Hi,
I am covering elementary predicate calculus and have just come across
the universal and existential quantifiers.
If I wanted to say ‘Some swans are black’ I am having difficulty seeing
the difference between (please excuse the existential quantifier being
the wrong way round)
(Ex) (Sx & Bx)
and
(Ex) (Sx -> Bx)
I can see why the former is so, but not the latter.
Also, I read that the latter is true if there are no swans; would anyone
be kind enough to explain why is this so? I’m confused 
Any help would be greatly appreciated!
Best wishes,
–
Craig Done
To e-mail me, please remove the ‘w’ from my address.
In article <atwFVFAtaJO3I…@cdone.freeserve.co.uk>, Craig Done
- Hide quoted text — Show quoted text -
<cr…@cdone.freeswerve.co.uk> wrote:
> Hi,
> I am covering elementary predicate calculus and have just come across
> the universal and existential quantifiers.
> If I wanted to say ‘Some swans are black’ I am having difficulty seeing
> the difference between (please excuse the existential quantifier being
> the wrong way round)
> (Ex) (Sx & Bx)
> and
> (Ex) (Sx -> Bx)
> I can see why the former is so, but not the latter.
> Also, I read that the latter is true if there are no swans; would anyone
> be kind enough to explain why is this so? I’m confused
> Any help would be greatly appreciated!
> Best wishes,
The former of these does imply the latter. If there exists some swan
which is black, then there exists a thing such that, if it is a swan, then
it is black. However the implication does not go the other way. Suppose
no swans existed. Then clearly the former of these statements would be
false. There would not exist anything which is a swan and black. However
(provided that there exist any things at all) there would exist things
which, if they are swans (which in all cases they are not) then they are
black. It all rests upon the fact that an implication is true whenever
the antecedent is false. In this case, in a world where there are no
swans, it is false to say of anything that it is a swan. Therefore, any
implication which uses as its antecedent the claim that something is a
swan will be true.
-Tyrrell McAllister
tm…@ucdavis.edu
http://www.sonic.net/~bzm/Tyrrell/index.html
The truth value of an implication proposition is a slippery subject to
explain
in natural language. Converting the implication Sx –> Bx to ~Sx \/ Bx
helps.
(~ is used as the logical NOT operator.)
The proposition is true when the object is NOT a Swan OR when the object is
a Black Swan. If the object IS a Swan then the truth of the statement
depends
on whether the object is Black. If the object IS NOT a Swan then the
statement
is trivially true since we are making a statement about a particular Swan.
The
statement can only be false if the Swan we are referring to IS NOT Black.
The truth of the proposition Sx /\ Bx depends on whether the object is a
Swan
AND on whether the object is Black. The proposition can only be true when
both conditions are satisfied. The proposition is false when the object is
NOT
a Swan OR when the object is NOT Black. (Apply DeMorgan’s theorem.)
Adding the existential quantifier extends the context of the proposition.
With
that addition the proposition now applies to a universe of objects, x,
instead of
a single object. The predicate (Ex)(Sx /\ Bx) states that a Black Swan
exists
somewhere in the universe of objects x. The predicate (Ex)(Sx –> Bx) states
that IF Swans exist THEN at least one of them is Black.
Chuck Simpson
Tyrrell McAllister <tm…@ucdavis.edu> wrote in message
news:tmcal-1105991743290001@iras-2-93.ucdavis.edu…
- Hide quoted text — Show quoted text -
> In article <atwFVFAtaJO3I…@cdone.freeserve.co.uk>, Craig Done
> <cr…@cdone.freeswerve.co.uk> wrote:
> > Hi,
> > I am covering elementary predicate calculus and have just come across
> > the universal and existential quantifiers.
> > If I wanted to say ‘Some swans are black’ I am having difficulty seeing
> > the difference between (please excuse the existential quantifier being
> > the wrong way round)
> > (Ex) (Sx & Bx)
> > and
> > (Ex) (Sx -> Bx)
> > I can see why the former is so, but not the latter.
> > Also, I read that the latter is true if there are no swans; would anyone
> > be kind enough to explain why is this so? I’m confused
> > Any help would be greatly appreciated!
> > Best wishes,
> The former of these does imply the latter. If there exists some swan
> which is black, then there exists a thing such that, if it is a swan, then
> it is black. However the implication does not go the other way. Suppose
> no swans existed. Then clearly the former of these statements would be
> false. There would not exist anything which is a swan and black. However
> (provided that there exist any things at all) there would exist things
> which, if they are swans (which in all cases they are not) then they are
> black. It all rests upon the fact that an implication is true whenever
> the antecedent is false. In this case, in a world where there are no
> swans, it is false to say of anything that it is a swan. Therefore, any
> implication which uses as its antecedent the claim that something is a
> swan will be true.
