Re: What isn't a tautology?

Chris Menzel wrote:
> On 20 Jul 2005 07:17:38 -0700, George Dance <georgedance04@xxxxxxxx> said:
> > Chris Menzel wrote:
> >> On 16 Jul 2005 19:26:47 -0700, George Dance <georgedance04@xxxxxxxx> said:
> >> > Chris Menzel wrote:
> >> >> On 16 Jul 2005 11:51:09 -0700, George Dance <georgedance04@xxxxxxxx> said:
> >> >> > Chris Menzel wrote:
> >> >> >> On 16 Jul 2005 09:32:31 -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.

> >> The short answer is that the syntactic type of a (nonatomic) sentence A
> >> of predicate logic is determined by the logical operator occurring in A
> >> that has the widest scope in A. The operator that has widest scope in
> >> the definition of "Prime" above is the universal quantifier. It is a
> >> simple exercise to show that every formula is of exactly one type.
> >
> > 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: above, when
formalizing "A prime number is a positive integer [etc.]" in PC; below,
to formalize "A prime number is a prime number" in PC; and, finally
(when you give an example of a statement that is a tautology in PC) to
dispense with English sentences entirely and translate a predicate
logic formula instead. 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.

> You wanted to know why I denied that the definition in
> question was a biconditional. And the answer is that every statement of
> predicate logic (which includes propositional logic) has a unique
> logical form, and the form of the definition in question was a universal
> quantification, not a biconditional.

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. Then I
wanted to know why you were translating the FOPL statement rather than
the English one; and your answer ("What makes you think I'm interested
in doing it?") wasn't particularly helpful.

>>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; 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).

> >> And just to be clear, this is no more my opinion than that it is my
> >> opinion that, say, no biconditional is a conjunction, or that "p v
> >> ~p" is a tautology. This is standard material covered in most any
> >> thorough text on mathematical logic.
> >
> > If it is a basic rule of mathematical logic that definition statements
> > (like "A prime number is a positive integer > 1 whose only positive
> > integer divisors are itself and 1") must be universally quantified
> > before being translated into propositional logic (PC) -
>
> Huh, wha? Rule? 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.

> > 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, 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.

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).

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

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

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

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

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

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

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

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

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

> > 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.

.

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