In terms of if conditions in programming languages. I do not remember this stuff so I am asking.
I know what (A OR B OR (C AND D)) means
but what does (A OR B OR C AND D) mean?
In general, AND takes precedence over OR. But like the comments said, it can depend on the language. For more information on order of operations, see: http://en.wikipedia.org/wiki/Order_of_operations
Another reference, that is Excel specific, also states AND precedes OR in the order of operations. See here: http://office.microsoft.com/en-us/excel-help/HV080557451.aspx
EDIT:
What this means is that your two statements are equal.
(A OR B OR C AND D)
is evaluated as:
(A OR B OR (C AND D))
Related
Let's imagine a simple database of genealogy facts where mother(M, C) and father(F, C) denotes that M/F is the mother/father of child C.
I've written a rule to find known parents of a child (zero, one or both):
parents(C, M, F) :-
(mother(M, C) -> true; true),
(father(M, C) -> true; true).
which binds M and F if they are known and leaves them unbound otherwise.
It works fine, which means that for a set of facts:
mother(m1, c1).
father(f1, c1).
mother(m2, c2).
a call to parents(c1, M, F) returns:
M = m1,
F = f1.
while parents(c2, M, F) returns:
M = m2.
but the use of the arrow operators seems a little strange to me. Am I missing something basic? Can the (X -> true ; true) calls be avoided/simplified?
Any help appreciated.
Cheers,
From a logical perspective, a major mistake in this program is its incompleteness.
Consider for example the most general query:
?- parents(X, Y, C).
X = c1,
Y = m1.
So, no solution for c2 is reported.
But such a solution exists, as can be seen with:
?- parents(c2, Y, C).
Y = m2.
So, which is it, is there a solution or not?
Such mistakes almost invariably arise if you use (->)/2 and other constructs that violate logical purity of your code. Please see logical-purity for more information.
Hence, from a logical perspective, I can only recommend to avoid such constructs, since they defeat the primary advantage of a logic programming language to begin with: The ability to reason logically about your programs.
Instead, focus on a clear description of the relations you want to describe, and state the conditions that make them true. This will allow you to use your Prolog programs in a sensible way.
EDIT: I see you prefer a botched program. For this purpose, I recommend ignore/1. ignore(Goal) calls Goal as once(Goal), and succeeds. You can use this to simplify your program and still ensure that it remains incomplete.
Prolog is a real down to earth programming language. It has a pure subset. Both have their place.
once( (A ; true) ) is the answer to the question "how can we simplify (A -> true; true)".
If you want more purity, you could write (A *-> true ; true ) with the "soft cut" *-> which admits all solutions from the successful A and only switches to the unsuccessful branch in case A didn't produce any. See also e.g. this answer of mine for more discussion.
Another variant is (A ; \+ A).
Having a little trouble understanding the core terms of the MicroKanren DSL. Section 4 says:
Terms of the language are defined by the unify operator. Here, terms of the language consist of variables, objects deemed identical under eqv?, and pairs of the foregoing.
But they never describe what the "pairs" actually mean. Are the pairs supposed to represent equality of two subterms, like so:
type 'a ukanren = KVar of int | KVal of 'a | KEq of 'a kanren * 'a kanren
So a term like:
(call/fresh (λ (a) (≡ a 7)))
Generates a pair for (≡ a 7)?
Edit: upon further thought, I don't think this is it. The mentions of "pair" in the paper seem to come much later, with extensions and refinements to the basic system, which would mean the pairs have no meaning in the terms for the basic intro. Is this correct?
In this context, "pair" just means a cons pair, such as (5 . 6) or (foo . #t). An example unification between two pairs is:
(call/fresh
(λ (a)
(call/fresh
(λ (b)
(≡ (cons a b) (cons 5 6))))))
which associates a with 5 and b with 6.
Sorry for the confusion and difficulty!! And thank you for the question!
You can think of the (typical) Kanren term language as having a single binary functor tag cons/2, and an infinite quantity of constants (the exact makeup changes from embedding to embedding).
Under the assumption that cons/2 is the only (n < 0)-ary functor tag, every compound term will have been built with it. If you look up standard presentations of unification (e.g. Martelli-Montanari) you'll usually see a step f(t0,...,tn) g(s0,...,sm) => fail if f =/= g or n =/= m. We handle clash-failure in unification when comparing ground atomic terms; for us, compound terms must necessarily be of the same arity and tag.
pair? recognizes conses. In Racket, in fact, they provide a cons? alias, to make this clearer!
