theorem Th126:
  for X1, X2 being non empty SubSpace of X for g being Function
  of X1 union X2,Y holds g = (g|X1) union (g|X2)
proof
  let X1, X2 be non empty SubSpace of X;
  let g be Function of X1 union X2,Y;
  now
    assume
A1: X1 meets X2;
    then
A2: X1 meet X2 is SubSpace of X2 by TSEP_1:27;
A3: X2 is SubSpace of X1 union X2 by TSEP_1:22;
A4: X1 is SubSpace of X1 union X2 by TSEP_1:22;
    X1 meet X2 is SubSpace of X1 by A1,TSEP_1:27;
    hence (g|X1)|(X1 meet X2) = g|(X1 meet X2) by A4,Th72
      .= (g|X2)|(X1 meet X2) by A2,A3,Th72;
  end;
  hence thesis by Def12;
end;
