theorem Th38:
  a 'nor' (b 'nor' c) = 'not' a '&' (b 'or' c)
proof
  thus a 'nor' (b 'nor' c) = 'not' (a 'or' (b 'nor' c)) by Th2
    .= 'not' (a 'or' 'not' (b 'or' c)) by Th2
    .= 'not' a '&' 'not' 'not' (b 'or' c) by BVFUNC_1:13
    .= 'not' a '&' (b 'or' c);
end;
