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.
Related
I'm just beginning to learn Coq via software foundations. One of the homework Theorems (with my successful proof elided) in Induction.v is:
Theorem plus_n_Sm : forall n m : nat,
S (n + m) = n + (S m).
Proof.
(* elided per request of authors *)
Qed.
Later, I noticed that the following similar "leftward" statement comes for free with the built-in tactic .simpl:
Example left_extract : forall n m : nat, S n + m = S (n + m).
Proof.
intros. simpl. reflexivity.
Qed.
I've perused the documentation and haven't been able to figure out why .simpl gives us one direction "for free" but the other direction requires a user-supplied proof. The documentation is over my head at this very early point in my learning.
I guess it has something to do with left-ness being built-in and right-ness being not, but the propositions seem to my childlike eyes to be of equal complexity and subtlety. Would someone be so kind as to explain why, and perhaps give me some guidance about what is going on with .simpl?
Why should I NOT be surprised by my finding?
What other good things can I expect from .simpl, so it surprises me less and so I can eventually predict what it's going to do and rely on it?
What's the best way to wade through the theory -- unfolding of iota reductions and what not -- to focus on the relevant bits for this phenomenon? Or do I have to learn all the theory before I can understand this one bit?
I believe your surprise stems from the fact that you are accustomed to think of addition as a primitive concept. It is not, it is a function that has been defined and other primitive concepts are used to explain its behavior.
The addition function is defined by a function with a name written with letters (not the + symbol) in a way that looks like this:
Fixpoint add (n m : nat) : nat :=
match n with
| 0 =>
| S p => S (add p)
end.
You can find this information by typing
Locate "_ + _".
The + notation is used for two functions, only one of them can be applied on numbers.
Coming back to the add function, its very description explains that add 0 m computes to m and add (S n) m computes to S (add m n), but it does not say anything when the second argument has the form S m, it is just not written anywhere. Still the theory of Coq makes it possible to prove the fact.
So the equality that is given for free comes directly from the definition of addition. There are a few other equality statements that are natural to the human eye, but not given for free. They can just be proved by induction.
Would it be possible to design the Coq system in such a way that both equality statements would be given for free? The answer is probably yes, but the theory must be designed carefully for that.
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 looks like such an easy problem but still can't figure it out. How do I prove ¬(¬a = a)?
No given premises.
I got this so far (in Fitch):
This is a subproof where I assume the negation of my goal and then try to reach the absurd/contradiction so I can state the negation of my assumption, which would be my goal.
Thanks in advance!
Looking at your screenshot I'd say your =Intro introduces a variable a (that is, a is an object of the domain, rather than a predicate).
I say this because
in all books I've read, the =Intro rule is used for objects rather than predicates, and
for predicates, equals is expressed as "if and only if" which is typically written as ↔ and not =.
So, in other words, the only sensible interpretation of ¬(¬a = a) is that = binds harder than ¬, and the whole formula should be interpreted as ¬(¬(a = a)).
Now you should be able to
introduce a = a
assume the contrary: ¬(a = a)
arrive at a contradiction, ⊥, based on 1. and 2.
Use ¬Intro on 2 and 3 to get ¬(¬(a = a)).
Given a set of constraints, I would like to efficiently generate the set of values.
Suppose I have a few constraints on my Thungus[1]:
goodThungus(X) :-
X > 100,
X < 1000.
sin(X) = 0.
Now, I can check a Thungus by asking:
goodThungus(500).
I would like to generate all good Thungi. I'm not sure how to do that; I'm really not sure about how to do it efficiently.
Note: this of course has to be a computable generation.
[1] Arbitrary object selected for this example.
What you are asking for can't be done in the full general case: imagine doing f(X) = 0 where f is a function for which the roots cannot be analytically determined, for example. Or suppose f(X) is the function "does the program X halt?". No computer is going to solve that for you.
Your options are basically to either:
Limit the set of constraints to things that you can reason about. e.g. inequalities are good because you can identify ranges, then do intersections and unions on ranges efficiently etc.
Limit the set of values to a small enough number that you can test them individually against each of the constraints
UPDATE: For the kind of constraints stated in the question (ranges of real values and real-valued functions that can be analytically solved and have a finite number of solutions within any range) I would suggest the following approach:
Write a generating function that can iteratively return solutions for you function within a given range... this will need to be done analytically e.g. exploiting the fact that sin(X)=0 implies X=n*pi where n is any integer.
Do interval arithmetic and bounding on your range constraints to work out the range(s) that need to be scanned (in the example you would want the range 100 < X < 1000)
Apply your generating function to each of the target ranges in order to create all of the possible solutions.
