What is the exact difference between Well-formed formula and a proposition in propositional logic?
There's really not much given about Wff in my book.
My book says: "Propositions are also called sentences or statements. Another term formulae or well-formed formulae also refer to the same. That is, we may also call Well formed formula to refer to a proposition". Does that mean they both are the exact same thing?
Proposition: A statement which is true or false, easy for people to read but hard to manipulate using logical equivalences
WFF: An accurate logical statement which is true or false, there should be an official rigorus definition in your textbook. There are 4 rules they must follow. Harder for humans to read but much more precise and easier to manipulate
Example:
Proposition : All men are mortal
WFF: Let P be the set of people, M(x) denote x is a man and S(x)
denote x is mortal Then for all x in P M(x) -> S(x)
It is most likely that there is a typo in the book. In the quote Propositions are also called sentences or statements. Another term formulae or well-formed formulae also refer to the same. That is, we may also call Well formed formula to refer to a preposition, the word "preposition" should be "proposition".
Proposition :- A statement which is either true or false,but not both.
Propositional Form (necessary to understand Well Formed Formula) :- An assertion which contains at least one propositional variable.
Well Formed Formula :-A propositional form satisfying the following rules and any Wff(Well Formed Formula) can be derived using these rules:-
If P is a propositional variable then it is a wff.
If P is a propositional variable,then ~P is a wff.
If P and Q are two wffs then,(A and B),(A or B),(A implies B),(A is equivalent to B) are all wffs.
Related
Context
I started teaching myself lambda calculus last night and I am trying to determine if what I understand so far is correct.
Understanding
SKK is equivalent to the Identity combinator, I.
Where L stands for lambda:
S = LxLyLz((xz)(yz))
K = LxLy(x)
K essentially takes the next 2 (lambda) terms and gives back the first of those. S seems a little more complicated in the untyped lambda calculus.
My Interpretation
SK(any-lambda-term) is also equivalent to I.
I.e. the application of the application of S to K to Any-lambda-term is equivalent to the Identity combinator:
((S K)(Any)) = I = S K K = ((S K)(K))
I am using the convention of “left-association” in my above notation, if that helps (And I tried to make that clear in the 4th term above with parentheses. Everything I have read so far seems to use this convention).
Reasoning
S K = LyLz((K z)(y z))
The next lambda term will be substituted for y, let the term be Y.
S K Y = Lz((K z)(Y z))
(Y z) is the application of Y to z, also a lambda term.
(K z)returns the constant function that returns z, given another term input: (Y z).
Is my interpretation true? If not, can you provide an explanation? I would greatly appreciate it. Particularly if a sort of order of operations can be explained—I regularly find myself confused when considering when to evaluate. Perhaps that will be refined with practice.
Your intuition is correct, but an intuition proves nothing (alas...)
So, how can we prove your statement? Simply by showing that SKK and SKS have the same behaviour. "Behaviour" is an informal notion, which is formally capture by "semantics": if SKK and SKS are equals, then they should always reduce to the same term, according to the SKI-calculus semantics.
Now, there is a deep question, which is: what are the SKI-calculus? Actually, there is not a single way to answer that. What you implicitly do in your question is that you express SKI in terms of λ terms and you rely on the semantics of the λ calculus. This is absolutly correct. An other way to do it could have been to define directly SKI semantics. For instance, if you look at the wikipedia page, you can see that the semantics are not defined with lambda terms (and the fact that it correspond to lambda term is a (nice and expected) side effect). In the rest of this answer, I'll take the same approach as you do, and convert SKI terms in λ terms. A good exercise for you is to redo the proof, using the proper SKI semantics.
So, let formalize your question: your question is whether, for any SKI term t, SKKt = SKSt? Well... Let's see.
SKKt is encoded as (λx.λy.λz.(xz)(yz))(λx.λy.x)(λx.λy.x)t in the λ-calculus. We now just have to reduce it to a normal form (I detail every step, each time I reduce the leftmost λ, even tho it is not the fastest strategy):
(λx.λy.λz.(xz)(yz))(λx.λy.x)(λx.λy.x)t
= (λy.λz.((λx.λy.x)z)(yz))(λx.λy.x)t
= (λz.((λx.λy.x)z)((λx.λy.x)z))t
= ((λx.λy.x)t)((λx.λy.x)t)
= (λy.t)((λx.λy.x)t)
= t
So, the encoding of SKKt in the λ calculus reduces to t (as a sidenote, we just proved that SKK is equivalent to I here). To conclude our proof, we have to reduce SKSt and see whether it also reduces to t.
