Discussion:
When are relations vacuously true?
(too old to reply)
Newberry
2009-09-10 16:07:49 UTC
Permalink
The formula

~(Ex)(Ey)[(x + y < 6) & (y = 8)] (1)

is vacuously true. The formula (2)

~(Ex)(x + 8 < 6)

is also true, but NOT vacuously true.

* * * * *
To see that (1) is vacuously true let us pick y = 8:

~(Ex)[(x + 8 < 6) & (8 = 8)]

This is equivalent to

(x)[(x + 8 < 6) -> ~(8 = 8)]

which is vacuously true.

* * * * *
Let's pick some y # 8, say, y = 4:

~(Ex)[(x + 4 < 6) & (4 = 8)]

This is equivalent to

(x)[(4 = 8) -> ~(x + 4 < 6)]

which is vacuously true
.
* * * * *
It is apparent that for any choice of y the corresponding sentence
will be vacuously true, hence (1) is vacuously true. This has some
interesting consequences.

http://www.scribd.com/doc/19206866/Vacuous-Relations

Comments appreciated.
Dodge
2009-09-10 16:23:30 UTC
Permalink
Post by Newberry
The formula
~(Ex)(Ey)[(x + y < 6) & (y = 8)] (1)
is vacuously true. The formula (2)
OK let's translate this from blither into English.

There do not exist x and y such that the sum of x and y is less than 6 AND y
=8.

Now let's substitute y:

x+8 < 6

and simplify

x < -2

So apparently "there do not exist the number x < -2. Thus proving
faster-than-light travel and Fermat's last theorem!!!!!!!

Then we notice that the poster is a google-poster! A google-poster! Who
can't possibly wrap his tiny brain around the concept that the sum of two
numbers can exist, and actually be less than 6 (imagine!), and even more,
that one of the numbers is given (8). The number 8 boggles the mind! The
sum confuses! The value of 6 causes fainting spells and nausea! How can it
possibly be?!?!? How can there not exist numbers? Let's cause a
meltdown - let's introduce a THIRD number. Let's say -3. Now we have -3+8
< 6, or 5 < 6. But, by the original statement, they don't exist. Does 8
not exist? Does -3 not exist? Then, of course, we remember. It's a
google-poster! Who doesn't exist! Whoes? Who's? What? Numbers?
Musatov? Archimedes Plutonium? Mathforum? Insanity Sean Hannity squared!
And factorialed! By rabid sea urchins! On Mars! In a gift-wrapped box!
Arturo Magidin
2009-09-10 16:39:45 UTC
Permalink
Post by Newberry
The formula
~(Ex)(Ey)[(x + y < 6) & (y = 8)]                                (1)
is vacuously true.
Ehr... it's true if we are considering only positive (or nonnegative)
integers. But I would *not* say it is "vacuously true"!

'Vacuously true' is usually used to refer to an implication in which
the antecedent is always false. Now, granted, one can translate this
statement into a universal statement with an implication; you get
either

For all x, for all y, if x+y < 6 then y=/=8;

or

For all x, for all y, if y=8, then x+y >= 6.

Neither of those is "vacuously true" in the world of nonnegative
integers, because in both cases, there *are* instances of x and y that
make the antecedent true.

This, as opposed to something like

For all x, for all y, if x^2 + y^2 < 0, then x=y

which *would be* vacuously true when considering x and y nonnegative
integers (or integers; or rationals; or real numbers). The reason
being that the antecedent, "x^2 + y^2 <0", is always false. That is,
the collection of all x and y that satisfy the antecedent is *empty*,
vacuous; that is why we say it is "vacuously true". Compare with the
two statements above: in both 'translations', the set of x and y that
satisfy the antecedent is *not* empty, so the statements, while true,
are not *vacuously* true.
Post by Newberry
The formula                                   (2)
~(Ex)(x + 8 < 6)
is also true, but NOT vacuously true.
Correct; not even an implication.

By the way: though you are correct in calling them "formulas", it is
more common to call them "sentences", as they have are logical
expressions with no free variables.
Post by Newberry
     * * * * *
Sorry, doesn't work that way.... At least, I know nobody who would
call it a "vacuously true" sentence.
Post by Newberry
     * * * * *
It is apparent that for any choice of y the corresponding sentence
will be vacuously true, hence (1) is vacuously true.
That's not what anyone I know means by "vacuously true" for a
universal sentence with more than one universally quantified
variable...

