Related
I'm struggling in order to understand the meaning of the following expression:
aᵢ ⊕ bᵢ = xᵢ ∧ yᵢ
I know the symbol ⊕ is actually an exclusive OR, and the ∧ is an and symbol.
But I cannot grasp the overall meaning. What does that mean in simple words?
The context is what is stated here.
Can someone help me?
Thanks a lot
The paper you reference uses notation a⊕b = x·y, but · and ∧ mean the same in this context: logical AND operation on single-bit variables.
This equality describes the requirement of the CHSH game. The game involves two players, Alice and Bob, who cannot communicate with one another. They are each given a single random bit (Alice gets X and Bob gets Y). Alice and Bob then output a single bit they choose independently based on their input bits (A from Alice and B from Bob) with the goal of satisfying the formula X · Y = A ⊕ B.
This game illustrates that quantum entanglement enables strategies that are dramatically better than the purely classical strategies. The best
classical strategy is for Alice and Bob to output 0 regardless of the input - this strategy wins the game 75% of the time. But a quantum strategy exists that allows them to win 85% of the time if they share an entangled pair of qubits before the start of the game.
You can read more on the CHSH game here.
Translate the following statements into FOL sentances
1) Alex likes John
Likes(alex, john) - I know this is correct
2) Each person is either a man or woman
AxAy( man(x) v woman(y) )
EDIT: Is this better??: Az(Person(z) -> man(x) v woman(y))
OR EDIT: Is this better??: Ax(Person(x) -> man(x) v woman(x))
3) No one is both man and woman
Ex( (man(x) ^ ¬woman(x)) v (¬man(x) ^ woman(x)) )
4) Alex likes a man who likes a woman
AxEy(Likes( man(x), woman(y) ) -> Likes(alex, man(x) ))
Thanks
Here is a screenshot of the background info
EDIT: For number 3, I have found this online
"The exclusive disjunction of p and q asserts that either p is true or q is true but not both. The natural, but long-winded, way to express exclusive disjunction, then, is (p | q) & ~(p & q)."
If this can apply, then I assume the correct answer is Ax( (man(x) v woman(x)) ^ ¬(man(x) & woman(x)) )
But now I am getting confused as to how 2 and 3 are different...
Hey I just wanted to know if these were correct
1. Testing the Correctness of Translation
One of the ways to test if a first order sentence agrees with an informal specification is to use a model.
To perform a test you need:
To determine how many relations are there in your first order sentence and list them. This list would play the role of the logical signature of your model.
Now take a set of individuals large enough to have all interesting combinations of properties assigned to at least one individual.
Assign the properties to individuals according to your understanding of the meaning of relations in your first order sentence.
Finally for each universally quantified variable try different assignments of individuals to variables and check if the property holds.
Consider one of the examples from the post.
Informal specification: Alex likes a man who likes a woman
We have
one constant symbol: Alex
two unary relations: Man(x) and Woman(x)
a binary relation: Likes(x,y)
1.1 Structure No. 1
Now consider a structure where we have an individual for Alex, an individual for a man who is not Alex, and an individual for a woman who is not Alex.
Let's start with three: p1, p2, p3.
Alex is p1
Man(p2)
Woman(p3)
Woman(p1)
Likes (p2,p3)
Likes (p1,p2)
The informal specification says that Alex likes someone who is a Man and Likes someone else who is a Woman. Our model satisfies this specification.
Now consider the following statement taken from the OP:
AxEy(Likes( man(x), woman(y) ) -> Likes(alex, man(x) ))
This statement is from a different language. Here man(x) and woman(y) are unary functions instead of unary relations, so we cannot check this sentence.
I would not speculate any further on the subject of woman(y) or man(x) being a function (Pun intended). Instead I would consider a different sentence.
AxEy(Man(x) & Woman(y) & Likes( x, y ) -> Likes(alex, x ))
Let's see what happens if x=p2 and y=p1. In our model the premise
Man(p2) & Woman(p3) & Likes( p2, p3 )
holds, and the conclusion
Likes(p1,p2)
also holds, therefore the formula is true under this variable assignment.
1.2 Structure No. 2
Now consider a different model. This time with four individuals: p1,p2,p3,p4.
Alex is p1
Man(p2)
Man(p4)
Woman(p3)
Woman(p1)
Likes (p2,p3)
Likes (p1,p2)
Likes (p4,p3)
This model also satisfies our informal specification as Alex likes p2 who is a Man and likes p3 who is a Woman.
