Hello,
I am trying gather information about first-order predicate logic and
syllogisms in the hopes of using predicate logic to explain syllogisms. I
know that I could use basic catergorical propositions to detail syllogisms
but I was wondering if I could use the more powerul predicate logic to
explain syllogisms.
If anybody has any books or web sites that would help me, please let me
know.
Thank you.
Simple minds may discover simple solutions — but solutions nevertheless!
Puerile minds may find puerile solutions — but what if they are solutions
nevertheless?!
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or
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I would be grateful of any help with the following problem:
Prove |- A V ¬A using the laws of natural deduction.
(Notation: |- = Turnstile, V = OR, ¬ = NOT)
Thanks
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Me and my friend have had a discussion about whether a proof derivation can
be infinite or not.
To be more explicit: Can a formal derivation in first order logic be
infinite?
For example:
If I have G |- P can I then conclude,
since a derivation only contains a finite number of formulas (which is the
question),
that G must be a finite set.
(Notation: |- = Turnstile)
Thanks Michael Gustafsson
THE PARADOXES OF PREDICATION
A predicate is on the level of language, a property on the level of
the world. For example, the concept "is red" is a predicate and is a
part of our language, while things in the world have the property of
redness. Because there seems to be a different relationship between
referring concept and referent than between predicate and property, we
say that the predicate "is red" *gives* (rather than *refers to*) the
property of redness. While we will say a thing *has* a property, we
say a thing *satisfies* a predicate.
Now consider the predicate
[C1] "Does not have oneself",
which predicates of properties. Reason into a paradox in the
following way. Let P be the property given by C1. We would say that
the property shortness has P, because shortness is not short (only
material objects are short). On the other hand, the property of being
a property does not have P. Does P have P? If it does, then it does
not have itself; and if it does not, then it has itself; hence a
paradox.
However, one has made an assumption in this argument, namely that C1
gives a property at all, or more generally that every one-place
predicate gives a property. For if C1 does not give a property, then
there is no P, and so no P to have or not to have P. In brief, by
*reductio ad absurdam*, we must conclude that C1 does not give any
property. And so the paradox disappears.
But, if not all one-place predicates give (unique) properties, then
which do? Should we not give some sort of criteria to help us
determine which do and which do not? Certainly this would be nice,
but there is no reason to suppose that it can be done. Quite possibly
there are no rules, based on our language, which determine whether a
particular predicate gives or does not give a property. After all,
there all no such rules which determine whether a thing exists
("unicorn" obeys our language rules just as well as "giraffe").
Indeed, it would be surprising if, by a mere fact of language, we
could be assured that certain properties exist.
Still, while the paradox is admittedly stymied by saying that C1 does
not give a property, this is not a complete explanation. After all,
there certainly is a one-place predicate, and any assertion about
properties can be reformulated in terms of predicates. For the person
who wants to understand "predicate" when he hears ‘property’ (and
"satisfies" when he hears ‘has’), C1 certainly does give a "property"
(itself!). So, if we are truly to provide a solution, we must also
solve the cousin paradox which uses only the concepts of predicate and
satisfaction.
I want to explain my opinion about translating a given statement
into a formula (well-formed formula).
and I want to know whether I am doing a mistake or not.
and is my interpretation right?
firstly, let me give some notations.
V_x : is read as "for all x"
E_x : is read as "there exists an x"
—> : (if …then)
<—>: iff
Suppose we are interested in the statement "There exist two
distinct elements of the array A that sum is equal to 0."
which one(s) of the following well-formed formulas do you suggest
I should use for the statement and why?
1) E_i E_j [ A(i)+A(j)=0]
for the array A={1,3,0,4}, let i and j be 3,
then, the predicate [ A(i)+ A(j)]=0] will be "true". Since the
array
A is not consist of two distinct elements that sum is equal to 0.,
I think that the given statement can not be represented by this
formula.
2) E_i E_j [ {i is not equal j } ---> {A(i)+A(j)=0}]
for the array A={1,3,0,4}, let i and j be 3,
then, the predicate [{i is not equal j} --->{A(i)+A(j)=0}] will be
"true".
[{3is not equal 3} --> {0+0=0}]
[ false ---> true] = true
because of the same reason, the statement can also not be
represented
by this formula.
3) E_i E_j [ {i is not equal j} and {A(i)+A(j)=0}]
for the array A={1,3,0,4}, let i and j be 3,
then, the predicate will be "false". [false and true]= false
but for the array A={-1,3,0,1}, let i be 1 and j be 4.
Now, the predicate will be "true".
therefore,we can use this formula for the given statement.
4) E_i E_j [{i is not equal j} <--> {A(i)+A(j)=0}]
I think this formula can be also used.
Mustafa EGE
e…@eti.cc.hun.edu.tr
THE PARADOXES OF PREDICATION (cntd.)
Like referring concepts and facts, whether a thing satisfies a
predicate may be determined by whether it satisfies another predicate.
For instance, consider
[C2] "satisfies the predicate C3"
[C3] "is red"
Red things satisfy C2; but if C3 had been a different predicate, then
they might not have. As with referring concepts, we can create
endless chains, and vicious predicates, such as:
[C4] "satisfies the predicate C4"
C4 is vicious, since to determine whether a thing satisfies it, we
must know whether the thing satisfies it. And thus we must conclude
that nothing ever satisfies it.
Now consider:
[C5] "Does not satisfy the predicate C5".
If a thing does not satisfy the predicate C5, then it seems to satisfy
it; and if it does satisfy it, then it seems not to. We have another
paradox.