--
Arturo Magidin
Newberry
2009-09-10 19:21:52 UTC
Permalink
Post by Arturo Magidin
Post by Newberry
The formula
~(Ex)(Ey)[(x + y < 6) & (y = 8)]                                (1)
is vacuously true.
Ehr... it's true if we are considering only positive (or nonnegative)
integers. But I would *not* say it is "vacuously true"!
'Vacuously true' is usually used to refer to an implication in which
the antecedent is always false. Now, granted, one can translate this
statement into a universal statement with an implication; you get
either
For all x, for all y, if x+y < 6 then y=/=8;
or
For all x, for all y, if y=8, then x+y >= 6.
Neither of those is "vacuously true" in the world of nonnegative
integers, because in both cases, there *are* instances of x and y that
make the antecedent true.
This, as opposed to something like
For all x, for all y, if x^2 + y^2 < 0, then x=y
which *would be* vacuously true when considering x and y nonnegative
integers (or integers; or rationals; or real numbers). The reason
being that the antecedent, "x^2 + y^2 <0", is always false. That is,
the collection of all x and y that satisfy the antecedent is *empty*,
vacuous; that is why we say it is "vacuously true". Compare with the
two statements above: in both 'translations', the set of x and y that
satisfy the antecedent is *not* empty, so the statements, while true,
are not *vacuously* true.
Post by Newberry
The formula                                   (2)
~(Ex)(x + 8 < 6)
is also true, but NOT vacuously true.
Correct; not even an implication.
By the way: though you are correct in calling them "formulas", it is
more common to call them "sentences", as they have are logical
expressions with no free variables.
Post by Newberry
     * * * * *
Sorry, doesn't work that way....
Why not?
Post by Arturo Magidin
At least, I know nobody who would
call it a "vacuously true" sentence.
Post by Newberry
     * * * * *
It is apparent that for any choice of y the corresponding sentence
will be vacuously true, hence (1) is vacuously true.
That's not what anyone I know means by "vacuously true" for a
universal sentence with more than one universally quantified
variable...
--
Arturo Magidin
Arturo Magidin
2009-09-10 22:38:40 UTC
Permalink
Post by Newberry
Post by Arturo Magidin
Post by Newberry
The formula
~(Ex)(Ey)[(x + y < 6) & (y = 8)]                                (1)
is vacuously true.
[...]
Post by Newberry
Post by Arturo Magidin
Post by Newberry
     * * * * *
Sorry, doesn't work that way....
Why not?
It just doesn't. Although the meaning of "vacuously true" is not
formal, it usually refers to universal sentences that involve
implications, and where the collection of all constants for which the
antecedent is true is empty.

So if you have a sentence of the form (For all x)(For all y) (P(x,y)--
Post by Newberry
Q(x,y)), then the standard meaning of "the sentence is vacuously
true" is that the the set {(x,y) : P(x,y)} is empty.

What you are doing instead is considering two families of
propositions: the first family of propositions is

R_y(x): (For all x) (P(x,y) -->Q(x,y))

and the second family of propositions is

S_y(x): (For all x)(not(Q(x,y)) --> not(P(x,y))).


Then you are saying that because for each constant y_0, EITHER R_{y_0}
(x) is vacuously true or S_{y_0}(x) is vacuously true, then you
conclude that the original sentence is vacuously true.

But note that neither R_y(x) is vacuously true for ALL y, NOR is S_y
(x) vacuously true for ALL y.

Now, granted, not every sentence of the form

(forall x)(forall y) (P(x,y) --> Q(x,y))

has this property (there are sentences in which for some constants
y_0, the truth set of P(x,y_0) is neither empty nor the entire model,
and likewise for Q(x,y_0) ). But that is not the property usually
associated with the phrase "vacuously true".

So it "doesn't work that way" because you are taking a common phrase
and using it in an uncommon way. You can define it that way if you
want, but remember that even if you call the tail a leg, a cat still
has four legs.

--
Arturo Magidin
Newberry
2009-09-10 23:38:51 UTC
Permalink
Post by Newberry
Post by Arturo Magidin
Post by Newberry
The formula
~(Ex)(Ey)[(x + y < 6) & (y = 8)]                                (1)
is vacuously true.
   [...]
Post by Newberry
Post by Arturo Magidin
Post by Newberry
     * * * * *
Sorry, doesn't work that way....
Why not?
It just doesn't. Although the meaning of "vacuously true" is not
formal, it usually refers to universal sentences that involve
implications, and where the collection of all constants for which the
antecedent is true is empty.
So if you have a sentence of the form  (For all x)(For all y) (P(x,y)-->Q(x,y)), then the standard meaning of "the sentence is vacuously
true" is that the the set {(x,y) : P(x,y)} is empty.
What you are doing instead is considering two families of
propositions: the first family of propositions is
R_y(x):   (For all x) (P(x,y) -->Q(x,y))
and the second family of propositions is
S_y(x): (For all x)(not(Q(x,y)) --> not(P(x,y))).
Then you are saying that because for each constant y_0, EITHER R_{y_0}
(x) is vacuously true or S_{y_0}(x) is vacuously true, then you
conclude that the original sentence is vacuously true.
But note that neither R_y(x) is vacuously true for ALL y, NOR is S_y
(x) vacuously true for ALL y.
Now, granted, not every sentence of the form
(forall x)(forall y) (P(x,y) --> Q(x,y))
has this property (there are sentences in which for some constants
y_0, the truth set of P(x,y_0) is neither empty nor the entire model,
and likewise for Q(x,y_0) ). But that is not the property usually
associated with the phrase "vacuously true".
So it "doesn't work that way" because you are taking a common phrase
and using it in an uncommon way. You can define it that way if you
want, but remember that even if you call the tail a leg, a cat still
has four legs.
Do you agree that
(x)[(x + 8 < 6) -> ~(8 = 8)]
is vacuously true?