SKSt is encoded as (λx.λy.λz.(xz)(yz))(λx.λy.x)(λx.λy.λz.(xz)(yz))t. Let reduce it. (I don't detail as much this time)
(λx.λy.λz.(xz)(yz))(λx.λy.x)(λx.λy.λz.(xz)(yz))t
= ((λx.λy.x) t)((λx.λy.λz.(xz)(yz)) t))
= (λy.t)((λx.λy.λz.(xz)(yz)) t))
= t
Hurrah! It also reduce to t, so indeed, SKS and SKK are equivalent. It seems that the third combinator is not important: that as soon as you have SK?, it is equivalent to I. As an exercise, you can easily prove it (same strategy, if it is the case, then for any terms t and s, SKts = s). As mentionned above, an other good exercise is to redo the proof without using the λ semantics, but the proper SKI semantics.
Finally, my answer should raise a new question to you: we have two semantics, one that encodes SKI terms into λ terms, and one that does not. The question you may have is: are the two semantics equivalent? What does it mean for two semantics to be equivalent? If you are only starting to teach yourself λ calculus, it may be a bit early to try to answer those questions right now, but you can keep it in a corner of your head for when you'll get more familiar with formal languages.
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.
I read the question asked in Herbrand universe, Herbrand Base and Herbrand Model of binary tree (prolog) and the answers given, but I have a slightly different question more like a confirmation and hopefully my confusion will be clarified.
Let P be a program such that we have the following facts and rule:
q(a, g(b)).
q(b, g(b)).
q(X, g(X)) :- q(X, g(g(g(X)))).
From the above program, the Herbrand Universe
Up = {a, b, g(a), g(b), q(a, g(a)), q(a, g(b)), q(b, g(a)), q(b, g(b)), g(g(a)), g(g(b))...e.t.c}
Herbrand base:
Bp = {q(s, t) | s, t E Up}
Now come to my question(forgive me for my ignorance), i included q(a, g(a)) as an element in my Herbrand Universe but from the fact, it states q(a, g(b)). Does that mean that q(a, g(a)) does not suppose to be there?
Also since the Herbrand models are subset of the Herbrand base, how do i determine the least Herbrand model by induction?
Note: I have done a lot of research on this, and some parts are well clear to me but still i have this doubt in me thats why i want to seek the communities opinion. Thank you.
From having the fact q(a,g(b)) you cannot conclude whether or not q(a,g(a)) is in the model. You will have to generate the model first.
For determining the model, start with the facts {q(a,g(b)), q(b,g(b))} and now try to apply your rules to extend it. In your case, however, there is no way to match the right-hand side of the rule q(X,g(X)) :- q(X,g(g(g(X)))). to above facts. Therefore, you are done.
Now imagine the rule
q(a,g(Y)) :- q(b,Y).
This rule could be used to extend our set. In fact, the instance
q(a,g(g(b))) :- q(b,g(b)).
is used: If q(b,g(b)) is present, conclude q(a,g(g(b))). Note that we are using here the rule right-to-left. So we obtain
{q(a,g(b)), q(b,g(b)), q(a,g(g(b)))}
thereby reaching a fixpoint.
Now take as another example you suggested the rule
q(X, g(g(g(X)))) :- q(X, g(X)).
Which permits (I will no longer show the instantiated rule) to generate in one step:
{q(a,g(b)), q(b,g(b)), q(a,g(g(g(b)))), q(b, g(g(g(b))))}
But this is not the end, since, again, the rule can be applied to produce even more! In fact, you have now an infinite model!
{g(a,gn+1(b)), g(b, gn+1(b))}
This right-to-left reading is often very helpful when you are trying to understand recursive rules in Prolog. The top-down reading (left-to-right) is often quite difficult, in particular, since you have to take into account backtracking and general unification.
Concerning your question:
"Also since the Herbrand models are subset of the Herbrand base, how do i determine the least Herbrand model by induction?"
If you have a set P of horn clauses, the definite program, then you can define
a program operator:
T_P(M) := { H S | S is ground substitution, (H :- B) in P and B S in M }
The least model is:
inf(P) := intersect { M | M |= P }
Please note that not all models of a definite program are fixpoints of the
program operator. For example the full herbrand model is always a model of
the program P, which shows that definite programs are always consistent, but
it is not necessarily a fixpoint.