Once again, thank you!
I have a Relation f defined as f: A -> B × C. I would like to write a firsr-order formula to constrain this relation to be a bijective function from A to B × C?
To be more precise, I would like the first order counter part of the following formula (actually conjunction of the three):
∀a: A, ∃! bc : B × C, f(a)=bc -- f is function
∀a1,a2: A, f(a1)=f(a2) → a1=a2 -- f is injective
∀(b, c) : B × C, ∃ a : A, f(a)=bc -- f is surjective
As you see the above formulae are in Higher Order Logic as I quantified over the relations. What is the first-order logic equivalent of these formulae if it is ever possible?
PS:
This is more general (math) question, rather than being more specific to any theorem prover, but for getting help from these communities --as I think there are mature understanding of mathematics in these communities-- I put the theorem provers tag on this question.
(Update: Someone's unhappy with my answer, and SO gets me fired up in general, so I say what I want here, and will probably delete it later, I suppose.
I understand that SO is not a place for debates and soapboxes. On the other hand, the OP, qartal, whom I assume is the unhappy one, wants to apply the answer from math.stackexchange.com, where ZFC sets dominates, to a question here which is tagged, at this moment, with isabelle and logic.
First, notation is important, and sloppy notation can result in a question that's ambiguous to the point of being meaningless.
Second, having a B.S. in math, I have full appreciation for the logic of ZFC sets, so I have full appreciation for math.stackexchange.com.
I make the argument here that the answer given on math.stackexchange.com, linked to below, is wrong in the context of Isabelle/HOL. (First hmmm, me making claims under ill-defined circumstances can be annoying to people.)
If I'm wrong, and someone teaches me something, the situation here will be redeemed.
The answerer says this:
First of all in logic B x C is just another set.
There's not just one logic. My immediate reaction when I see the symbol x is to think of a type, not a set. Consider this, which kind of looks like your f: A -> BxC:
definition foo :: "nat => int × real" where "foo x = (x,x)"
I guess I should be prolific in going back and forth between sets and types, and reading minds, but I did learn something by entering this term:
term "B × C" (* shows it's of type "('a × 'b) set" *)
Feeling paranoid, I did this to see if had fallen into a major gotcha:
term "f : A -> B × C"
It gives a syntax error. Here I am, getting all pedantic, and our discussion is ill-defined because the notation is ill-defined.
The crux: the formula in the other answer is not first-order in this context
(Another hmmm, after writing what I say below, I'm full circle. Saying things about stuff when the context of the stuff is ill-defined.)
Context is everything. The context of the other site is generally ZFC sets. Here, it's HOL. That answerer says to assume these for his formula, wich I give below:
Ax is true iff x∈A
Bx is true iff x∈B×C
Rxy is true iff f(x)=y
Syntax. No one has defined it here, but the tag here is isabelle, so I take it to mean that I can substitute the left-hand side of the iff for the right-hand side.
Also, the expression x ∈ A is what would be in the formula in a typical set theory textbook, not Rxy. Therefore, for the answerer's formula to have meaning, I can rightfully insert f(x) = y into it.
This then is why I did a lot of hedging in my first answer. The variable f cannot be in the formula. If it's in the formula, then it's a free variable which is implicitly quantified. Here's the formula in Isar syntax:
term "∀x. (Ax --> (∃y. By ∧ Rxy ∧ (∀z. (Bz ∧ Rxz) --> y = z)))"
Here it is with the substitutions:
∀x. (x∈A --> (∃y. y∈B×C ∧ f(x)=y ∧ (∀z. (z∈B×C ∧ f(x)=z) --> y = z)))
In HOL, f(x) = f x, and so f is implicitly, universally quantified. If this is the case, then it's not first-order.
Really, I should dig deep to recall what I was taught, that f(x)=y means:
(x,f(x)) = (x,y) which means we have to have (x,y)∈(A, B×C)
which finally gets me:
∀x. (x∈A -->
(∃y. y∈B×C ∧ (x,y)∈(A,B×C) ∧ (∀z. (z∈B×C ∧ (x,z)∈(A,B×C)) --> y = z)))
Finally, I guess it turns out that in the context of math.stackexchange.com, it's 100% on.
Am I the only one who feels compulsive about questioning what this means in the context of Isabelle/HOL? I don't accept that everything here is defined well enough to show that it's first order.
