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.
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 question regarding Knights and Knaves and logical proposition. If I want to solve the puzzle and I assume I have two kinds of citizens: Knights, who always tell the truth, and knaves, who always tell lies. On the basis of utterances from some citizens, I must decide what kind they are.
There are three kinds of citizens: a, b and c, who are talking about themselves:
a says: ”All of us are knaves.”
b says: ”Exactly one of us is a knight.”
To solve the puzzle I should determine: What kinds of citizens are a, b and c? I should solve the puzzle by modelling the two utterances above using propositional logic, and I assume that I can use p to describe a knight and ¬p to describe a knave. How would I go about doing that? Any hint for someone who hasn't done any noticeable discrete mathematics in college?
A and C are Knaves. B is a Knight.
If A is a Knight, "All of us are knaves" is true. So, A would also be a Knave. This is a contradiction. Hence, A is a Knave.
If B is a Knave, then "Exactly one of us is a knight." is false. Meaning that 2 or more are Knights. But neither A nor B is a Knight. How can possibly be 2 or more Knights (since C is the only one with a possibility of being a Knight). This is also a contradiction. So, B is a Knight.
We just showed that B is a Knight. So, he himself is the only Knight he is talking about. So, C is a Knave.
Now, I don't think you can model this argument in the Propositional Logic. For one, notice the universal and existential quantifiers ("All" and "Exactly One") in the statements "All of us are knaves" and "Exactly one of us is a knight.". For another one, notice that A and B are talking about themselves. Modeling situations like this is one of the hardest problems in the history (not kidding!). Look at the following links for more info:
https://en.wikipedia.org/wiki/Liar_paradox
https://en.wikipedia.org/wiki/Self-reference
you can create a Truth table,
by first look at it i can say A must be knave, and B is a knight. because if A is a knight he can't say he's a knave(lie), also he can't be right about that all are knaves (can't say the truth) so B is a knight(if B Knave he can't say the truth that makes A a liar and he must be one) and then C is a Knave.
In lambda calculus, if a term has normal form, normal order reduction strategy will always produce it.
I just wonder how to prove the above proposition strictly?
The result you mention is a corollary of the so called standardization theorem, stating that for any reduction sequence M->N there is another "standard" one between the same terms M and N, where you execute redexes in leftmost outermost order. The proof is not so trivial, and there are several different approaches in the literature. I add a short bibliography below.
The recent proof by Kashima 5 (see also 1) has the advantage of not using the notion of residual and of being based on purely inductive techniques. It is also good for formalization 2, but unless you are not already confident with the subject, it is not particularly instructive.
The general idea behind standardization is the following.
Suppose to have two redexes R and S, where S is in leftmost outermost position with respect to R, and consider the following reduction:
R S
M -> P -> N
Then, you can start firing S, instead, but in this way you can possibly duplicate (or erase) the redex R. These redexes, that are essentially what remains of R after firing S, are called residuals, and are usually indicated as R/S (read: residuals of R after S).
So, the basic lemma is that
R S = S (R/S)
In order to use it for standardization, we need to generalize R to an arbitrary sequence ρ (that we may assume to be standard, with no redex in leftmost outermost position w.r.t. S). It is still true that
(*) ρS = S (ρ/S)
but what is not so evident is the standardization of (ρ/S). To this aim,
let us observe that ρ was performed before firing S = C[\x.M N], that
essentially splits the term in three disconnected regions: the context C, M, and N.
This induces a repartition of ρ in three consecutive sequences:
ρ1 inside M
ρ2 inside N
ρ3 inside C
(remember that no redex was in leftmost outermost position w.r.t. S).
The only part that can be duplicated (or erased) is ρ2, and the residuals
ρ2-0 ... ρ2-k are easily ordered according to the different positions of
the k copies of N created by the firing of S. So
S ρ1 ρ2-0 ... ρ2-k ρ_3
is the standard version of (*).
Basic bibliography.
1 A.Asperti, JJ. Levy. The cost of usage in the lambda-calculus. LICS 2013.
3 H. P. Barendregt. The Lambda Calculus, North-Holland (1984).
4 G.Gonthier, JJ. Levy, PA. Mellies. An abstract standardisation theorem. LICS ’92.
2 F.Guidi. Standardization and Confluence in Pure Lambda-Calculus
Formalized for the Matita Theorem Prover. Journal of Formalized Reasoning
5(1):1–25, 2012.
5 R.Kashima. A proof of the standardization theorem in lambda-calculus.
Technical Report C-145,, Tokyo Institute of Technology, 2000.
[6] JW. Klop. Combinatory Reduction Systems. PhD thesis, CWI,
Amsterdam, 1980.
[7] G.Mitschke. The standardisation theorem for the lambda-calculus.
Z. Math.Logik. Grundlag. Math, 25:29–31, 1979
[8] M.Takahashi. Parallel reductions in lambda-calculus.
Information and Computation 118, pp.120-127, 1995.
[9] H. Xi, Upper bounds for standardizations and an application. Journal of Symboloc Logic 64, pp.291-303, 1999.
I want to know why the order of quantifiers are important in a logic formula?
When I read books about logic programming, such points are mentioned, but did not say why.
Is there any one could explain with some examples?
Also, how can we determine order of quantifiers from a given logic formula?
Thanks in advance!
You would be well advised to read a book about first-order logic before the books about
logic programming.
Consider the true statement:
1. Everybody has a mother
Let's formalize it in FOL. To keep it simple, we'll say
that the universe of discourse is the set of people, i.e.
our individual variables x, y, z... range over people. Then
1 becomes:
1F. (x)(Ey)Mother(y,x)
which we can read as: For every person x there exists
some person y such that y is the mother of x.
Now let's swap the order of the universal quantifier (x) and existential
quantifier (Ey):
2F. (Ey)(x)Mother(y,x)
That reads: There is some person y such that for every person x,
y is the mother of x. Or in plain English:
2. There is somebody who is the mother of everybody
You see that swapping the quantifiers changes the meaning of the statement,
taking us from the true statement 1 to the false statement 2. Indeed, to the absurdly false statement
2, which entails that somebody is their own mother.
That's why the order of quantifiers matters.
how can we determine order of quantifiers from a given logic formula?
Well, in 1F and 2F, for example, all the variables are already bound by quantifiers,
so there's nothing to determine. The order of the quantifiers is what you see,
left to right.
Suppose one of the variables was free (not bound), e.g.
3F. (Ey)Mother(y,x)
You might read that as: There is someone who is the mother of x, for variable person x.
But this formula really doesn't express any statement. It expresses a unary predicate of persons, the predicate Someone is the mother of x. If you free up the remaining variable:
4F. Mother(x,y)
then you have the binary predicate, or relation: x is the mother of y.
A formula with 1,2,...,n free variables expresses a unary, binary,...,n-ary predicate.
Given a predicate, you can make a statement by binding free variables with quantifiers and/or substituting individual constants for the free variables. From 4F you can make:
(x)(y)Mother(x,y) (Everybody is everybody's mother)
(Ex)(y)Mother(x,y) (Somebody is everybody's mother)
(Ex)(Ey)Mother(x,y) (Somebody is somebody's mother)
(x)Mother(x,Arnold) (Everybody is the mother of Arnold)
(x)Mother(Bernice,x) (Bernice is the mother of everybody)
Mother(Arnold,Bernice) (Arnold is the mother of Bernice)
...
...
and so on ad nauseam.
What this should make clear is that if a formula has free variables, and therefore expresses
a predicate, the formula as such does not imply any particular way of quantifying
the free variables, or that they should be quantified at all.
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.