On the other hand each fixpoint of the program operator is a model of the
definite program. Namely if you have T_P(M) = M, then one can conclude
M |= P. So that after some further mathematical reasoning(*) one finds that
the least fixpoint is also the least model:
lfp(T_P) = inf(P)
But we need some further considerations so that we can say that we can determine
the least model by a kind of computation. Namely one easily observes that the
program operator is contiguous, i.e. preserves infinite unions of chains, since
horn clauses do not have forall quantifiers in their body:
union_i T_P(M_i) = T_P(union_i M_i)
So that again after some further mathematical reasoning(*) one finds that we can
compute the least fixpoint via iteration, witch can be used for simple
induction. Every element of the least model has a simple derivation of finite
depth:
union_i T_P^i({}) = lpf(T_P)
Bye
(*)
Most likely you find further hints on the exact mathematical reasoning
needed in this book, but unfortunately I can't recall which sections
are relevant:
Foundations of Logic Programming, John Wylie Lloyd, 1984
http://www.amazon.de/Foundations-Programming-Computation-Artificial-Intelligence/dp/3642968287
I want to use universal quantifier in the body of a predicate rule, i.e., something like
A(x,y) <- ∀B(x,a), C(y,a).
It means that only if for each a from C(y, a), B(x,a) always has x to match (x,a), then A(x,y) is true.
Since in Datalog, every variable bounded in rule body is existential quantifier by default, the a would be an existential quantifier too. What should I do to express universal quantifier in the body of a predicate rule?
Thank you.
P.S. The Datalog engine I am using is logicblox.
The basic idea is to use the logical axiom
∀x φ(x) ⇔ ¬∃x ¬φ(x)
to put your rules in a form where only existential quantifiers are required (along with negation). Intuitively, this usually means computing the complement of your answer first, and then computing its complement to produce the final answer.
For example, suppose you are given a graph G(V,E) and you want to find the vertices which are adjacent to all others in the graph. If universal quantification were allowed in a Datalog rule body, you might write something like
Q(x) <- ∀y E(x,y).
To write this without the universal quantifier, you first compute the vertices which are not adjacent to all others
NQ(x) <- V(x), V(y), !E(x,y).
then return its complement as the answer
Q(x) <- V(x), !NQ(x).
The same kind of trick can be used in SQL, which also lacks universal quantifiers.
I was wondering what sort of sentences can't you express in Prolog? I've been researching into logic programming in general and have learned that first-order logic is more expressive compared to definite clause logic (Horn clause) that Prolog is based on. It's a tough subject for me to get my head around.
So, for instance, can the following sentence be expressed:
For all cars, there does not exist at least 1 car without an engine
If so, are there any other sentences that CAN'T be expressed? If not, why?
You can express your sentence straightforward with Prolog using negation (\+).
E.g.:
car(bmw).
car(honda).
...
car(toyota).
engine(bmw, dohv).
engine(toyota, wenkel).
no_car_without_engine:-
\+(
car(Car),
\+(engine(Car, _))
).
Procedure no_car_without_engine/0 will succeed if every car has an engine, and fail otherwise.
The most problematic definitions in Prolog, are those which are left-recursive.
Definitions like
g(X) :- g(A), r(A,X).
are most likely to fail, due to Prolog's search algorithm, which is plain depth-first-search
and will run to infinity and beyond.
The general problem with Horn Clauses however is, that they're defined to have at most one positive element. That said, one can find a clause which is limited to those conditions,
for example:
A ∨ B
As a consequence, facts like ∀ X: cat(X) ∨ dog(X) can't be expressed directly.
There are ways to work around those and there are ways to allow such statements (see below).
Reading material:
These slides (p. 3) give an
example of which sentence you can't build using Prolog.
This work (p. 10) also explains Horn Clauses and their implications and introduces a method to allow 'invalid' Horn Clauses.
Prolog is a programming language, not a natural language interface.
The sentence you show is expressed in such a convoluted way that I had hard time attempting to understand it. Effectively, I must thanks gusbro that took the pain to express it in understandable way. But he entirely glossed over the knowledge representation problems that any programming language pose when applied to natural language, or even simply negation in first order logic. These problems are so urgent that the language selected is often perceived as 'unimportant'.
Relating to programming, Prolog lacks the ability to access in O(1) (constant time) any linear data structure (i.e. arrays). Then a QuickSort, for instance, that requires access to array elements in O(1), can't be implemented in efficient way.
But it's nevertheless a Turing complete language, for what is worth. Then there are no statements that can't be expressed in Prolog.
So you are looking for sentences that can't be expressed in clausal logic that can be expressed in first order logic.
Strictly speaking, there are many, simply because clausal logic is a restriction of FOL. So that's true by definition.
What you can do though is you can rewrite any set of FOL sentences into a logic program that is not equivalent but with good properties. So for example if you want to know if p is a consequence of your theory, you can use equivalently the transformed logic program.
A few notes on the other answers:
Negation in Prolog (\+) is negation as failure and not first order logic negation
Prolog is a programming language, as correctly pointed out, we should be talking about clausal logic instead.
Left recursion is not a problem. You can easily use a different selection rule, or some other inference mechanism.