The De Morgan Equivalences state that
(1) ~ (p v q) is equivalent to (~p & ~q);
(2) ~ (p & q) is equivalent to (~p v ~q).
As I've noticed before, you can do propositional logic with categorical syllogisms if you assume a singleton universe; that is, as Tom nicely puts it, you can treat standard propositional logic as a "term logic with a single individual". It's verbally convenient to take this individual as 'the world' (in whatever way one chooses at any given time to understand that). Since we can do this, we should be able to prove the De Morgan Equivalences by categorical means. It is, in fact, fairly easy to do so, and so I post the arguments for fun.
The First Equivalence. We start with p or q, which we will be negating:
(p v q)
This has the following Welton diagram:
| | | | X |
This says that No nonp-world is a nonq-world, which is the categorical translation of the disjunction.
The contradictory of this is:
Some nonp-world is a nonq-world.
Now, given that our universe is singleton (there is only one world), we can conclude the universal from the particular (A from I and E from O); I will call this 'propositional concession'. Propositional concession gives us:
Every nonp-world is a nonq-world.
This is diagrammed as:
| X | X | X | |
This diagram pictures the following things: Every nonp-world is a nonq-world; Some nonp-world is a nonq-world; ~(p v q); and (~p & ~q). And, in fact, (~p & ~q) is the propositional translation of the categorical Every nonp-world is a nonq-world.
Thus our proof:
(1) ~ (p v q)
(2) It is not the case that ~pE~q.
(3) ~pI~q
(4) ~pA~q
(5) (~p & ~q)
(2) is a translation of (1). (3) takes "It is not the case that" as a marker for contradiction. (4) is from (3) by propositional concession. (5) is the translation of (4).
We can actually simplify this proof somewhat by taking propositional negation as contrariety rather than contradiction. (As far as I can tell this seems to be a good idea in general for reasons of convenience; but in a singleton universe there is no significant difference between contrariety and contradiction.) The contrariety-based proof is as follows:
(1) ~ (p v q)
(2) It is not the case that ~pE~q
(3) ~pA~q
(4) (~p & ~q)
(2) is from (1) by translation. (3) is from (2) by taking "It is not the case that" to indicate contrariety. (4) is from (3) by translation.
The Second Equivalence. We start with p and q, which we will be negating:
(p & q)
This has the following Welton diagram:
| | X | X | X |
This says two things at once:
Every p-world is a q-world.
Every q-world is a p-world.
It doesn't matter which of these we take. I'll take the first. The contradictory of this is:
Some p-world is not a q-world.
With propositional concession we get:
No p-world is a q-world.
We can diagram this as:
| X | | | |
This diagram is equivalent to three things: No p-world is a q-world; Some p-world is not a q-world; ~ (p & q); and (~p v ~q). The last of the four is what we wished to prove. And, in fact, the categorical translation of (~p v ~q) is No p-world is a q-world.
So the proof goes:
(1) ~ (p & q)
(2) It is not the case that pAq
(3) pOq
(4) pEq
(5) (~p v ~q)
(2) is the categorical translation of (1). (3) is what you get by reading "It is not the case that" as contradiction. (4) you get by propositional concession. (5) is the propositional translation of (4).
This assumes, of course, that propositional negation is contradiction. If we assume that it is contrariety, we get the following, simpler, result.
(1) ~ (p & q)
(2) It is not the case that pAq
(3) pEq
(4) (~p v ~q)