Do you agree that the above is equivalent to
~(Ex)[(x + 8 < 6) & (8 = 8)]

Do you agree that
(x)[(4 = 8) -> ~(x + 4 < 6)]
is vacuously true?

Do you agree that the above is equivalent to
~(Ex)[(x + 4 < 6) & (4 = 8)]

Do you agree that for ANY n
~(Ex)[(x + n < 6) & (n = 8)]
is vacuously true?

[BTW, obviously we are dealing with natural numbers only.]
Arturo Magidin
2009-09-11 00:39:17 UTC
Permalink
Post by Newberry
Post by Newberry
Post by Arturo Magidin
Post by Newberry
The formula
~(Ex)(Ey)[(x + y < 6) & (y = 8)]                                (1)
is vacuously true.
   [...]
Post by Newberry
Post by Arturo Magidin
Post by Newberry
     * * * * *
Sorry, doesn't work that way....
Why not?
It just doesn't. Although the meaning of "vacuously true" is not
formal, it usually refers to universal sentences that involve
implications, and where the collection of all constants for which the
antecedent is true is empty.
So if you have a sentence of the form  (For all x)(For all y) (P(x,y)-->Q(x,y)), then the standard meaning of "the sentence is vacuously
true" is that the the set {(x,y) : P(x,y)} is empty.
What you are doing instead is considering two families of
propositions: the first family of propositions is
R_y(x):   (For all x) (P(x,y) -->Q(x,y))
and the second family of propositions is
S_y(x): (For all x)(not(Q(x,y)) --> not(P(x,y))).
Then you are saying that because for each constant y_0, EITHER R_{y_0}
(x) is vacuously true or S_{y_0}(x) is vacuously true, then you
conclude that the original sentence is vacuously true.
But note that neither R_y(x) is vacuously true for ALL y, NOR is S_y
(x) vacuously true for ALL y.
Now, granted, not every sentence of the form
(forall x)(forall y) (P(x,y) --> Q(x,y))
has this property (there are sentences in which for some constants
y_0, the truth set of P(x,y_0) is neither empty nor the entire model,
and likewise for Q(x,y_0) ). But that is not the property usually
associated with the phrase "vacuously true".
So it "doesn't work that way" because you are taking a common phrase
and using it in an uncommon way. You can define it that way if you
want, but remember that even if you call the tail a leg, a cat still
has four legs.
Do you agree that
(x)[(x + 8 < 6) -> ~(8 = 8)]
is vacuously true?
Do you agree that the above is equivalent to
~(Ex)[(x + 8 < 6) & (8 = 8)]
Do you agree that
(x)[(4 = 8) -> ~(x + 4 < 6)]
is vacuously true?
Do you agree that the above is equivalent to
~(Ex)[(x + 4 < 6) & (4 = 8)]
Did you even *bother* to try to understand what I wrote?

As I noted before, yes, it is the case that for each constant y_0,
your sentence has the property that either

(for all x) (P(x,y_0) --> Q(x,y_0) )

is vacuously true, OR that

(for all x)(not(Q(x,y_0)) --> not(P(x,y_0)))

is vacuously true.

What I do *not* agree with is your assertion that this special
property is what is equivalent to saying that

(for all y)(for all x) (P(x,y) --> Q(x,y)) is vacuously true
Post by Newberry
Do you agree that for ANY n
~(Ex)[(x + n < 6) & (n = 8)]
is vacuously true?
No. And I thought I was very clear about WHY I do not agree with
that. I explained very clearly what it is that, in my experience, is
refered to when one says a proposition is "vacuously true", and I also
explained very clearly why the sentence that *you* claim is "vacuously
true" does not have the refered to property.

Do you disagree with the fact that your sentence does *not* have the
property I outlined? That of being a universal statement in the form
of an implication in which the antecedent is always false ? There are
many ways of translating your two-universally-quantified-variable
sentence into a sentece of that form, but *none* of those translations
has the refered to property. Instead, you are trying to extend the
informal terminology to a class of sentences that is not normally
considered to be covered under that terminology.

Simply put, while everyone agrees that "Z" means either "for all x P
(x)" or "for all x Q(x)" , you want it to *also* mean "for all x (P(x)
or Q(x))", and the two are not equivalent (though one implies the
other).

Again, you are free to call the situation you are describing anything
you want, but calling the tail a leg does not make it a leg.

