On Nov 3, 1:48 pm, Gary Forbis <forbisga
...@msn.com> wrote:
> On Friday, November 2, 2012 1:14:26 PM UTC-7, Graham Cooper wrote:
> > ALL(X) is a difficult function to define as it's an INTRA FUNCTION, it
> > has access to the arbitrary particular string which is how we define
> > functions with arbitrary symbols.
> That wasn't too helpful to me. What do you mean by "INTRA FUNCTION"?
Hi Gary, sorry I missed your post before.
PROLOG cannot see or examine the string itself of it's lines of code.
equals( S1, S2 ) :- e( A, S1 ) , e( A, S2 ).
This is the same as:
equals( X, Y ) :- e( Z, X ) , e (A , Y ).
You could substitute the 2nd version for the 1st version and it
wouldn't make a difference.
------
In 2nd order logic and axiom writing, you can use VARS for FORMULAS.
So if version 1 was P
and version 2 was Q
and you wanted to ADD A QUANTIFIER
A(A) P
A(A) Q
those 2 formulas are different because A() cannot see what variables
it is referring to inside P and Q.
The variable lettering is arbitrary on a line by line basis.
-------------------
A(A) P
A(A) equals( S1, S2 ) :- e( A, S1 ) , e( A, S2 ).
THIS IS AN AXIOM OF SET EQUALITY
--------------------
A(A) Q
A(A) equals( X , Y ) :- e( Z, X ) , e( Z, Y ).
THIS IS AN AXIOM OF SET INTERSECTION
and the A(A) is meaningless
------------------
That is why A(X) needs to see INSIDE the formula.
A(x) ..... P(...x...)
is like
{ x | P(...x...) }
it is a SYNTACTIC LEVEL NOTATION
> The best I could find was fromhttp://www.cs.rochester.edu/u/xiaoming/csc254-intra-function-code-gen...
> Grammar modification for control flow: (i.e. all intra-function control flow is changed to gotos and if-gotos.) <if-statement> --> if left_parenthesis <operand> <comparison op> <operand> right_parenthesis goto LABEL semicolon
> <goto-statement> --> goto LABEL semicolon
> <label-statement> --> LABEL colon
> LABEL is just another name for ID token
> > e.g.
> > ALL(X) E(Y) Y = X + 1
> > is the same function as
> > ALL(A) E(B) B = A + 1
> For all A there exists a B where B = A + 1.
> OK, you are defining A and B as the free variables
> in the expression B = A + 1. I think I understand.
> > ------------------
> > So if you use P = [X = Y + 1]
> Why would you do that? What does P respresent?
It's a 2OL VAR which is a PREDICATE
> > How can you Quantify over P by treating it as a function argument?
> > ALL(A) P
> > has forgotten what the variable names are inside P and is meaningless!
> It would be helpful if you stuck as close as possible to a single
> syntax unless you are responding to some one who uses a different
> syntax and seems less willing to adjust.
> > -------------------
> > BORROWING off PROLOG SYNTAX, or HORN CLAUSES
> > but we will only examine an atomic predicate.
> > *******************
> > PROLOG ATOMS
> > term
> > VAR
> > predicate(.....)
> > # here I use the word predicate for the predicate name, not including
> > the arguments.
> I'm not sure what you mean by a predicate here. Sure, prolog has
> a loose meaning but as far as I know for prolog a predicate is a
> term with or without qualificaiton. That is "true" is a built in
> predicate as is "false" as is "assert(X)". Heck I've just (re)learned
> "!" and "fail" are predicates.
Not really they have side effects so by definition are not pure
functions.
They use the Environment not their parameters list to determine their
'result'.
> > *******************
> > A term is written in lowercase such as tom, fred, 0, 1, set1, nums,
> > reals,
> > A VAR or VARIABLE is written in uppercase.
> > *******************
> > Prolog uses the DOUBLE VARIABLE INSTANTIATION RULE.
> > Here is a Prolog predicate that returns true iff the X [c]ordinates are
> > the same.
> > vertical( point(X,Y) , point(X,Z) ).
> I'm still relearning prolog just for you.
> How do I know when vertical returns
> YES, NO, or vertical(point(1,3), point(1,5))?
if the X values are the same it should!
?- vertical( point( 1,2 ) , point(1,9) ).
YES
> > ********************
> > FORALL vs FORANY
> > Let's examine the AXIOM OF EXTENSIONALITY
> > ALL(S1) ALL(S2) ALL(X) (XeS1 <-> XeS2) <-> (S1=S2)
> I still don't understand XeS1. I think you mean X is an element of S1.
> is that correct or incorrect?
> > **********************
> > Procedurally the ALL(S1) ALL(S2) are not required.
> > This works equally as well.
> > ANY(S1) ANY(S2) ALL(X) (XeS1 <-> XeS2) <-> (S1=S2)
> Now I'm confused. You used E(X) above to mean exists, or so
> I thought. This seems to imply "ANY" is distinct from "E" (exists).
