Chapter 7, Tutorial 2

A Special Case: Derivations in Categorical Logic

In 7.1 we used an interpretation as guidance only. Let's be clear that we can do derivations without thinking about content, meaning, or interpretations. We can just think about rules of form. So, at any time on a derivation one can move from 'Ca' to '(%x)Cx': that is just %I. This is formal logic after all!

But it can *help* to know that %I makes sense. Suppose that Ann is a member of the group under consideration, i.e., is in our "universe of discourse". Then %I is just a PL version of the reasoning from "Ann is a carnivore" to "someone is a carnivore".

Let's continue to have an interpretation around for intuitive guidance.

Universe of Discourse: *All* things

Ax: x is an animal; Cx: x is a carnivore; Mx: x is a mammal, Rx: x is a reptile; Tx: x is a tiger.

Now, let's see that the rules of PD we've already developed work well for categorical logic's equivalences. And we'll also see that our PD is not yet complete. First the good news: Categorical logic's equivalences can be understood with the rules we already have.

Conversion via Commutation

We have seen that commutation gives a logical equivalence for existential and negative categorical sentences.

"Some S are P" is logically equivalent to "Some P are S".

and

"No S are P" is logically equivalent to "No P are S".

For examples,

(*) "Some mammals are carnivores" is logically equivalent to "Some carnivores are mammals".

and

(**) "No mammals are reptiles " is logically equivalent to "No reptiles are mammals".

OK? Now, if these are really equivalent, then their symbolizations in PL should be equivalent too. Let's see that they are.

First, to symbolize the sentences in (*) we would write these:

(* in PL)

(%x)(Mx&Cx)

(%x)(Cx&Mx)

Now, are these two logically equivalent? The answer is *yes*. All we need to do is show how to derive one from the other.

The __derivation test for equivalence of two sentences__ is to show that if either one is taken as premise, then the other can be derived as conclusion.

But to deriving one from the other is a snap, it's just **CM**:

Notice that this is just our old way of doing two derivations on one page to show logical equivalence. We've derived one from the other, so

(%x)(Mx&Cx)

and

(%x)(Cx&Mx)

are logically equivalent: there is no way for one to be true without the other.

(Note: These are just two separate derivations. You could do them on separate pages if you prefer.)

Now, let's see if we can do the same thing with

"No S are P" is logically equivalent to "No P are S".

and

(**) "No mammals are reptiles " is logically equivalent to "No reptiles are mammals".

First we symbolize.

(* in PL)

~(%x)(Mx&Rx)

~(%x)(Rx&Mx)

There **is** an alternative symbolization..

...for "No S are P": that is '(^x)(Sx>~Px)'. We will get a hint how to move back and forth between this and "No P are S"='(^x)(Px>~Sx)' in the next section of this tutorial. It's really very easy and it's 7.1b.

Now, suppose we try to do the two derivations...what __one__ rule will fit the question marks at lines 2 and 5 to show equivalence?