Hopefully this reminds the reader of the Liar’s Paradox, and indeed
our solution will follow the same lines. Call the schema
x satisfies the predicate "P" if and only if Px
*Tarski’s Satisfaction Principle*, or TSP for short. For instance,
snow satisfies the predicate "is white" if and only if
snow is white.
As with TTP, we assert that TSP cannot be universally correct, that C5
is a counter-example, and that there *cannot* be a concept "plok" such
that
x ploks the predicate "P" if and only if Px.
We would explain this as we did with ‘truth’. For instance, if we
consider "satisfies" as an abbreviation, it is well-defined precisely
in situations where it can be made redundant. In
"satisfies the predicate C5",
‘satisfies’ cannot be eliminated, so the predicate is vicious and
*nothing* satisfies it. In particular,
C5 does not satisfy the predicate C5.
Because we do not universally accept TSP, we cannot infer that C5
satisfies C5, and so there is no contradiction.
Now consider
[C6] "does not satisfy oneself".
Clearly, C6 is vicious when applied with itself, i.e. C6 is vicious in
"C6 satisfies C6". So C6 does not satisfy C6. On the other hand, "is
not a predicate" satisfies C6. So viciousness may depend on the
argument to which the predicate applies.
Formally, the assertion
[C7] (x)(Sat(P,x) iff Px),
where Sat(P,x) means "x satisfies P", is fallacious for certain P, for
instance C5 or ~Sat(x,x).
Remark that here we are essentially using a logic which does not
distinguish between predicate and non-predicate thing variables. That
is,
(x)(Sat(P,x) iff ~Sat(x,x))
is contradictory, because by substituting a big letter for a little,
‘P’ for ‘x’, one can infer that
Sat(P,P) iff ~Sat(P,P).
This is normal, because there is no reason to distinguish between
predicates and other things (as is done in "normal" second-order
logic). After all, one can refer to the predicate "is red" just as
well as one can refer to snow.
Now rewrite ‘Sat(P,x)’ as ‘P[x]‘. Then C7 becomes
[C8] (x)(P[x] iff Px).
Next introduce, for every predicate P, a term in the language ‘{y |
Py}’, which is to be read "the predicate of those y such that Py."
Then C8 becomes
[C9] (x)({y | Py}[x] iff Px).
Finally, instead of writing Px and Py, let phi be any one-place
predicate whose free variable is x (and no y variable), and let phi*
be phi with y replaced for every free occurrence of x. Then C9
becomes
[C10] (x)({y | phi*}[x] iff phi).
C10 is just a manner of writing TSP, and so it is fallacious. For
instance, phi as ‘~x[x]‘ leads to a contradiction.
In order for C10 to hold, we must restrict phi to certain predicates,
where all uses of "satisfies" can be made redundant. A simple rule
depending only on syntax is preferred, even if it is too restrictive,
i.e. it not only forbids all phi for which C10 does not hold, but also
some for which it does.
Now "satisfies" cannot in general be made redundant if there is a
variable in the predicate place, on the left-handside of "[". So the
rule is: only constant symbols are allowed in this place.
For instance, the (universal) predicate {y | y = y} is such that
(x)({y | y = y}[x] iff x = x).
On the other hand, we cannot assert that
(x)({y | ~y[y]}[x] iff ~x[x]),
since the variable ‘x’ appears in the predicate place. (There is
still, of course, a predicate {y | ~y[y]}. It is only that it fails
TSP and, so to speak, is not a "well-behaved" predicate.)
Let ‘N’ stand for "is a natural number." Then
(x)({y | N[y] & ~y = 0}[x] iff N[x] & ~x = 0)
is a correct application of C10, since only the constant ‘N’ appears
in the predicate place.
On the other hand,
(x)({y | (z)(z = N => z[y] & ~y = 0)}[x]
iff (z)(z = N => z[x] & ~x = 0))
is not, because the variable ‘z’ is in the predicate place. As we
said, our rule may be too restrictive, since (z)(z = N => z[x] & ~x =
0) is equivalent to N[x] & ~x = 0 and so there is arguably nothing
wrong with it. But nothing is lost–the logical system is not
weakened–, because, if need be, it is possible in any case simply to
assert that
(x)({y | N[y] & ~y = 0}[x] iff N[x] & ~x = 0),
which again is a correct application of C10.
Finally, there is admittedly one subtlety in all this. Even certain
constants in the predicate place cannot be made redundant, e.g. C5 or
{z | ~z[z]}. For instance, not
(x)({y | ~C5[y]}[x] iff ~C5[x]),
since after all {y | ~C5[y]} = C5. Obviously, then, such constants
must not be allowed. So, for instance, primitive predicate constants
(such as "N," if one is doing arithmetic) must be reducible to a
predicate without "satisfies" (e.g. "N[x]" reduces to "x is a natural
number").
All this scales up to n-place predicates. The technical solution to
the paradoxes, in its full generality, can be stated as follows. Let
x and y be vectors of n arguments. Let phi contain x as free variables
(and no y variables), and let phi* be phi with the y substituted in
for the x. Then
(x)({y | phi*}[x] iff phi)
provided phi does not contain any variables in its predicate places.
Hi,
Two persons, A and B are running a race in the
same direction. The distance between A and B is 1
km with B ahead. A’s speed is 1 km/hr and B’s
speed is half of A’s i.e. 1/2 km/hr.
Now when A reaches B’s position, B would have
covered some distance. When again A covers this
distance B will again have covered some more
distance.When again A ……..
Thus this means that there will always be a
finite distance between A and B. So this menas
that A will never catch up with B.
But we know that since B’s speed is less than
A’s,eventually A will catch up with B.So what’s
wrong with the logical reasoning above?
thanks,
jay.
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