> I don't know of any formal quantifiers other than universal and
> existential. In informal language we might say "any one who bites
> on that is a fool. We'd typically translate that to:
> ALL(X)(bites(X,that) -> fool(X))
> Why have you introduce a new quantifier?
RIGHT!
Sometimes you can covert LOGIC to PROLOG
LOGIC
ALL(x) bites(x,that) -> fool(x).
PROLOG
bites(X,that) -> fool(X).
just by capitalising the x in ALL(x)
and changing it to *ANY* solution for X.
-----
Other times you have to DO ALL THE CALCULATIONS
for EVERY SOLUTION to X.
Like the AXIOM OF SET EQUALITY - PROLOG cannot do that ALL(elements)
with simple native clauses!
> > **********************
> > But note:
> > ANY(S1) ANY(S2) ANY(X) (XeS1 <-> XeS2) <-> (S1=S2)
> > ANY(X) is ambiguous, does the equation hold for ANY X
> > or is there any X for which the equation CAN hold?
> > meaning the same as EXIST(X)!
> Yes, I would agree your use of ANY was ambiguous. More
> so, I took it as a variety of existential quantification
> until I remembered your use of E(X) above. I like your
> use of EXIST(X) as a quantifier better and would prefer
> you used ELEMENT_OF(x,S1) as well. I'll use it here:
> ESISTS(S1) EXISTS(S2) ALL(X)(
> (ELEMENT_OF(X,S1) <-> ELEMENT_OF(X,S2)) <-> (S1=S2))
> is a pretty weak claim. It leaves open the possibility
> that there exists S1 and S2 for which S1 and S2 contain
> exactly the same elements but we wouldn't say the sets
> are equal.
That's what Set Theory uses to define set equality, since
MEMBER e SET
'e' is the only relation used to determine set membership.
so it follows by extension.
It's actually called AXIOM OF EXTENSIONALITY
> > In PROLOG, S1 and S2 work just by using the double variable
> > instantiation rule to match to any (1) value of sets in the current
> > model under examination.
> Google didn't turn up any definition for "double variable
> instantiation rule". I'll have to try to determine it from
> context since I don't know what you mean here.
haha! I read someone else use the term... must have jumbled up the
words..
> > ***************************
> > TOWARDS PROLOG
> > ALL(X) (XeS1<->XeS2) <-> (S1=S2)
> > The ALL(X) is MANDATORY! It cannot be processed using double
> > variable instantiation rule.
> I don't get what you are saying but I have a suggestion. I
> don't know if it will work.
> is_predicate(man).
> is_predicate(human).
> is_predicate(woman).
> man([joe,jim,john]).
> woman([emma,ella,elizabeth]).
> human(X):- man(X).
> human(X):- woman(X).
> elementof(X,Y):- is_predicate(X),call(X(Y)).
> ?elementof(X,john)
> if it works should return
> elementof(man,john)
> elementof(human,john)
> or
> man(john)
> human(john)
> I don't know. I don't have a prolog compiler
> or interpretor and don't remember prolog.
I avoid list processing for my work.
> (must later I came back here. found
http://www.complang.tuwien.ac.at/ulrich/Prolog-inedit/ISO-Hiord
> )
> > *************************************
> > HIGH ORDER PROLOG ATOMS
> > *************************************
> > term
> > VAR
> > predicate(.....)
> > ALL(......)
> > *************************************
> > e.g.
> > ALL(n) p(n)
> > is just
> > N(p(N))
> have you implemented that?
> I would have thought you implemented
> p(X) to return all X in p. For instance, using the above facts:
> ?man(X).
> would return
> man(joe)
> man(jim)
> man(john)
> > DEFINITION:
> > ALL( string ) is a predicate type that returns the conjunction
> > of all boolean values of the predicate string argument for all
> > solutions of the variable ALL.
> More work needed. ALL returns a list of TRUE and FALSE?
> > e.g. INDUCTION FORMULA
> > p(0) & ALL(n) p(n)->p(s(n))
> > -> ALL(n) p(n)
> I don't get this at all. If you assert the fact:
> natnum(0).
> Then this function:
> natnum(X):-
> X > 0,
> Xp is X - 1,
> natnum(Xp).
> should have the following behavior
> ?natnum(5)
> YES
> ?natnum(5.2)
> NO
> I don't know what would be returned by
> ?natnum(X)
> I'm thinking I've defined a very inefficient
> routine where using the mod function would be
> much faster.
In Peano arithmetic there is only the digit 0
and s(0) = 1 s(s(0)) = 2 ...
num(0).
num(s(X)) :- num(X).
> > Using HIGH ORDER PROLOG method.
> > p(0) ^ N(p(N)->p(s(N))
> > -> N(p(N))
> What does "HIGH ORDER PROLOG" reference?
> Certainly nothing that looks like what you wrote.
> > This notation provides a direct computational method for
> > backward chaining formula to their axiomatic derivation
> > by using double variable instantiation on the ALL type predicate
> > inside the predicate and branching into width first (conjunction)
> > solution search mode in a modified
...
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