Re: What isn't a tautology?

On 21 Jul 2005 06:52:59 -0700, George Dance <georgedance04@xxxxxxxx> said:
Chris Menzel wrote:
>> Consider, e.g., "A prime number is a positive integer > 1
>> whose only positive integer divisors are itself and 1", i.e.:
>> (x)(Prime(x) <-> Integer(x) & x > 1 & (y)(Integer(y) & y > 1
>> & Divides(y,x) -> (y = 1 v y = x))). This definition has a
>> complex logical form in predicate logic, but its
>> propositional form is simply that of an atomic sentence P;
>> it's not a negation, conjunction, disjunction, conditional,
>> or biconditional.
>
> snip
>
>> Its formalization in propositional logic would be: P. Granted, all of
>> its *instances* would be biconditionals, but they don't do much better
>> in capturing the content of the original definition; for that you need
>> the apparatus of quantification.
>
> The original definition, "A prime number is a positive integer [etc.]"
> asserts nothing about all the objects of the universe; while your
> translation of it into FOPL does. It's obvious that translating the
> statement into FOPL added content.

Well, I'm afraid most any logician or linguist (working in semantics)
would tell you that the indefinite article is serving semantically as a
universal quantifier. And I guess I myself don't see a bit of
difference in content between "A prime number is a positive integer such
that ..." and "For any object x in the universe, that object x is a
prime number iff it's a positive integer ...". Both tell us what it is
that sets prime numbers apart from everything else in the universe; the
latter just brings in everything else more overtly.

>> > Well, that's all well and good; but one normally doesn't use
>> > propositional logic to formalize sentences of predicate logic, does
>> > one? What would be the point of doing that?
>>
>> I haven't the slightest idea. Why would you think I'm interested in
>> doing it?
>
> Because that is what you've been doing consistently...

Er, well, yes, but *only* to show that the definition is not a
tautology, as it lacks the appropriate propositional logical form.
That's it.

> Which was my first objection to your demonstration: I don't see why it
> would be necessary to translate an English sentence into FOPL before
> translating it into PC.

This was never the procedure. I never translated *first* into FOPL. I
provided the FOPL translation *and* a PC translation -- in the latter
case, once again, only to show that the definition was not a tautology.

> I wanted to know why "A prime number is a positive integer [etc.]" was
> not a biconditional - your answer was that (x)(Prime(x) <-> Integer(x)
> & x > 1 & (y)(Integer(y) & y > 1 & Divides(y,x) -> (y = 1 v y = x)))
> is not a biconditional.

Which it isn't.

> Then I wanted to know why you were translating the FOPL statement
> rather than the English one;

But I wasn't. But just to get past this silly sticking point, I hereby
translate the English definition "A prime number is a positive integer
...." directly into the language of PC: "P".

> From what you've said, I infer that you're saying that your two
> definitions (the one in English and the one in FOPL) are the exact same
> proposition;

I am saying that the FOPL version is the correct rendering of the
English sentence into the language of predicate logic; but, yes, I
suppose I would say that they express the same proposition.

> that it's absurd and incoherent to believe one and disbelieve the
> other. I'm not convinced of that, but I've said more than enough
> about why up above - so, I'd suggest, let's just drop that point (and
> stipulate, for the balance of the discussion, that statements in
> English and their FOPL translations are (for our purposes) the same
> assertion).

I can't imagine what your reasons are, but I'm happy to move past this
point as well.

>> ...It's just a fact that definitions like the one in
>> question *are* universally quantified. The point about their coarse and
>> inadequate representation in propositional logic was meant only to show
>> that such definitions are not tautologies.
>
> Which would imply that (at least some) tautologies are not tautologies,
> either; because it would be 'just a fact' that at least some
> tautological statements in English, being true in every conceivable
> state of affairs (and therefore true wrt everything) *are* universally
> quantified as well.

Completely lost here; again, probably best to let it go.

>> > and hence all must be translated into PC as non-tautological atomic
>> > constants - then that rule must apply to tautological statements
>> > (like "A prime number is a prime number") as well:
>>
>> That is not a tautology, its a universal quantification; it's of the
>> form "(x)(Px -> Px)".
>
> Thank you. That's a consistent answer,

Like every other answer I've given.

