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) ?

  • On what ground are you asserting that "either p or ~p = p & ~p" ? The first one is TRUE while the second one is FALSE. – Mauro ALLEGRANZA Nov 11 '19 at 16:21
  • I do know that according to classical logic 'either p or ~p' is a tautology and always true, 'p & ~p' a contradiction and always false but if you read the whole text it should be clear that there seems to be a paradox following the tenets of classical logic if we assume that p & ~p is equivalent to ~p & ~~p then p & ~p seems to be equal to either p or ~p. Lets use unconventional thinking skills for a minute ;) – Noname Nov 11 '19 at 16:41
  • If you derive a FALSE conclusion, either you have made some mistake in using the logical rules or you have used a FALSE premise in your argument. – Mauro ALLEGRANZA Nov 11 '19 at 18:34
  • Thats seems to be a false conclusion you have there, since "this statement is false" leads us to the fact that statement (Liar sentence) is true if and only if it is false. using logical rules without false premises ( at least seemingly) which is a FALSE conclusion. We have trouble pointing out faulty application of rules of logic or a faulty premise so that seems to be a counterexample, just like my example above – Noname Nov 11 '19 at 19:04
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    I added a lot of punctuation, in an attempt to clarify the rationale. I hope I was able to follow your line of thought. – Mark Andrews Nov 11 '19 at 21:20

p&~p is equivalent to ~p&~~p, but p&~p and ~p&~~p are not equivalent to pv~p -- in fact, they are contradictory to each other. This can be easily seen by taking a look at the truth table:

p | ~p | ~~p | p&~p | ~p&~~p | pv~p
1 | 0  |  1  |  0   |   0    |  1  
0 | 1  |  0  |  0   |   0    |  1

p&~p and ~p&~~p are equivalent to each other and contradictions (= statements that are always false) -- there is no interpretation under which both p and ~p, or both ~p and ~~p, can be true at the same time, hence they have 0 in all rows of the truth table.
pv~p, on the other hand, is a tautology (= a statement that is always true): It is true under every possible interpretation, and correspondingly, in every row of the truth table.

Since p&~p and ~p&~~p have the same truth value in each row, they are logically equivalent.
Since p&~p/~p&~~p and pv~p have different truth values in at least one row, p&~p/~p&~~p are not logically equivalent to pv~p. Since they even have different truth values in all rows, they are not just inequivalent, but contradictory to each other.

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  • Yes I know that but if 'p & ~p' (which violates the law of non-contradiction BUT NOT the law of excluded middle) is equivalent to '~p & ~~p' (which violates the law of excluded middle BUT NOT the law of non-contradiction) then p,~p seems also to be not both not the case (via the violation of law of non-contradiction), but also not both the case (via violation of excludes middle) but that follows tge definition of excluded middle , hence we have also either p or ~p , but then p & ~p seems to also express the proposition either p or ~p , hence p&~p= either p or ~p – Noname Nov 11 '19 at 16:53
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    ~p&~~p violates the law of non-contradiction. – lemontree Nov 11 '19 at 16:54
  • not if the law of non-contradiction is stated as ~(p & ~p) since ~p & ~~p states that p,~p are both NOT the case (they are both false) , while the law of non-contradiction states that they cannot be both the case (they cannot be both true) on the other hand it does violate the law of excluded middle : p or ~p which states that p,~p cannot be both not the case (that they cannot be both false) – Noname Nov 11 '19 at 16:59
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    The law of non-contradiction is ~(phi&~phi), for arbitrary instances of formulas phi. ~p&~~p is such an instance with phi=~p that violates this principle. – lemontree Nov 11 '19 at 17:06
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    ... But this doe not in any way change the fact that neither of these are equivalent to pv~p. Look at the truth table and you got your answer, there's not more to it than that. – lemontree Nov 11 '19 at 17:07

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