--
Arturo Magidin
Newberry
2009-09-11 02:15:33 UTC
Permalink
Post by Arturo Magidin
Post by Newberry
Post by Newberry
Post by Arturo Magidin
Post by Newberry
The formula
~(Ex)(Ey)[(x + y < 6) & (y = 8)]                                (1)
is vacuously true.
   [...]
Post by Newberry
Post by Arturo Magidin
Post by Newberry
     * * * * *
Sorry, doesn't work that way....
Why not?
It just doesn't. Although the meaning of "vacuously true" is not
formal, it usually refers to universal sentences that involve
implications, and where the collection of all constants for which the
antecedent is true is empty.
So if you have a sentence of the form  (For all x)(For all y) (P(x,y)-->Q(x,y)), then the standard meaning of "the sentence is vacuously
true" is that the the set {(x,y) : P(x,y)} is empty.
What you are doing instead is considering two families of
propositions: the first family of propositions is
R_y(x):   (For all x) (P(x,y) -->Q(x,y))
and the second family of propositions is
S_y(x): (For all x)(not(Q(x,y)) --> not(P(x,y))).
Then you are saying that because for each constant y_0, EITHER R_{y_0}
(x) is vacuously true or S_{y_0}(x) is vacuously true, then you
conclude that the original sentence is vacuously true.
But note that neither R_y(x) is vacuously true for ALL y, NOR is S_y
(x) vacuously true for ALL y.
Now, granted, not every sentence of the form
(forall x)(forall y) (P(x,y) --> Q(x,y))
has this property (there are sentences in which for some constants
y_0, the truth set of P(x,y_0) is neither empty nor the entire model,
and likewise for Q(x,y_0) ). But that is not the property usually
associated with the phrase "vacuously true".
So it "doesn't work that way" because you are taking a common phrase
and using it in an uncommon way. You can define it that way if you
want, but remember that even if you call the tail a leg, a cat still
has four legs.
Do you agree that
(x)[(x + 8 < 6) -> ~(8 = 8)]
is vacuously true?
Do you agree that the above is equivalent to
~(Ex)[(x + 8 < 6) & (8 = 8)]
Do you agree that
(x)[(4 = 8) -> ~(x + 4 < 6)]
is vacuously true?
Do you agree that the above is equivalent to
~(Ex)[(x + 4 < 6) & (4 = 8)]
Did you even *bother* to try to understand what I wrote?
As I noted before, yes, it is the case that for each constant y_0,
your sentence has the property that either
(for all x) (P(x,y_0) --> Q(x,y_0) )
is vacuously true, OR that
(for all x)(not(Q(x,y_0)) --> not(P(x,y_0)))
is vacuously true.
What I do *not* agree with is your assertion that this special
property is what is equivalent to saying that
(for all y)(for all x) (P(x,y) --> Q(x,y)) is vacuously true
Post by Newberry
Do you agree that for ANY n
~(Ex)[(x + n < 6) & (n = 8)]
is vacuously true?
No.  And I thought I was very clear about WHY I do not agree with
that. I explained very clearly what it is that, in my experience, is
refered to when one says a proposition is "vacuously true", and I also
explained very clearly why the sentence that *you* claim is "vacuously
true" does not have the refered to property.
Do you disagree with the fact that your sentence does *not* have the
property I outlined? That of being a universal statement in the form
of an implication in which the antecedent is always false ?
Do you agree that for any n the formula is a universal statement in
the form of an implication in which the antecedent is always false?
Post by Arturo Magidin
There are
many ways of translating your two-universally-quantified-variable
sentence into a sentece of that form, but *none* of those translations
has the refered to property. Instead, you are trying to extend the
informal terminology to a class of sentences that is not normally
considered to be covered under that terminology.
Simply put, while everyone agrees that "Z" means either "for all x P
(x)" or "for all x Q(x)" , you want it to *also* mean "for all x (P(x)
or Q(x))", and the two are not equivalent (though one implies the
other).
Again, you are free to call the situation you are describing anything
you want, but calling the tail a leg does not make it a leg.
Actually I could not care less what you call the property I am
describing. My point is that

~(Ex)(Ey)[(x + y < 6) & (y = 8)]

has a property X, which

~(Ex)(x + 8 < 6)

does not.
Post by Arturo Magidin
--
Arturo Magidin- Hide quoted text -
- Show quoted text -
Jesse F. Hughes
2009-09-11 10:08:58 UTC
Permalink
Post by Newberry
Actually I could not care less what you call the property I am
describing. My point is that
~(Ex)(Ey)[(x + y < 6) & (y = 8)]
has a property X, which
~(Ex)(x + 8 < 6)
does not.
Out of curiosity, do you plan on claiming that the former statement is
neither true nor false because statements with property X are neither
true nor false?
--
Jesse F. Hughes
"Ultimately, I can bring the entire mathematical establishment to its
knees... Live in a fantasy world if you wish, but to me that's just
an expression of your intellectual inferiority." --James Harris
Newberry
2009-09-11 16:36:24 UTC
Permalink
Post by Jesse F. Hughes
Post by Newberry
Actually I could not care less what you call the property I am
describing. My point is that
~(Ex)(Ey)[(x + y < 6) & (y = 8)]
has a property X, which
~(Ex)(x + 8 < 6)
does not.
Out of curiosity, do you plan on claiming that the former statement is
neither true nor false because statements with property X are neither
true nor false?
Onviously. P -> Q is truth-functionally equivalent to ~(P & ~Q). It
does not matter if you restrict the term to expressions with arows.
Newberry
2009-09-11 18:38:20 UTC
Permalink
Post by Jesse F. Hughes
Post by Newberry
Actually I could not care less what you call the property I am
describing. My point is that
~(Ex)(Ey)[(x + y < 6) & (y = 8)]
has a property X, which
~(Ex)(x + 8 < 6)
does not.
Out of curiosity, do you plan on claiming that the former statement is
neither true nor false because statements with property X are neither
true nor false?
BTW, Strawson makes it explicit that he considers

