Drexel dragonThe Math ForumDonate to the Math Forum


Math Forum
 Ask Dr. Math
 Discussions
 Internet Newsletter
 MathTools
 Problems of the Week
 Teacher2Teacher
 Teacher Exchange
 Workshops

Search All of the Math Forum:
Browse our Internet Mathematics Library

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
Reply to this Topic Reply to this Topic

Back to Topic List Back to Topic List Jump to Tree View Jump to Tree View   Messages: [ Previous | Next ]
Graham Cooper

Posts: 4,031
Registered: 5/20/10
Re: PREDICATE CALCULUS 2
Posted: Nov 27, 2012 9:29 PM
  Click to see the message monospaced in plain text Plain Text   Click to reply to this topic Reply

>
> {(Y1,Y2)|r(X,Y1)^r(X,Y2)} C {(Y1,Y2)|r(X,Y1)^r(X,Y2)->Y1=Y2}
> DETECTING r is a FUNCTION without QUANTIFIERS!
>

This simplifies to..

{(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.

So it switches from Quantifier ALL(X) P(...) HOLDS
to ALL VALUES OF X SATISFYING { X | P(...) }

which is a different type of result, a SET
but SET1 C SET2
then returns the same Boolean type as Quantifier ALL(X)


Herc

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

[Privacy Policy] [Terms of Use]

© Drexel University 1994-2013. All Rights Reserved.
The Math Forum is a research and educational enterprise of the Drexel University School of Education.