> which lets me bring in my second objection to your proof method.
>
>> Granted, the unquantified part has the form of a tautology, those
>> variable occurrences need to be quantified to capture the logical
>> form of the example.
>
> And therefore, by your argument, "A prime number is a prime number"
> the 'logical form of the example' must have the 'coarse and inadequate
> representation in propositional logic' of an atomic proposition, P.

That is correct.

> Which is my second objection to your proof: even granting that "A prime
> number is a prime number" has the inherent form (x)(Px->Px) (P: "-is a
> prime number") - that they're the exact same proposition, merely in two
> different languages - it does not follow that "A prime number is a
> prime number" is not a tautology; because you've not shown that (x)(Px
> -> Px) is not a tautology (merely said that it was not a tautology
> because it's universally quantified).

Well, sorry you don't recognize it, but in fact that *is* a
demonstration that it's not a tautology.

> I'd argue that (x)(Px -> Px) is a tautology, as follows:
>
> 1. If two statements are logically equivalent (ie, have the exact same
> truth value), ...

Goodness me. I hope this is just a slip. Do you really think that,
say, "Bush is President" and "2 is prime" are logically equivalent just
because they are both true? Two statements are logically equivalent if
they have the same truth value *under all interpretations*.

> ...then, if one is a tautology, so is the other.

Well, that is just so not true on the (completely standard and common)
notion of "tautology of predicate logic" that I clarified in my previous
post (quoted below), and which you appeared to agreed with. On that
notion, it is trivially provable that no universally quantified sentence
is a tautology. Period. There's absolutely no way to replace
subformulas with propositional constants in a quantified sentence to
yield a PC tautology. It just can't be done.

> 2. If the propositional form of any statement is a tautology, then
> that statement is a tautology.

Well, there isn't in general such a thing as *the* propositional form of
a sentence of predicate logic, but I take this to be an informal way of
expressing, or at least agreeing with, the notion of tautologousness in
question.

> 3. (x)(Px->Px) is logically equivalent to ~(Ex)(Px & ~Px).

True.

> 4. ~(Ex)(Px & ~Px) is logically equivalent to ~(Pa & ~Pa).

True, because both are logical truths, i.e., true in all
interpretations.

> 5. The propositional form of ~(Pa & ~Pa) is ~(P & ~P).

The most "fine-grained" propositional form, yes.

> 6. ~(P & ~P) is a tautology.

Yep.

> 7. ~(Pa & ~Pa) is a tautology. (2,6)

Yep.

> 8. ~(Ex)(Px & ~Px) is a tautology. (1,7)

Nope -- (1) is false, as noted, so the inference is unsound.

> 9. (x)(Px -> Px) is a tautology. (1,8)

Sorry.

>> > meaning that no tautological statement could be translated into PC as
>> > a tautology, either.
>>
>> No. The notion of tautology can be extended to predicate logic in a
>> natural way. One way to do this, roughly, is to say that a sentence S
>> of predicate logic is a tautology if there is way of uniformly replacing
>> subformulas of S with propositional constants in such a way that the
>> result of the substitution is a tautology of propositional logic. So,
>> for instance, "(x)(Px v Qx) -> ((x)Rx -> (x)(Px v Qx))" is a tautology
>> because we can replace "(x)(Px v Qx)" with "p" and "(x)Rx" with "q" and
>> result is the propositional tautology "p -> (q -> p)".
>
>> Agreed. At the same time, we'd disagree about whether (x)(((Px v Qx)
>> -> (Rx -> (Px v Qx))) is a tautology.

Perhaps, but you'd be wrong -- it is demonstrably not a tautology of
predicate logic on the above definition. Perhaps you have some other
meaning of the word "tautology" in mind, but if it entails that
"(x)(((Px v Qx) -> (Rx -> (Px v Qx)))" is a "tautology", then you are
talking about another concept entirely. We can talk about that concept,
and perhaps it is an interesting one, but the fact will remain that
"(x)(((Px v Qx) -> (Rx -> (Px v Qx)))" is not a tautology, as the term
is used in mathematical logic.

Chris Menzel

.

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