In classical logic,
(1) p & ~p is equivalent to (2) ~p & ~~p;
if we read 'p & ~p' as p, ~p are both true/the case, and if we read '~p & ~~p' as p, ~p are both false/not the case (because their negations are true).
Or equivalently: if we read (1) as violating the law of non-contradiction ~(p & ~p) , and (2) as violating the law of excluded middle p or ~p , although they are equivalent in Classical logic notably through DeMorgans Law, there is a subtle difference which becomes even more apparent in non-classical logics, which have, e.g., the law of non-contradiction but not excluded middle.
That is: strictly speaking, the law of excluded middle and its instances have an inclusive disjunction. It is the law of non-contradiction that makes the disjunction of LeM an exclusive one; in classical logic, conversely, strictly speaking, the conjunction ~p & ~~p does not violate the form ~(p & ~p); in classical logic, it is because of the excluded middle p or ~p , that it is equivalent to violating ~(p & ~p).
That being said, it becomes clear that:
Following the convention of reading ' p & ~p' as p,~p being both true/ the case , we should also read is as NOT being both false /not the case, since, strictly speaking, it does not violate p or ~p.
Therefore we should also read ' ~p & ~~p' as p, ~p NOT being both true/the case, since, strictly speaking, it does violate p or ~p (and strictly speaking it does not violate , ~p or ~~p ).
But then since in (1), (2) (which are intersubstitutable according to classical logic) p,~p are ALSO not both the case & not both not the case; Then we have also either p or ~p , where either/or, is an exclusive disjunction (meaning p,~p are not both the case, but also not both not the case).
But then we have a paradox, since we seem to have ( either p or ~p) = (p & ~p ) (meaning at the same time in the same sense : p,~p are not both the case , neither both not the case, but also both the case , and both not the case); and according to classical logic (either p or ~p) ≠ (p & ~p).
Because when we apply the same scheme to (3) 'either p or ~p' and (4) 'either ~p or ~~p', we can read (3) as: p,~p are not both the case, neither both not the case. ~(p & ~p) & (p or ~p). We can read the (4) as: p,~p both being not the case is not the case , neither both being not not the case ~(~p & ~~p) & (~p or ~~p).
It is clear to see that (3) and (4) do not violate the law of non-contradiction, neither the law of excluded middle.
But then how can (either p or ~p) = (p & ~p) ?