Really, qartal, your notation should be specific to a particular logic.
First answer
With Isabelle, I answer the question based on my interpretation of your
f: A -> B x C, which I take as a ZFC set, in particular a subset of the
Cartesian product A x (B x C)
You're sort of mixing notation from the two logics, that of ZFC
sets and that of HOL. Consequently, I might be off on what I think you're
asking.
You don't define your relation, so I keep things simple.
I define a simple ZFC function, and prove the first
part of your first condition, that f is a function. The second part would be
proving uniqueness. It can be seen that f satisfies that, so once a
formula for uniqueness is stated correctly, auto might easily prove it.
Please notice that the
theorem is a first-order formula. The characters ! and ? are ASCII
equivalents for \<forall> and \<exists>.
(Clarifications must abound when
working with HOL. It's first-order logic if the variables are atomic. In this
case, the type of variables are numeral. The basic concept is there. That
I'm wrong in some detail is highly likely.)
definition "A = {1,2}"
definition "B = A"
definition "C = A"
definition "f = {(1,(1,1)), (2,(1,1))}"
theorem
"!a. a \<in> A --> (? z. z \<in> (B × C) & (a,z) \<in> f)"
by(auto simp add: A_def B_def C_def f_def)
(To completely give you an example of what you asked for, I would have to redefine my function so its bijective. Little examples can take a ton of work.)
That's the basic idea, and the rest of proving that f is a function will
follow that basic pattern.
If there's a problem, it's that your f is a ZFC set function/relation, and
the logical infrastructure of Isabelle/HOL is set up for functions as a type.
Functions as ordered pairs, ZFC style, can be formalized in Isabelle/HOL, but
it hasn't been done in a reasonably complete way.
Generalizing it all is where the work would be. For a particular relation, as
I defined above, I can limit myself to first-order formulas, if I ignore that
the foundation, Isabelle/HOL, is, of course, higher-order logic.
How would I write the following in Prolog?
a -> b V c
In English that would be a implies that b or c (or both)
The clause
a => (b ; c) % ';' means 'or'
is not a Horn clause and hence cannot be represented in (pure) Prolog (see e.g Wikipedia). On the other hand (b ; c) => a is a Horn clause and can obviously be represented by two Prolog rules.
I'm not entirely sure what you want to do with with this implies statement. But I would have thought the following would suffice (bear in mind this is SICStus not swi, but at this low level I think it's all the same).
predicate(a, b).
predicate(a, c).
?- predicate(a, Then).
Then = b ;
Then = c ;
no
?- predicate(x, Then).
no
You could do more complicated checks to make sure a is never an unbound value (to prevent predicate(If, b). being true), but unless you're making a huge application then I'm sure good documentation would suffice.
Logically, "b or c" is the same thing as "b or c (or both)"
You can read about logical operators in Prolog here: http://rigaux.org/language-study/syntax-across-languages-per-language/Prolog.html
Can you explain a bit more please what you're trying to do with 'implies'?
I've got a library that I have to interface with which acts basically as a data source. When retrieving data, I can pass special "filter expressions" to that library, which later get translated to SQL WHERE part. These expressions are pretty limited. They must be in conjunctive normal form. Like:
(A or B or C) and (D or E or F) and ...
This of course isn't very comfortable for programming. So I want to make a little wrapper which can parse arbitrary expressions and translate them to this normal form. Like:
(A and (B or C) and D) or E
would get translated to something like:
(A or E) and (B or C or E) and (D or E)
I can parse an expression to a tree with the Irony library. Now I need to normalize it, but I don't know how... Oh, also, here's the twist:
The final expression may not contain the NOT operator. However, I can inverse the individual terms by replacing the operators with the inverse operators. So, this is OK:(not A or not B) AND (not C or not D)but this is not:
not (A or B) and not (C or D)
I would like to make the expression as simple as possible, because it will be translated to a practically identical SQL WHERE statement, so a complex statement would most likely slow down execution speed.
I'd use two iterations over the tree, although it's probably possible in one.
First iteration: get rid of your NOT Nodes by walking through the tree and using De Morgan's law (wikipedia link) and remove double negation wherever applicable.
Second iteration (the NOT are now only directly before a leaf node)
Go through your tree:
Case "AND NODE":
fine, inspect the children
Case "OR NODE":
if there is a child which is neither a Leaf nor a NOT node
apply the distributive law.
start from parent of current node again
else
fine, inspect children
After that you should be done.