> -Tyrrell McAllister
> tm…@ucdavis.edu
> http://www.sonic.net/~bzm/Tyrrell/index.html
- Hide quoted text — Show quoted text -
On Wed, 12 May 1999, Earthlink wrote:
> The truth value of an implication proposition is a slippery subject to
> explain in natural language. Converting the implication Sx –> Bx to
> ~Sx \/ Bx helps. (~ is used as the logical NOT operator.)
> The proposition is true when the object is NOT a Swan OR when the object
> is a Black Swan. If the object IS a Swan then the truth of the statement
> depends on whether the object is Black. If the object IS NOT a Swan then
> the statement is trivially true since we are making a statement about a
> particular Swan. The statement can only be false if the Swan we are
> referring to IS NOT Black.
> The truth of the proposition Sx /\ Bx depends on whether the object is a
> Swan AND on whether the object is Black. The proposition can only be
> true when both conditions are satisfied. The proposition is false when
> the object is NOT a Swan OR when the object is NOT Black. (Apply
> DeMorgan’s theorem.)
> Adding the existential quantifier extends the context of the
> proposition. With that addition the proposition now applies to a
> universe of objects, x, instead of a single object. The predicate
> (Ex)(Sx /\ Bx) states that a Black Swan exists somewhere in the universe
> of objects x. The predicate (Ex)(Sx –> Bx) states that IF Swans exist
> THEN at least one of them is Black.
Not quite! It says that if all things are swans, then at least one of them
is black. If there is a non-swan, it will automatically satisfy the
condition.
Bennett Standeven <stand…@SLU.EDU> wrote in message
news:Pine.PMDF.3.96.990512225053.551880980A-100000@SLU.EDU…
- Hide quoted text — Show quoted text -
> On Wed, 12 May 1999, Earthlink wrote:
> > The truth value of an implication proposition is a slippery subject to
> > explain in natural language. Converting the implication Sx –> Bx to
> > ~Sx \/ Bx helps. (~ is used as the logical NOT operator.)
> > The proposition is true when the object is NOT a Swan OR when the object
> > is a Black Swan. If the object IS a Swan then the truth of the statement
> > depends on whether the object is Black. If the object IS NOT a Swan then
> > the statement is trivially true since we are making a statement about a
> > particular Swan. The statement can only be false if the Swan we are
> > referring to IS NOT Black.
> > The truth of the proposition Sx /\ Bx depends on whether the object is a
> > Swan AND on whether the object is Black. The proposition can only be
> > true when both conditions are satisfied. The proposition is false when
> > the object is NOT a Swan OR when the object is NOT Black. (Apply
> > DeMorgan’s theorem.)
> > Adding the existential quantifier extends the context of the
> > proposition. With that addition the proposition now applies to a
> > universe of objects, x, instead of a single object. The predicate
> > (Ex)(Sx /\ Bx) states that a Black Swan exists somewhere in the universe
> > of objects x. The predicate (Ex)(Sx –> Bx) states that IF Swans exist
> > THEN at least one of them is Black.
> Not quite! It says that if all things are swans, then at least one of them
> is black. If there is a non-swan, it will automatically satisfy the
> condition.
I think you have mixed the scope of your interpretation. If all things are
Swans
then a non-Swan is out of scope. If you look at the dual statement,
~(Ax)(Sx /\ ~Bx) you see that a nonSwan or a BlackSwan make this statement
true. However, the truth of the statement does not force all objects to have
a Swan attribute. Instead it says that a subset of objects that are Swans
and
Black will make the statement true as will a subset of objects that are not
Swans.
If you limit the scope of x to the subset of Swans, i.e., make x of type
Swan,
then the first sentence of your interpretation is true, but the second
sentence
then makes no sense because the type=Swan does not admit a nonSwan.
In my interpretation, Swan is an attribute of object x as is its color.
In a programming language you would use the dual statement to construct a
loop similar to the following to implement the statement:
Boolean blackSwanFunction( ObjectList objects ) {
// A function that tests for the existence of Black Swans
while ( x = getNextObject(objects) ) { // (Ax)
if( x.species == Swan && x.color == Black ) {
return TRUE; // (Sx /\ Bx)
}
x = nextObject(x)
}
return FALSE; // (Ax)(~Sx \/ ~Bx)
} // ~(Ax)(Sx /\ ~Bx)
In this function the universe of objects, the x’s are members of the list
provided
as the input parameter of the function. Each object has attributes ‘species’
and ‘color’.
Both attribute values must be used to determine the truth of the statement.
Chuckl Simpson
Dear Craig,
>If I wanted to say ‘Some swans are black’ I am having difficulty seeing
>the difference between (please excuse the existential quantifier being
>the wrong way round)
>(Ex) (Sx & Bx)
>and
>(Ex) (Sx -> Bx)
>I can see why the former is so, but not the latter.