(x)(Fx -> Gx)

and

~(Ex)(Fx & ~Gx)

the same thing. I.e. they are both neither true nor false if

~(Ex)Fx

[Actually he does not use formalization but English.]
Jesse F. Hughes
2009-09-11 21:23:19 UTC
Permalink
Post by Newberry
Post by Jesse F. Hughes
Post by Newberry
Actually I could not care less what you call the property I am
describing. My point is that
~(Ex)(Ey)[(x + y < 6) & (y = 8)]
has a property X, which
~(Ex)(x + 8 < 6)
does not.
Out of curiosity, do you plan on claiming that the former statement is
neither true nor false because statements with property X are neither
true nor false?
BTW, Strawson makes it explicit that he considers
(x)(Fx -> Gx)
and
~(Ex)(Fx & ~Gx)
the same thing. I.e. they are both neither true nor false if
~(Ex)Fx
[Actually he does not use formalization but English.]
You didn't answer my question.

Oh, and to be sure, those comments about Strawson are irrelevant in
this case. As you know, (Ey)(x + y < 6) and (Ey)(y = 8) are both
true.
--
"After years of arguing I realize that your intellects are too limited
to fully grasp my work. [...] Still, no matter how child-like your
minds are, [...] since you have language, [...] there's a chance that
I'll be able to find something that your minds can handle." --JSH
©2009 Martin Musatov. All Rights Reserved In Perpetuity.
2009-09-11 21:43:17 UTC
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* Y8 (redirect from Y-8)
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to ...
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Arturo Magidin
2009-09-11 16:10:29 UTC
Permalink
Post by Newberry
Post by Arturo Magidin
Post by Newberry
Post by Newberry
Post by Arturo Magidin
Post by Newberry
The formula
~(Ex)(Ey)[(x + y < 6) & (y = 8)]                                (1)
is vacuously true.
   [...]
Post by Newberry
Post by Arturo Magidin
Post by Newberry
     * * * * *
Sorry, doesn't work that way....
Why not?
It just doesn't. Although the meaning of "vacuously true" is not
formal, it usually refers to universal sentences that involve
implications, and where the collection of all constants for which the
antecedent is true is empty.
So if you have a sentence of the form  (For all x)(For all y) (P(x,y)-->Q(x,y)), then the standard meaning of "the sentence is vacuously
true" is that the the set {(x,y) : P(x,y)} is empty.
What you are doing instead is considering two families of
propositions: the first family of propositions is
R_y(x):   (For all x) (P(x,y) -->Q(x,y))
and the second family of propositions is
S_y(x): (For all x)(not(Q(x,y)) --> not(P(x,y))).
Then you are saying that because for each constant y_0, EITHER R_{y_0}
(x) is vacuously true or S_{y_0}(x) is vacuously true, then you
conclude that the original sentence is vacuously true.
But note that neither R_y(x) is vacuously true for ALL y, NOR is S_y
(x) vacuously true for ALL y.
Now, granted, not every sentence of the form
(forall x)(forall y) (P(x,y) --> Q(x,y))
has this property (there are sentences in which for some constants
y_0, the truth set of P(x,y_0) is neither empty nor the entire model,
and likewise for Q(x,y_0) ). But that is not the property usually
associated with the phrase "vacuously true".
So it "doesn't work that way" because you are taking a common phrase
and using it in an uncommon way. You can define it that way if you
want, but remember that even if you call the tail a leg, a cat still
has four legs.
Do you agree that
(x)[(x + 8 < 6) -> ~(8 = 8)]
is vacuously true?
Do you agree that the above is equivalent to
~(Ex)[(x + 8 < 6) & (8 = 8)]
Do you agree that
(x)[(4 = 8) -> ~(x + 4 < 6)]
is vacuously true?
Do you agree that the above is equivalent to
~(Ex)[(x + 4 < 6) & (4 = 8)]
Did you even *bother* to try to understand what I wrote?
As I noted before, yes, it is the case that for each constant y_0,
your sentence has the property that either
(for all x) (P(x,y_0) --> Q(x,y_0) )
is vacuously true, OR that
(for all x)(not(Q(x,y_0)) --> not(P(x,y_0)))
is vacuously true.
What I do *not* agree with is your assertion that this special
property is what is equivalent to saying that
(for all y)(for all x) (P(x,y) --> Q(x,y)) is vacuously true
Post by Newberry
Do you agree that for ANY n
~(Ex)[(x + n < 6) & (n = 8)]
is vacuously true?
No.  And I thought I was very clear about WHY I do not agree with
that. I explained very clearly what it is that, in my experience, is
refered to when one says a proposition is "vacuously true", and I also
explained very clearly why the sentence that *you* claim is "vacuously
true" does not have the refered to property.
Do you disagree with the fact that your sentence does *not* have the
property I outlined? That of being a universal statement in the form
of an implication in which the antecedent is always false ?
Do you agree that for any n the formula is a universal statement in
the form of an implication in which the antecedent is always false?
Are you planning to actually address what I wrote, instead of just
repeating the same thing over and over in the hope that I will
eventually either tire of it or agree with you? I outlined my
reasoning twice now, you've ignored it twice now and simply repeated
your original claim.

