I need to merge two individuals with OWL API (I mean, given two individuals A, B, adding all axioms and annotation involving B in A). Let axiomsSet be the set of axioms involving B. Is there any simple way to replace B with A in each axiom in axiomsSet?
Thank you.
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Let A and B be Types, and f : A -> B and g : B -> A be functions inverse to each other. In other words A and B are isomorphic types.
Is it true that one cannot prove that A <> B?
Can I safely add the axiom that A = B?
Is this axiom compatible with the other axioms of the standard library of Coq? And, if yes, why isn't it in the standard library?
These kind of questions are the domain of predilection of homotopy type theory but here is a tentative synthetic answer to your questions.
In CIC (± the ideal theory underlying Coq), we say that two types A and B are equivalent if there is a function f : A -> B with left inverse g : B -> A (that is f o g = id_B) and right inverse h : B -> A (that is h o f = id_B).
You can use an isomorphism from A to B to show that they are equivalent.
Homotopy type theory is concerned with the addition of an axiom to CIC (or rather MLTT) called univalence that roughly states that equivalence between types and equality coincide (to be more precise, there is a map id_to_equiv : A = B -> equiv A B and univalence says that this map is itself an equivalence).
Univalence is compatible with CIC in the sense that there exist models of CIC that validate this axiom.
So to answer your questions:
Yes, one cannot prove A <> B in CIC because any such proof would contradict univalence (hence no model of CIC + univalence could exist)
You won't be able to prove False since it is a special case of univalence that does have models, however the computational content of Coq will be partially lost (as it is always the case with the addition of any axiom).
Univalence is compatible with some axioms of the standard library of Coq (excluded-middle for instance) but not with others (Uniqueness of identity proofs, a.k.a. UIP, univalence states that there are 2 proofs of Bool = Bool). Concerning the absence of univalence from the standard library, it stems from the fact that the standard library defines equality in Prop, which is incompatible with the existence of multiple witnesses of identity. However the axiom that you mention might be formulated in a way that's compatible with UIP (there have been some work by Bauer and Winterhalter on a so called cardinal model of type theory). Whether it is compatible with the other axioms from the standard library would have to be checked.
It seems to me that Bernstein's synthesis / 3NF synthesis always yields BCNF subrelations, but that's apparently not true.
When one uses 3NF synthesis, one will have subrelations as a result, and they will each consist of either:
just one functional dependency along with all attributes of the schema, so the left side of the lone functional dependency will be a superkey, and that subrelation will therefore be in BCNF.
multiple functional dependencies each of which have the same left side, so they're each superkeys, and that subrelation will therefore be in BCNF.
no functional dependency where the schema includes the attributes making up the primary key of the original / non-decomposed relation, which would satisfy BCNF vacuously because of there being no functional dependencies.
What is an example of the 3NF synthesis algorithm yielding a non-BCNF decomposition and why it is so?
Bernstein's algorithm returns (one or more) components in EKNF, which lies between 3NF & BCNF.
Your claims of "that subrelation will therefore be in BCNF" are wrong. The FDs that hold in a component are all the ones in the closure of the original relation whose attributes are all in the component. So FDs could hold in a component that are not out of its superkeys. (Which by definition of BCNF is just another way of saying a component could be not in BCNF. Obviously--since we are told that the algorithm doesn't always give BCNF.)
Since your reasoning is unsound, finding a counterexample seems moot. But just about any presentation of BCNF gives an example non-BCNF 3NF relation, which it then decomposes to BCNF. You can join the non-BCNF 3NF relation with a projection on attributes of one of its CKs extended by a fresh non-prime attribute, and Bernstein's algorithm can decompose back to the 2 tables.
Chris Date's classic An Introduction to Database Systems has a non-BCNF 3NF schema R(S, J, T) with minimal/irreducible cover
{S, J} -> T
{T} -> J
CKs are {S, J} & {T, J}. Berstein gives component (S, J, T)--non-BCNF 3NF input R--in which both given FDs hold--plus redundant component (T, J).
For an example with an additional non-redundant component, extend the cover by {T} -> X. CKs are the same. {S, J} -> T again gives (S, J, T)--non-BCNF--plus component (T, J, X).
So, could someone please give me an example of the 3NF synthesis algorithm yielding a non-BCNF decomposition and tell why it is so?
A better "So, [...]" would be, So, what is wrong with my reasoning? You would do well to examine the assumptions you made about what FDs could hold in a component. (That article happens to point out (with reference) that "A 3NF table that does not have multiple overlapping candidate keys is guaranteed to be in BCNF.")
