There are logical relations between some of the Aristotelian forms.

**Contradictories:** sentences that say the opposite of each other. If one is T the other is F, and vice versa.

For example, some of them contradict each other: they say the opposite of each other.

That means if you put them in a set together, it would be a contradictory set.

Let’s see if you can figure out which ones.

Take All Ps are Q.

If it is true that All Ps are Q, then it must be false that Some Ps are not Q.

It is tempting to think that No Ps are Q contradicts All Ps are Q, but here we have to recall the vacuous possibility: if there are no Ps, then both could be true.

And that means they don’t contradict each other.

Since All Ps are Q and Some Ps are not Q are contradictories, the negation of one is equivalent to the other.

Ax(P(x)->Q(x)) ⇔ ~Ex(P(x)&~Q(x))

~Ax(P(x)->Q(x)) ⇔ Ex(P(x)&~Q(x))

Do the other two Aristotelian forms also make a pair of contradictories?

If No Ps are Q, then it must be false that Some Ps are Q. And vice versa.

So they are also a contradictory pair.

Hence:

Ax(P(x)->~Q(x)) ⇔ ~Ex(P(x)&Q(x))

~Ax(P(x)->~Q(x)) ⇔ Ex(P(x)&Q(x))

The two translations of **No Ps are Q **reveal the contradictories:

Ax(P(x)->~Q(x))

~Ex(P(x)&Q(x))

Remember how we said that there are two equally good translations of No Ps are Q?

Now we can see better why both ways are good: one of them is just the negation of Some Ps are Q.

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