theorem Th50: :: Lemma 5
  \delta (x _1 + x _2, x) = x _0
proof
  thus x _0 = Expand (x _3, x _0) by Th36
    .= \delta (x _1 + x _2, \delta(x _3, x _0)) by Th43
    .= \delta (x _1 + x _2, x) by Th45;
end;