There is no "why" in mathematics. We assume things ("assumptions", "axioms", "premises") & other things follow. We can ask for a proof of something, but the proof does not say "why" the something is so, it's a demonstration that it is so. "Why" might be used trying to ask for a proof or for steps that you got wrong in or are missing from whatever almost-proof you have in mind.
PS Such a ubiquitous non-BCNF 3NF relation is Today's Court Bookings in the Wikipedia article on BCNF as I write. But beware that that particular example has perhaps unintuitive FDs. Indeed beware that almost every relational model Wikipedia page--including that one--has errors & misconceptions. So do many, many textbooks, especially re normalization.
The answer of philipxy is correct. Since you are asking for an example, here there are a couple of them.
The relation (with a cover of the functional dependencies):
R (A B C D)
A B → C
C → D
D → B
through the synthesis algorithm is decomposed in:
R1 (A B C)
R2 (C D)
R3 (B D)
and R1 is not in BCNF for the dependency C → B (the candidate key is AB). Note that C → B is not present in the original cover, but is a dependency implied from it.
Here is another (classical) example:
Phones (AreaCode, PhoneNumber, Subscriber, Town, Street)
AreaCode, PhoneNumber → Town
AreaCode, PhoneNumber → Subscriber
AreaCode, PhoneNumber → Street
Town → AreaCode
The Bernsteins’s synthesis algorithm produces two subschemas:
R1 (AreaCode, PhoneNumber, Subscriber, Town, Street)
AreaCode, PhoneNumber → Town
AreaCode, PhoneNumber → Subscriber
AreaCode, PhoneNumber → Street
and:
R2 (Town, AreaCode)
Town → AreaCode
since R2 is included in R1, the algorithm eliminates the second relation. The resulting relation is in 3NF but not in BCNF, since the relation has two candidate keys, (AreaCode, PhoneNumber) and (PhoneNumber, Town) and the functional dependency Town → AreaCode violates the BCNF.
Consider a relation R with five attributes ABCDE. Now
assume that R is decomposed into two smaller relations ABC and CDE.
Define S to be the relation (ABC NaturalJoin CDE).
a) Assume that the above decomposition is lossless join. What is the
dependency that guarantees the lossless join property.
b) Give an additional FD such that “dependency preserving” property is
violated by this decomposition.
c) Give two additional FD's that would be preserved by this
decomposition.
Question seems different to me because there is no FD given and its asking:
a)
R1=(A,B,C) R2=(C,D,E) R1∩R2 =C (how can i control dependency now)
F1' = {A->B,A->C,B->C,B->A,C->A,C->B,AB->C,AC->B,BC->A...}
F2' = {C->D,C->E,D->E....}
then i will find F' ??
b,c) how do i check , do i need to look for all possible FD's for R1 and R2
The question is definitely assuming things it hasn't said clearly. ABCDE could be subject to the JD *{ABC,CDE} while not being subject to any nontrivial FDs at all.
But suppose that the relation is subject to some FDs and isn't subject to any JDs other than ones that they imply. If C is a CK then the join is lossless. But then C -> ABCDE holds, because a CK determines all attributes, and C -> ABDE holds, because a CK determines all other attributes. No other FD holding would imply that the join is lossless, although that requires tedium (by looking at every possible case of CK) or inspiration to show.
Both these FDs guarantee losslessness. Although if one of these holds the other holds, and they express the same condition. So the question is sloppy. Or the question might consider that the two expressions express the same FD in the sense of a condition, but a FD is an expression and not a condition, so that would also be sloppy.
I suspect that the questioner really just wanted you to give some FD whose holding would guarantee losslessness. That would get rid of the complications.
I'm rather confused about the definition of 3NF.
Let R be a relation with attribute set X.
Suppose Y -> A is a functional dependency where A is a non-prime attribute and Y is a subset of X.
If Y is a proper subset of any candidate key for R, then the relation is not in 3NF (and not even in 2NF) because this is a partial dependency, which is not permitted in 2NF (and by extension 3NF).
If Y is a non-prime attribute, the relation is not in 3NF because this is a transitive dependency of the non-prime attribute A on any candidate key through the non-prime attribute Y.
But what if Y is a set containing both prime and non-prime attributes? What if A is a subset of Y? What if Y contains only prime attributes, but those prime attributes come from different keys of R so that Y is not a proper subset of any particular key of R? What if Y contains only, but multiple non-prime attributes? Which of these cases violates the requirements of 3NF and why?
TL;DR Get definitions straight.
To know whether a case violates 3NF you have to look at the criteria used in some definition.
