theorem Th65:
  a 'imp' (b 'nor' c) = 'not' (a '&' (b 'or' c))
proof
  thus a 'imp' (b 'nor' c) =a 'imp' 'not' (b 'or' c) by Th2
    .='not' a 'or' 'not' (b 'or' c) by BVFUNC_4:8
    .='not' (a '&' (b 'or' c)) by BVFUNC_1:14;
end;