Consider the assignment x=p4 and y=p3.
Once again the premise holds in our model
Man(p4) & Woman(p3) & Likes( p4, p3 )
however the conclusion
Likes(p1,p4)
does not hold in our model. Since x is universally quantified this might be a problem as the formula does not hold under this assignment, however y is under an existential quantifier. Let's see if we can find an assignment for y that would turn our formula into a true statement with x=p4.
This is indeed possible. Let y=p2.
Man(p4) & Woman(p2) & Likes( p4, p2 )
does not hold, therefore the formula
Man(p4) & Woman(p2) & Likes( p4, p2 ) -> Likes(p1,p4)
is true. This sounds better yet to satisfy the formula we took an assignment that is far from the intended meaning of the informal specification. The original specification clearly was not speaking about a man who likes another man. So we are not done yet.
1.3 Structure No. 3
Consider a model with just two individuals: p1 and p2.
Alex is p1
Man(p2)
Woman(p1)
The informal specification does not hold in this model, so our sentence also should be false. There are four possible assignments of variables
x=p1,y=p2
x=p1,y=p1
x=p2,y=p1
x=p2,y=p2
Since Like(x,y) is always false in our model, the premise fails therefore according to the rules of implication, the formula is true. So our sentence is also true in a model where it should not hold. Once again, our first-order formalization does not hold against the informal specification.
This seems to be a very complex process that we have taken to test the translation. It assumes a certain skill of finding counterexamples and always keeping in mind the intended meaning of the informal specification.
2. How to Come up with a Correct Translation
Let's look back at our informal specification
Informal specification: Alex likes a man who likes a woman
and reformulate it in a way that is easier to translate
There exists a man whom Alex likes, and this man likes some woman
There are two parts in this sentence connected with an and
Ex (Man(x) & Likes(Alex,x))
Ey (Woman(y) & Likes(x,y))
So we have
Ex (Man(x) & Likes(Alex,x) & Ey (Woman(y) & Likes(x,y))).
You like prenex normal forms where all quantifiers are collected in one quantifier prefix. We can apply logical equivalence to get
ExEy (Man(x) & Likes(Alex,x) & Woman(y) & Likes(x,y)).
Now let's check if this statement agrees with the specification in each of the structures from the previous section.
2.1 Structure 1
Since the variables are existentially quantified and the specification holds, we need to find only one satisfying assignment for the formula.
Consider
Alex is p1
x = p2
y = p3
The conjunction holds under this assignment.
2.2 Structure 2
The same assignment as in the previous subsection can be used. In fact
Structure 1 is an substructure of Structure 2. For an existentially quantified statement we know that if it is true in a substructure, it is also true in the whole structure.
2.3 Structure 3
Since there is no pair of elements (x,y) such that Likes(x,y) in our structure, the conjunction
Man(x) & Likes(Alex,x) & Woman(y) & Likes(x,y)
cannot be true, so the statement is false. We also know that Structure 3 does not satisfy our informal specification, so our formula has passed our tests.
Our testing procedure is by no means complete. However it gives us some assurance that the translation is indeed correct.
I was reading Learn Prolog Now! and I'm confused about their use of logic and I was hoping someone can clarify the ambiguity of arguments in rules.
For example, on Chapter 1, knowledge base 4 they reference the following rule:
loves(vincent,mia).
They then provide a query:
loves(marsellus,X), woman(X).
and then the make the English statement:
"Now, remember that , means "and", so this query says: is there any individual X such that Marsellus loves X and X is a woman ?"
The key word being, Marsellus loves X.
This is worded in a way that makes me think there is some "logic" tied to the arrangement of arguments, such as to say,
Is it true that loves(vincent,mia). "logically" means:
vincent loves mia?
BUT is it false that
mia loves vincent?
That's what I originally thought but then on Chapter 3, example 2: Descendant they have the following example of:
child(bridget,caroline).
Then the English statement: "That is, Caroline is a child of Bridget"
But if #1 above is true (vincent loves mia), then shouldn't this be read as Bridget is a child of Caroline?
Or does the order of arguments not matter and it's your additional programming that will determine the logic of the arguments?
loves(X, Y) intuitively would mean X loves Y but not Y loves X. But as a programmer, you could assume it's symmetrical. It just wouldn't be as intuitive. It just depends upon how you choose to define it. Prolog Now chose the most intuitive meaning, so loves(X, Y) means X loves Y, but not Y loves X.