I agree that for *each* n there is a way to translate the sentence
into an equivalent sentence which happens to be an implication in
which the antecedent is false.

However, for each n, the "translation" is to a DIFFERENT sentence.
Moreover, there is no single sentence with one free variable which is
both (i) equivalent to your for every n; and (ii) vacuously true for
every n.


Because of that, once you quantify the n, you end up with a sentence
that is *not* vacuously true.

At a very basic level, are saying:

(*) (For all n)(Exists S) V(S,n)

No problem there.

Unfortunately, the usual meaning of "vacuously true" would be that

(**) (Exists S)(For all n) V(S,n).

(*) and (**) are *not* equivalent. You have an instance of (*), but
want to claim it as an instance of (**).
Post by Newberry
Post by Arturo Magidin
There are
many ways of translating your two-universally-quantified-variable
sentence into a sentece of that form, but *none* of those translations
has the refered to property. Instead, you are trying to extend the
informal terminology to a class of sentences that is not normally
considered to be covered under that terminology.
Simply put, while everyone agrees that "Z" means either "for all x P
(x)" or "for all x Q(x)" , you want it to *also* mean "for all x (P(x)
or Q(x))", and the two are not equivalent (though one implies the
other).
Again, you are free to call the situation you are describing anything
you want, but calling the tail a leg does not make it a leg.
Actually I could not care less what you call the property I am
describing.
Yes, I can see that. You prefer to talk in your own private language
and then complain when people tell you the words you are using don't
usually mean what you are using them to mean.
Post by Newberry
My point is that
~(Ex)(Ey)[(x + y < 6) & (y = 8)]
has a property X, which
~(Ex)(x + 8 < 6)
does not.
Your "point" seems to be that you want "vacuously true" to refer to a
lot of sentences that are not normally called "vacuously true". As I
pointed out, "vacuously true" is an *informal* term, so I can
certainly not prove to you formally that you are incorrect. The best I
can do is tell you, as I have done three times now, that in my
personal experience very few, if any, mathematician would agree with
you, for the reasons I outlined. But since you couldn't care less, I
guess there's no problem.

