Re: What isn't a tautology?



Chris Menzel wrote:
> On 21 Jul 2005 06:52:59 -0700, George Dance <georgedance04@xxxxxxxx> said:
> Chris Menzel wrote:
>
['silly sticking point' snipped]

> 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".
>
> 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.
>
> 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.

Perhaps; but not without one kick at the can. Fa v ~Fa is a tautology;
so it's true of everything; so, it needs to be quantified as (x)(Fa v
~Fa) "to capture the logical form of the example"; at which point, by
your reasoning, it would turn out to not be a tautology after all.

> >> > 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.

Yes, it was; I should have said "always" have the same truth value.

> Do you really think that,
> say, "Bush is President" and "2 is prime" are logically equivalent just
> because they are both true?

Obviously not; they weren't both true (or both false) in 1999, and
won't be both true (or both false) in 2009.

> 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.

And which I do agree with: "roughly ... a sentence S of predicate
logic is a tautology if there is [a] 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."

> 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.

That's an interesting counter-argument to 1:

1'. "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." [A->B]
2'. "There's absolutely no way to replace subformulas with
propositional constants in a quantified sentence to yield a PC
tautology." [~A]
--------------
3'. "no universally quantified sentence is a tautology." [~B]

See anything wrong with it? (hint: I don't disagree with any of your
premises)

> > 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,

True; one could transfer every sentence of a predicate logic proof, eg,
as P, Q, R etc. Read that as 'a propositional form.'

> but I take this to be an informal way of
> expressing, or at least agreeing with, the notion of tautologousness > in question.

Yes.

> > 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.

No, not just because both are logical truths; they're logically
equivalent for the same reason that ~(Ex)(Px & ~Qx) and ~(Pa & ~Qa),
neither of which are logical truths, are logically equivalent.

> > 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.

I don't think your counterargument (as written) actually refutes 1.

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

> >>
> >> 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.

I don't see the above as being a 'definition' at all (notice that it's
not a biconditional), but a method for identifying tautologies. I'm
not disagreeing with that, but arguing that there's another method:
that a sentence S of predicate logic is a tautology of propositional
logic if there is both (a) a sentence S* that is logically equivalent
to S, and (b) a 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.

> 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.

No, I'm talking about the same concept, and using the same reasoning,
as I did above: that, since (x)((Px v Qx) -> (Rx -> (Px v Qx))) is
logically equivalent to ~((Pa v Qa) & (Ra & ~(Pa v Qa))), and ~((Pa v
Qa) & (Ra & ~(Pa v Qa))) is a tautology, then (x)((Px v Qx) -> (Rx ->
(Px v Qx))) is a tautology as well.

> 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.

I suspect that, if I looked on the Web, I could find one counterexample
(re how the term is used in mathematical logic. Do you see that as
worth the effort?

.

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