KindiakMath

Sometimes, our predicates would be defined in terms of two propositions and . This requires the use of double quantifiers, and it turns out that the quantifiers all have distinct meanings.

The simplest case arises when both quantifiers are universal. Namely, we want to consider the nested proposition

where is a predicate over (we will properly define this later on when discussing sets). This nested proposition means that for any fixed , the proposition holds. Observe that we can get a few flavours of such a nested proposition:

This explains why we cannot, in general, just interchange and , unless we have good reason to. We can, rather carefully, interchange pairs of , by analysing that the following sequences are equivalent:

• Proving that for any , .
• Proving that for any , .

In both cases, we see that we exhaust all possible values of such that holds. Hence, we can remove the nested brackets and denote

This line of reasoning also works with double . However, we need to be careful when our nested propositions are of different types. Nevertheless, the two are connected via negation.

Theorem 3. Let be a predicate on the set . Then

We abbreviate by removing the brackets, remembering to note the ordering of the quantifiers, to obtain the following equivalence:

To properly wrap up our discussion on predicate logic, it behooves us to formally define the cartesian product, in the next post.

—Joel Kindiak, 27 Nov 24, 2230H