Your question is rather like asking, I know an even number is one that is divisible by 2 or one whose decimal representation ends in 0, 2, 4, 6 or 8, but what if it's three times a square? Well, you have to use the definition--show that the given conditions imply that it's divisible by two or that its decimal representation ends in one of those digits. Why do you even care about other properties than the ones in the definition?
When some FDs (functional dependencies) hold, others must also hold. We say the latter are implied by the former. So when given FDs hold usually tons of others also hold. So one or more arbitrary FDs holding doesn't necessarily tell you anything about any normal forms might hold. Eg when U is a superset of V, U → V must hold; such FDs are called trivial because they are implied by any collection of FDs. Eg when U → V, every superset of U determines every subset of V. Armstrong's axioms are some rules that can be mechanically applied to find all FDs that hold. There are algorithms to find a canonical/minimal/irreducible cover for a given set, a set of FDs that imply all those in it with no proper subset that does. There are also algorithms to determine whether a relation satisfies certain NFs (normal forms), and to decompose them into components with higher NFs when they're not.
Sometimes we think there is a case that the definition doesn't handle but really we have got the definition wrong.
The definition you are trying to refer to for a relation being in 3NF actually requires that there be no transitive functional dependence of a non-prime attribute on a candidate key.
In your non-3NF example you should say there is a transitive FD, not "this is a transitive FD", because the violating FD is of the form CK → A not Y → A. Also, U → V is transitive when there is an X where U → X AND X → V AND NOT X → V. It doesn't matter whether X is a prime attribute.
PS It's not very helpful to ask "why" something is or isn't so in mathematics. We describe a situation in terms of some givens, and a bunch of things follow. We can say that if certain of the givens weren't so then that thing wouldn't be so. But if certain other givens weren't so then it might also not be so. We can give a proof that something is or isn't so as "why" but it's not the only proof.
I'm taking an artificial intelligence class and we are working with Answer Set Programming (Clingo specifically). We're talking mostly theory at the moment and I am having some trouble differentiating between models and least models. I have the following definitions:
Satisfying rules, models, least models and answer sets of definite
program
A program is called definite if it does not have “not” in the body of its rules.
A set S is said to satisfy a rule of the form a :- b1, …, bm, not c1, …, not cn. if its body is satisfied by S (I.e., b1 … bm are in S
and none of c1 ... cn are in S) implies that its head must be
satisfied by S (i..e, a is in S).
A set S is said to satisfy a program if it satisfies all rules of that program.
A set S is said to be an answer set of a definite program P if (a) S satisfies P (also referred to as S is a model of P) and (b) No
strict subset of S satisfies P (i.e., S is the least model of P).
With the question (pulled from the lecture slides, not homework):
P is defined as:
a :- b,e.
b.
c :- d,b.
d.
Which of the following are models and least models?
{}, {b}, {b,d}, {b,d,c}, {b,d,c,e}, {b,d,c,e,a}
Can anyone let me know what the answer to the above question is? I can probably figure out the difference from there, although if someone could explain the difference in common speak (rather than text-book definition), that would be wonderful. I'm not sure which forum to post this question under - please let me know if it should be posted somewhere else.
Thanks
First of all, note that this section of your slides is talking about the answer sets of a positive program P (also called a definite program), even though it mentions also not. The positive program is the simple case, as for a positive program P there always exists a unique least model LM(P), which is the intersection of all it's models.
Allowing not rules in the rule bodies makes things more complex. The body of a rule is the right side of :-.
The answer to the question would be, set by set:
S={} is not a model, since b and d are facts b. d.
S={b} is not a model, since d is a fact d.
S={b,d} is not a model, since c is implied by c :- d,b. and c is not in S
S={b,d,c} is a model
S={b,d,c,e} is not a model, since a is implied by a :- b,e. and a is not in S
S={b,d,c,e,a} is a model
So what is the least model? It's S={b,c,d} since no strict subset of S satisfies P.
We can arrive to the least model of a positive program P in two ways:
Enumerate all models and take their intersection (here {b,c,d}∩{a,b,c,d,e}={b,c,d}).
Starting with the facts (here b. d.) and iteratively adding implied atoms (here c :- b,d.) to S, repeating until S is a model and stopping at that point.
Like your slides say, the definition of an answer set for a positive program P is: S is an answer set of P if S is the least model of P. To be stricter, this is actually if and only if, since the least model LM(P) is unique.
As a last note, so you are not later confused by them, a constraint :- a, b is actually just shorthand for x :- not x, a, b. Thus programs containing constraints are not positive programs; though they might look like they are at first, since the body of a constraint seemingly doesn't contain a not.