Whether child(X, Y) means X is a child of Y or Y is a child of X is up to the programmer. child used by itself as the name of the fact/predicate in this context is a bit ambiguous here (it's not a verb and isn't part of a phrase) and you just have to roll with whatever the Prolog Now site chose as their meaning. When you view the functor as a verb, the arguments are interpreted left to right as expected. So, better, would be to use child_of(X, Y) to mean X is a child of Y so that it's clearer.
Prolog as a language wouldn't care whether you define love(X,Y) as X loves Y or Y loves X. It's probably just a convention that we tend to follow in first order logic.
As to why we'd choose the ordering convention we usually follow, It's probably because the predicate defines a relation between two terms. So the order would depend on how you define the relation.
A predicate r(a,b) means that (a,b) is in the relation r:A->B.
The way i would clear my doubts of the case you've given in the question is:
loves(a,?) ? should anything which a loves. And
child(a,?) ? should be anything which is a child of a.
It would depend on how you define the relation 'child'.
As far as i know ( which isn't far ), There's no hard rule in logic that stops you from defining predicates the way you want. You could define love(X,Y) as 'X loves Y' or 'X is loved by Y' as long as you're consistent.
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.
This is the statement:
All birds can fly except for penguins and ostriches or unless they
have a broken wing.
Here is my attempt:
∀x birds(x)→ fly(x)^((birds(x, penguins)^birds(x,
ostriches))˅broken(wing)→¬fly(x))
is my attempt correct?
how do we present "except" in predicate logic?
thanks
On understanding "except" ...
When we say "A except B" we normally mean that A and B are mutually exclusive. Either A is the case or B is the case but not both.
If you think in terms of sets then
All birds can fly except for penguins and ostriches or unless they have a broken wing
could be re-written as
In the universe of birds, there are exactly two distinct sets -- one in which every member of the set can fly and the other in which you find penguins and ostriches and birds with broken wings.
(In passing, note the way the words "and" and "or" in English often need to be adjusted in a symbolic expression.)
Birds
+-----------+------------+
| | |
| Fly | Exceptions |
| | |
+-----------+------------+
Representing mutual exclusion in predicate logic is most easily handled by exclusive-or (XOR). We want to say fly XOR exceptions.
In systems that allow quantifiers to limit the universe of discourse, we could write:
∀x∊birds (fly(x) XOR (penguin(x) v ostrich(x) v brokenWing(x)))
If quantifiers are unlimited, then:
∀x (bird(x) → (fly(x) XOR (penguin(x) v ostrich(x) v brokenWing(x))))
And if XOR is not in the set of allowed operators, then you might have to use the equivalence:
p XOR q ≡ ((p v q) & -(p & q))
There are a couple of other implications hiding in the English sentence that are not fully expressed in the suggestions above.
The sentence in predicate logic allows the case that there are
no birds, whereas the English sentence probably implies that there is
at least one bird.
"A except B" in English normally implies that there are at
least some instances of the exception. Not only is there at least one
bird, but there is at least one penguin that cannot fly. That could
be added to the predicate sentence via appropriate use of existential
quantifiers.
"A except B" in English nearly always
implies that A is the most common case and B is the exception. In the
absence of other evidence, we would assume A. In the universe of
birds, most can fly and only the listed exceptions cannot fly. There
is no easy construct in predicate logic to capture the sense of a
majority case.
No, your attempt is incorrect. It says that all birds fly and also some birds don't fly, so it's a contradiction. Also note that broken(wing) doesn't mention x at all.
As a hint, it should look like
∀x (bird(x) ^ ¬<conditions under which birds don't fly>) → fly(x)
There is a "prolog" tag to your question. In Prolog it can be:
fly(X, WingCondition) :-
bird(X),
X \= penguin,
X \= ostrich,
WingCondition \= broken.
So all birds and not birds that are penguins, ostriches and birds with broken wings can fly
∀x (birds(x) ^ ¬ (birds(x, penguins) ^ birds(x, ostriches) ^ broken_wing(x))) → fly(x))
or this maybe
∀x (birds(x) ^ ¬ (birds(x, penguins) ^ birds(x, ostriches) ^ birds(x,broken_wing))) → fly(x))