--
Arturo Magidin
Newberry
2009-09-11 16:42:25 UTC
Permalink
Post by Arturo Magidin
Post by Newberry
Post by Arturo Magidin
Post by Newberry
Post by Newberry
Post by Arturo Magidin
Post by Newberry
The formula
~(Ex)(Ey)[(x + y < 6) & (y = 8)]                                (1)
is vacuously true.
   [...]
Post by Newberry
Post by Arturo Magidin
Post by Newberry
     * * * * *
Sorry, doesn't work that way....
Why not?
It just doesn't. Although the meaning of "vacuously true" is not
formal, it usually refers to universal sentences that involve
implications, and where the collection of all constants for which the
antecedent is true is empty.
So if you have a sentence of the form  (For all x)(For all y) (P(x,y)-->Q(x,y)), then the standard meaning of "the sentence is vacuously
true" is that the the set {(x,y) : P(x,y)} is empty.
What you are doing instead is considering two families of
propositions: the first family of propositions is
R_y(x):   (For all x) (P(x,y) -->Q(x,y))
and the second family of propositions is
S_y(x): (For all x)(not(Q(x,y)) --> not(P(x,y))).
Then you are saying that because for each constant y_0, EITHER R_{y_0}
(x) is vacuously true or S_{y_0}(x) is vacuously true, then you
conclude that the original sentence is vacuously true.
But note that neither R_y(x) is vacuously true for ALL y, NOR is S_y
(x) vacuously true for ALL y.
Now, granted, not every sentence of the form
(forall x)(forall y) (P(x,y) --> Q(x,y))
has this property (there are sentences in which for some constants
y_0, the truth set of P(x,y_0) is neither empty nor the entire model,
and likewise for Q(x,y_0) ). But that is not the property usually
associated with the phrase "vacuously true".
So it "doesn't work that way" because you are taking a common phrase
and using it in an uncommon way. You can define it that way if you
want, but remember that even if you call the tail a leg, a cat still
has four legs.
Do you agree that
(x)[(x + 8 < 6) -> ~(8 = 8)]
is vacuously true?
Do you agree that the above is equivalent to
~(Ex)[(x + 8 < 6) & (8 = 8)]
Do you agree that
(x)[(4 = 8) -> ~(x + 4 < 6)]
is vacuously true?
Do you agree that the above is equivalent to
~(Ex)[(x + 4 < 6) & (4 = 8)]
Did you even *bother* to try to understand what I wrote?
As I noted before, yes, it is the case that for each constant y_0,
your sentence has the property that either
(for all x) (P(x,y_0) --> Q(x,y_0) )
is vacuously true, OR that
(for all x)(not(Q(x,y_0)) --> not(P(x,y_0)))
is vacuously true.
What I do *not* agree with is your assertion that this special
property is what is equivalent to saying that
(for all y)(for all x) (P(x,y) --> Q(x,y)) is vacuously true
Post by Newberry
Do you agree that for ANY n
~(Ex)[(x + n < 6) & (n = 8)]
is vacuously true?
No.  And I thought I was very clear about WHY I do not agree with
that. I explained very clearly what it is that, in my experience, is
refered to when one says a proposition is "vacuously true", and I also
explained very clearly why the sentence that *you* claim is "vacuously
true" does not have the refered to property.
Do you disagree with the fact that your sentence does *not* have the
property I outlined? That of being a universal statement in the form
of an implication in which the antecedent is always false ?
Do you agree that for any n the formula is a universal statement in
the form of an implication in which the antecedent is always false?
Are you planning to actually address what I wrote, instead of just
repeating the same thing over and over in the hope that I will
eventually either tire of it or agree with you? I outlined my
reasoning twice now, you've ignored it twice now and simply repeated
your original claim.
Your objection is terminological. You may be right. I do not consider
it relevant.
Post by Arturo Magidin
I agree that for *each* n there is a way to translate the sentence
into an equivalent sentence which happens to be an implication in
which the antecedent is false.
Actualy this is my whole point.
Post by Arturo Magidin
However, for each n, the "translation" is to a DIFFERENT sentence.
Moreover, there is no single sentence with one free variable which is
both (i) equivalent to your for every n; and (ii) vacuously true for
every n.
But all the three sentences are equivalent, and at any time one of
them is vacuously true. If you do not want to call the other two
sentences "vacuously true", fine, let's call them "X".
Post by Arturo Magidin
Because of that, once you quantify the n, you end up with a sentence
that is *not* vacuously true.
(*) (For all n)(Exists S) V(S,n)
No problem there.
Unfortunately, the usual meaning of "vacuously true" would be that
(**) (Exists S)(For all n) V(S,n).
(*) and (**) are *not* equivalent. You have an instance of (*), but
want to claim it as an instance of (**).
Post by Newberry
Post by Arturo Magidin
There are
many ways of translating your two-universally-quantified-variable
sentence into a sentece of that form, but *none* of those translations
has the refered to property. Instead, you are trying to extend the
informal terminology to a class of sentences that is not normally
considered to be covered under that terminology.
Simply put, while everyone agrees that "Z" means either "for all x P
(x)" or "for all x Q(x)" , you want it to *also* mean "for all x (P(x)
or Q(x))", and the two are not equivalent (though one implies the
other).
Again, you are free to call the situation you are describing anything
you want, but calling the tail a leg does not make it a leg.
Actually I could not care less what you call the property I am
describing.
Yes, I can see that. You prefer to talk in your own private language
and then complain when people tell you the words you are using don't
usually mean what you are using them to mean.
Post by Newberry
My point is that
~(Ex)(Ey)[(x + y < 6) & (y = 8)]
has a property X, which
~(Ex)(x + 8 < 6)
does not.
Your "point" seems to be that you want "vacuously true" to refer to a
lot of sentences that are not normally called "vacuously true". As I
pointed out, "vacuously true" is an *informal* term, so I can
certainly not prove to you formally that you are incorrect. The best I
can do is tell you, as I have done three times now, that in my
personal experience very few, if any, mathematician would agree with
you, for the reasons I outlined. But since you couldn't care less, I
guess there's no problem.
--
Arturo Magidin- Hide quoted text -
- Show quoted text -- Hide quoted text -
- Show quoted text -
Arturo Magidin
2009-09-11 16:55:22 UTC
Permalink
Post by Newberry
Post by Arturo Magidin
Are you planning to actually address what I wrote, instead of just
repeating the same thing over and over in the hope that I will
eventually either tire of it or agree with you? I outlined my
reasoning twice now, you've ignored it twice now and simply repeated
your original claim.
Your objection is terminological. You may be right. I do not consider
it relevant.
Yes you seem to be very keen on getting me to agree with *your*
terminology...
Post by Newberry
Post by Arturo Magidin
I agree that for *each* n there is a way to translate the sentence
into an equivalent sentence which happens to be an implication in
which the antecedent is false.
Actualy this is my whole point.
Then, as I pointed out, your whole point is both moot and
terminologically incorrect.
Post by Newberry
Post by Arturo Magidin
However, for each n, the "translation" is to a DIFFERENT sentence.
Moreover, there is no single sentence with one free variable which is
both (i) equivalent to your for every n; and (ii) vacuously true for
every n.
But all the three sentences are equivalent, and at any time one of
them is vacuously true. If you do not want to call the other two
sentences "vacuously true", fine, let's call them "X".
And that was the whole thing I said to begin with, but you insist on
complaining about it. And "for each n there is an x(n) I can do" is
not the same thing as "there is an x I can do for every n".

