Drexel dragonThe Math ForumDonate to the Math Forum



Search All of the Math Forum:

Views expressed in these public forums are not endorsed by Drexel University or The Math Forum.


Math Forum » Discussions » sci.math.* » sci.math.independent

Topic: PREDICATE CALCULUS 2
Replies: 23   Last Post: Nov 29, 2012 1:13 AM

Advanced Search

Back to Topic List Back to Topic List Jump to Tree View Jump to Tree View   Messages: [ Previous | Next ]
Dan Christensen

Posts: 2,236
Registered: 7/9/08
Re: PREDICATE CALCULUS 2
Posted: Nov 28, 2012 10:13 AM
  Click to see the message monospaced in plain text Plain Text   Click to reply to this topic Reply

On Nov 28, 1:35 am, Graham Cooper <grahamcoop...@gmail.com> wrote:
> On Nov 28, 2:26 pm, Dan Christensen <Dan_Christen...@sympatico.ca>
> wrote:

> > On Nov 27, 8:59 pm, Graham Cooper <grahamcoop...@gmail.com> wrote:
>
> > > There are 2 ALLs which is more complicated but you can format it as a
> > > SUBSET using a cartesian product of the 2 X values with a common Y.

>
> > > isfunction(r)  <-  ALL(Y1) ALL(Y2) r(X,Y1)^r(X,Y2) -> Y1=Y2
>
> > There is more to functionality than this. I may not fully understand
> > your unusual notation (PROLOG?), but it would seem you have left out
> > the requirement that FOR ALL elements of some domain set, THERE EXISTS
> > a unique image in a codomain set. (This is where I think quantifiers
> > become indispensable).

>
> > > {(Y1,Y2)|r(X,Y1)^r(X,Y2)} C {(Y1,Y2)|r(X,Y1)^r(X,Y2)->Y1=Y2}
>
> > [snip]
>
> > The same comment applies... I think.
>
> > Anyway, I am still waiting for proofs of the following:
>
> > 1. {(x,y) | x in S, y=x}  is a function mapping the set S onto itself
> > 2. {x | ~x in x} cannot exist
> > 3. {x | x=x} cannot exist

>
> > You really need to address these fundamental results.
>
> > For what it is worth, and from what little I know about PROLOG, it
> > doesn't seem to be capable of all that is required to do mathematical
> > proofs in general. It may be able to model some interesting and useful
> > aspects of predicate logic and set theory, but, for your purposes,
> > important pieces of the puzzle seem to be missing.

>
>
> Nope, this is exactly the definition of function.
>
> {(Y1,Y2) | r(X,Y1)^r(X,Y2)}  C  {(Y,Y) | r(X,Y)}
>
> which simply guarantees only 1 Y value for any X value.
>


This is a common mistake. According to this erroneous view, every set
{(x,y)} is a function for any objects x and y. The functionality of a
set of ordered pairs is always defined in terms of a domain and
codomain set. Here is a typical formal(ish) definition of a function
from Wiki (my comments in []'s):

"A function f from X [the domain of f] to Y [the codomain of f] is a
subset of the Cartesian product X × Y subject to the following
condition: every element of X is the first component of one and only
one ordered pair in the subset.[3] In other words, for every x in X
there is exactly one element y such that the ordered pair (x, y) is
contained in the subset defining the function f."
http://en.wikipedia.org/wiki/Function_(mathematics)#Definition

Example: Let X=Y={0,1}.

Then {(0,0), (1,1)} is function from X to Y, while {(0,1)} is not.

Dan
Download my DC Proof 2.0 software at http://www.dcproof.com



Point your RSS reader here for a feed of the latest messages in this topic.

[Privacy Policy] [Terms of Use]

© Drexel University 1994-2014. All Rights Reserved.
The Math Forum is a research and educational enterprise of the Drexel University School of Education.