>Also, I read that the latter is true if there are no swans; would anyone
>be kind enough to explain why is this so? I’m confused
A major source of confusion is the fact that Sx -> Bx (where Sx means that x is
a swan and Bx means that x is black) does *not* mean that if x is a swan then x
is black. English "if – then" is not the same as material implication; it is
much richer. A good translation into English would be "x is a swan only if x is
black."
If x does not exist, it is neither black nor a swan; no problem. It could be a
swan if it were black; but it is not.
(Ex) (Sx -> Bx) can be read, "There is something that is a swan only if it is
black." That is not exactly what we mean by the English statement, "Some swans
are (swan is) black."
Best regards,
Bill Spight
P. S. The handle is just a pun. Remove "jiji" to reply via email.
On 21 May 1999 00:34:51 GMT, aunties…@aol.comjiji (AuntieSpm2)
wrote:
=>Dear Craig,
=>
=>>If I wanted to say ‘Some swans are black’ I am having difficulty
seeing
=>>the difference between (please excuse the existential quantifier
being
=>>the wrong way round)
=>>
=>>(Ex) (Sx & Bx)
=>>
=>>and
=>>
=>>(Ex) (Sx -> Bx)
=>>
=>>I can see why the former is so, but not the latter.
=>>Also, I read that the latter is true if there are no swans; would
anyone
=>>be kind enough to explain why is this so? I’m confused
=>
=>A major source of confusion is the fact that Sx -> Bx (where Sx
means that x is
=>a swan and Bx means that x is black) does *not* mean that if x is a
swan then x
=>is black. English "if – then" is not the same as material
implication; it is
=>much richer. A good translation into English would be "x is a swan
only if x is
=>black."
=>
=>If x does not exist, it is neither black nor a swan; no problem. It
could be a
=>swan if it were black; but it is not.
=>
=>(Ex) (Sx -> Bx) can be read, "There is something that is a swan only
if it is
=>black." That is not exactly what we mean by the English statement,
"Some swans
=>are (swan is) black."
In the same vein (or artery if you prefer), material implication may
be stated in any of the following ways:
1. A implies B
2. if A then B
3. A only if B
4. B if A
5. B provided that A
6. A is a sufficient condition for B
7. B is a necessary condition for A
Confusin’, ain’t it! <g>
C’ya,
RudeJohn
"I’m rude. It’s a job."
Hi,
Just a quick note to thank you for all for your help.
Best wishes,
—
Craig Done
Please remove the w from my address to reply.
Does anybody apart from me find it slightly offensive to suggest that
the flow of money through an economic system has some sort of ‘energy’.
Are our lives so desperate that we have to suggest that economic flows
have been so reified that they take on some sort of metaphysical
quality? Money and energy have nothing to do with each other. The
latter, if an incomprehensible notion, has reference to our status as
moral agents qua human beings. The former has nothing to do with our
essences. To suggest that it does is to hold that without money, we are
less human and our lives less authentic. That’s absurd.
This is not to argue for some sort of anti-economic Marxism, but rather
a plea for sanity.
Max Lewis.
————————————
Max Lewis wrote:
> Does anybody apart from me find it slightly offensive to suggest that
> the flow of money through an economic system has some sort of ‘energy’.
> Are our lives so desperate that we have to suggest that economic flows
> have been so reified that they take on some sort of metaphysical
> quality? Money and energy have nothing to do with each other. The
> latter, if an incomprehensible notion, has reference to our status as
> moral agents qua human beings. The former has nothing to do with our
> essences. To suggest that it does is to hold that without money, we are
> less human and our lives less authentic. That’s absurd.
> This is not to argue for some sort of anti-economic Marxism, but rather
> a plea for sanity.
No. It’s not really offensive, because it’s true.
What is offensive is that clueless economists
are the people who run the economy.
In article <Pine.SOL.3.95q.990513105446.15405C-100…@sis.bris.ac.uk>,
Max Lewis <ml8…@bris.ac.uk> wrote:
>Does anybody apart from me find it slightly offensive to suggest that
>the flow of money through an economic system has some sort of ‘energy’.
That’s not what was suggested. The article wasn’t saying that the flow of
money has "energy," but that the value of money depends ultimately on the
amount of useful energy (or products of the use of energy) that it can
purchase.
It went on to say that if money stops flowing, so will the energy.
Presumably this is because in a capitalist society, the acquisition of money
is people’s primary motivation to apply the energy they have available to
them. I did not detect any implication that money itself possesses energy,
only that the use of available energy in an economy is regulated by the flow
of money.
–
Bitwise, Andrew. &