```Date: Nov 28, 2012 10:13 AM
Author: Dan Christensen
Subject: Re: PREDICATE CALCULUS 2

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 aset of ordered pairs is always defined in terms of a domain andcodomain set. Here is a typical formal(ish) definition of a functionfrom Wiki (my comments in []'s):"A function f from X [the domain of f] to Y [the codomain of f] is asubset of the Cartesian product X × Y subject to the followingcondition: every element of X is the first component of one and onlyone ordered pair in the subset.[3] In other words, for every x in Xthere is exactly one element y such that the ordered pair (x, y) iscontained in the subset defining the function f."http://en.wikipedia.org/wiki/Function_(mathematics)#DefinitionExample: Let X=Y={0,1}.Then {(0,0), (1,1)} is function from X to Y, while {(0,1)} is not.DanDownload my DC Proof 2.0 software at http://www.dcproof.com
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