I'll preface my suggestion by stating that I'm no expert in using numerical constraint logic programming systems, but here goes...
On the surface, I'd think that solving this kind of problem in PROLOG would be best suited to a numerical constraint logic programming system, perhaps such as CLP(R) (for reals) in SWI-PROLOG; unfortunately, the specific problem you've asked for is seeking to solve for a set of constraints including a non-linear constraint, which seems to be not well or widely supported amongst PROLOG implementations; instead, they seem to deal mainly with linear constraints and often have limited support for non-linear constraints such as X = sin(Y), for example.
Take SWI-PROLOG's CLP(R) library, and the following example program:
:- use_module(library(clpr)).
report_xsq_zeros :-
findall(X, {0 = (X * X) - 10}, Results),
write_ln(Results).
report_sin_zeros :-
findall(X, {0 = sin(X)}, Results),
write_ln(Results).
Now, executing report_xsq_zeros gives us:
?- report_xsq_zeros.
[3.16228, -3.16228]
true.
Here, the system correctly computed the zeros of the quadratic x^2 - 10, which are indeed approximately 3.16228 and -3.16228, where the range of X was unbounded. However, when we execute report_sin_zeros, we get:
?- report_sin_zeros.
[0.0]
true.
We see that the system only computed a single zero of the function sin(X), even though the range of X was indeed also unbounded. Perhaps this is because it is recognized that there are an infinite number of solutions here (though I'm only guessing...). If we were to program what you've asked for:
report_sin_zeros :-
findall(X, {X > 100, X < 1000, 0 = sin(X)}, Results),
write_ln(Results).
We get no results, as the underlying system only computed a single zero for sin(X) as shown earlier (i.e., binding X to 0.0 which lies outside the stated range):
?- report_sin_zeros.
[]
true.
I conclude that I've either not demonstrated proper usage of SWI-PL CLP(R) (I suggest you look into it yourself), or it won't solve your specific (non-linear) problem. Other CLP(R) implementations may behave differently to SWI-PROLOG CLP(R), but I don't have them installed so I can't check, but you could try SICSTUS CLP(R) or others; the syntax looks similar.
He is searching any X in [100..1000] for that sin(x) = 0. But this is a pure mathematical problem, and not meant for relational logical deduction / backtracking. simple Prolog is not suited for this?
I am having problems for converting following algo in ocaml To implement this algorithm i used Set.Make(String) functor actualy in and out are 2 functions Can any one give me percise code help in ocaml for following
This is actually Algo for Live Variables[PDF] ..Help would be appreciated greatly
for all n, in[n] = out[n] = Ø
w = { set of all nodes }
repeat until w empty
n = w.pop( )
out[n] = ∪ n’ ∈ succ [n] in[n’]
in[n] = use[n] ∪ (out[n] — def [n])
if change to in[n],
for all predecessors m of n, w.push(m)
end
for all n, in[n] = out[n] = Ø
w = { set of all nodes }
repeat until w empty
n = w.pop( )
out[n] = ∪ n’ ∈ succ [n] in[n’]
in[n] = use[n] ∪ (out[n] — def [n])
if change to in[n],
for all predecessors m of n, w.push(m)
end
It's hard for me to tell what is exactly going on here. I think there is some alignment issues with your text --repeat until w empty should be repeating the next 5lines, right? And how are in and out functions, they look like arrays to me? Aside from those deficiencies I'll tackle some general rules I have followed.
I've had to translate a number of numerical methods in C and Fortran algorithms to functional languages and I have some suggestions for you.
0) Define the datatypes being used. This will really help you with the next step (spoiler: looking for functional constructs). Once you know the datatypes you can more accurately define the functional constructs you will eventually apply.
1) Look for functional constructs. fold, recursion, and maps, and when to use them. For example, the for all predecessors m is a fold (unsure if that it would fold over a tree or list, but none-the-less, a fold). While loops are a good place for recursion --but don't worry about making it a tail call, you can modify the parameters later to adhere to those requirements. Don't worry about being 100% pure. Remove enough impure constructs (references or arrays) to get a feel for the algorithm in a functional way.
2) Write any part of the algorithm that you can. Leaving functions blank, add dummy values, and just implement what you know --then you can ask SO better, more informed questions.
3) Re-write it. There is a good chance you missed some functional constructs or used an array or reference where you now realize you can use a list or set or by passing an accumulator. You may have defined a list, but later you realize you cannot randomly access it (well, it would be pretty detrimental to), or it needs to be traversed forward and back (a good place for a zipper). Either way, when you finally get it you'll know, and you should have a huge ear-to-ear grin on your face.