But, since you consider my points irrelevant, and since I will *not*
agree to your misuse of "vacuously true", and since you've said you
will use different terminology then, it would appear that we have
reached a happy conclusion: you don't care, but you will stop misusing
the terminology.

--
Arturo Magidin
Stephen J. Herschkorn
2009-09-11 03:17:44 UTC
Permalink
Post by Newberry
Do you agree that
(x)[(x + 8 < 6) -> ~(8 = 8)]
is vacuously true?
Yes.
Post by Newberry
Do you agree that the above is equivalent to
~(Ex)[(x + 8 < 6) & (8 = 8)]
Yes.
Post by Newberry
Do you agree that
(x)[(4 = 8) -> ~(x + 4 < 6)]
is vacuously true?
Yes.
Post by Newberry
Do you agree that the above is equivalent to
~(Ex)[(x + 4 < 6) & (4 = 8)]
Yes.
Post by Newberry
Do you agree that for ANY n
~(Ex)[(x + n < 6) & (n = 8)]
is vacuously true?
No. As A. Magadin has been trying to explain to you, the terminology
"vacuously true" is a characterization of *implications*. A statement
is by definition vacuously true if and only if it is of the form P ->
Q where ~P is a tautology, or of the form Ax (P -> Q) where Ax
~P is a theorem of your model. The term does not apply to
nonimplications, even they are equivalent to some implication which are
vacuously true.

Your second and fourth examples above are not vacuously true. As for
your fifth stament, even the equivalent implication

An Ax [(n = 8) -> (x + n >= 6)]

is not vacuously true.
Post by Newberry
[BTW, obviously we are dealing with natural numbers only.]
--
Stephen J. Herschkorn ***@netscape.net
Math Tutor on the Internet and in Central New Jersey
Stephen J. Herschkorn
2009-09-11 03:25:51 UTC
Permalink
Post by Newberry
Do you agree that for ANY n
~(Ex)[(x + n < 6) & (n = 8)]
is vacuously true?
As for [the above statement], even the equivalent implication
An Ax [(n = 8) -> (x + n >= 6)]
is not vacuously true.
Nor is the implication

An Ax [(x + n < 6) -> (n != 8)]

vacuously true.
Post by Newberry
<>[BTW, obviously we are dealing with natural numbers only.]
--
Stephen J. Herschkorn ***@netscape.net
Math Tutor on the Internet and in Central New Jersey
Newberry
2009-09-11 16:34:33 UTC
Permalink
Post by Newberry
Do you agree that
(x)[(x + 8 < 6) -> ~(8 = 8)]
is vacuously true?
Yes.
Post by Newberry
Do you agree that the above is equivalent to
~(Ex)[(x + 8 < 6) & (8 = 8)]
Yes.
Post by Newberry
Do you agree that
(x)[(4 = 8) -> ~(x + 4 < 6)]
is vacuously true?
Yes.
Post by Newberry
Do you agree that the above is equivalent to
~(Ex)[(x + 4 < 6) & (4 = 8)]
Yes.
Post by Newberry
Do you agree that for ANY n
~(Ex)[(x + n < 6) & (n = 8)]
is vacuously true?
No.  As A. Magadin has been trying to explain to you, the terminology
"vacuously true"  is a characterization of *implications*.  A statement
is by definition vacuously true if and only if it is of the form  P ->
Q  where  ~P  is a tautology,   or of the form  Ax (P -> Q)  where Ax
~P  is a theorem of your model.  The term does not apply to
nonimplications, even they are equivalent to some implication which are
vacuously true.
Well, in some systems '->' is not even a primary symbol and it is
defined as '~(P & ~Q). You seem to be objecting to applying the term
"vacuously true" to expressions that do not actually contain the
arrow. I am OK with that. What if I called it "truth-functionally
equivalent to vacuously true"?
Your second and fourth examples above are not vacuously true.  As for
your fifth stament, even the equivalent implication
An Ax [(n = 8) -> (x + n >= 6)]
is not vacuously true.
Post by Newberry
[BTW, obviously we are dealing with natural numbers only.]
--
Math Tutor on the Internet and in Central New Jersey
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