reserve A, B for non empty set,
  A1, A2, A3 for non empty Subset of A;
reserve X for TopSpace;
reserve X for non empty TopSpace;
reserve X1, X2 for non empty SubSpace of X;
reserve X0, X1, X2, Y1, Y2 for non empty SubSpace of X;
reserve X, Y for non empty TopSpace;
reserve f for Function of X,Y;
reserve X,Y,Z for non empty TopSpace;
reserve f for Function of X,Y,
  g for Function of Y,Z;
reserve X, Y for non empty TopSpace,
  X0 for non empty SubSpace of X;
reserve f for Function of X,Y;
reserve f for Function of X,Y,
  X0 for non empty SubSpace of X;
reserve X, Y for non empty TopSpace,
  X0, X1 for non empty SubSpace of X;
reserve f for Function of X,Y,
  g for Function of X0,Y;
reserve X0, X1, X2 for non empty SubSpace of X;
reserve f for Function of X,Y,
  g for Function of X0,Y;
reserve X for non empty TopSpace,
  H, G for Subset of X;
reserve A for Subset of X;
reserve X0 for non empty SubSpace of X;
reserve X, Y for non empty TopSpace;
reserve X1, X2 for non empty SubSpace of X;
reserve X, Y for non empty TopSpace;

theorem Th128:
  for X1, X2 being non empty SubSpace of X st X1 meets X2 for f1
being Function of X1,Y, f2 being Function of X2,Y holds (f1 union f2)|X1 = f1 &
  (f1 union f2)|X2 = f2 iff f1|(X1 meet X2) = f2|(X1 meet X2)
proof
  let X1, X2 be non empty SubSpace of X such that
A1: X1 meets X2;
  let f1 be Function of X1,Y, f2 be Function of X2,Y;
  thus (f1 union f2)|X1 = f1 & (f1 union f2)|X2 = f2 implies f1|(X1 meet X2) =
  f2|(X1 meet X2)
  proof
A2: X1 meet X2 is SubSpace of X2 & X2 is SubSpace of X1 union X2 by A1,
TSEP_1:22,27;
    assume that
A3: (f1 union f2)|X1 = f1 and
A4: (f1 union f2)|X2 = f2;
    X1 meet X2 is SubSpace of X1 & X1 is SubSpace of X1 union X2 by A1,
TSEP_1:22,27;
    then (f1 union f2)|(X1 meet X2) = f1|(X1 meet X2) by A3,Th72;
    hence thesis by A2,A4,Th72;
  end;
  thus thesis by Def12;
end;
