theorem
  a 'nand' (a 'imp' b) = 'not' (a '&' b)
proof
  thus a 'nand' (a 'imp' b) = ('not' a 'or' a) '&' 'not' (a '&' b) by Th24
    .= I_el(Y) '&' 'not' (a '&' b) by BVFUNC_4:6
    .= 'not' (a '&' b) by BVFUNC_1:6;
end;
