theorem Th35:
  (A \+\ B)"" = (A"") \+\ (B"")
proof
  thus (A \+\ B)"" = ((A\B)"") \/ ((B\A)"") by Th32
    .= ((A"")\(B"")) \/ ((B\A)"") by Th34
    .= (A"") \+\ (B"") by Th34;
end;
