theorem th11:
  a 'nand' (b 'or' c) = 'not' (a '&' b) '&' 'not' (a '&' c)
proof
  thus a 'nand' (b 'or' c) = 'not' (a '&' (b 'or' c)) by th1
    .= 'not' (a '&' b 'or' a '&' c) by BVFUNC_1:12
    .= 'not' (a '&' b) '&' 'not' (a '&' c) by BVFUNC_1:13;
end;
