
theorem Th42:
  for f being non-empty Function, X, Y being non empty set, 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
proof
  let f be non-empty Function, X, Y be non empty set, i, j be object;
  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));
  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
  proof
    assume A6: i <> j;
    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;
  end;
  thus X = (f +* (j,Y)).i by A1, A4, FUNCT_7:31
    .= Y by A1, A5, FUNCT_7:31;
end;
