Logic question (universal and existential quantifications) - math

I have a logical statement that says "If everyone plays the game, we will have fun".
In formal logic we can write this as:
Let D mean the people playing.
Let G be the predicate for play the game.
Let F be the predicate for having fun.
Thus [VxeD, G(x)] -> [VyeD, F(y)]
V is the computer science symbol for universal quantification. E below is the existential quantifier.
I'm looking for a way to write a similar statement using only existential quantifiers. My best guess would be that we simply need to find a way to find the counter-example where it doesn't happen, thus negate the above.
The problem is negating it doesn't make sense. We get:
[VxeD, G(x)] ^ [EyeD, !L(y)]
It's not a proper statement since the universal is still in there though it is also equivalent. Thus I need to re-fabricate my statement to something like: VxeD, VyeD, G(x) ^ F(y) I would get ExeD, EyeD, !G(x) v !F(y) which would mean "There exists someone who doesn't learn or someone else who doesn't have fun" which doesn't seem correct to me.
Some guidance or clarification would be fantastic :-)
Thanks!

I don't understand your ^ symbol, but I believe you are looking for the contrapositive. In your example, if the original statement is:
[VxeD, G(x)] -> [VyeD, F(y)]
then the contrapositive is
[ExeD, !F(x)] -> [EyeD, !G(y)]
meaning "if there is someone who is not having fun, then there exists someone not playing the game." Note that this is different than the statement in your comment above: it may well be the case that everyone is having fun, but not everyone is playing.
In general, p -> q is equivalent to !q -> !p.
(Of course I may not have understood your notation correctly.)

I'm having trouble reading your notation. I'll use A for the universal quantifier, E for the existence quantifier, F for the predicate 'having fun', G for the predicate 'playingng learned the game', then
AxL(x) -> AxF(x)
Now, you can just apply the usual gymnastics:
<==> !AxL(x) <- !AxF(x)
<==> Ex!G(x) <- Ex!F(x)
<==> Ex!F(x) -> Ex!G(x)
so, indeed, when someone's not having fun, it means not everybody played the game.

Related

How to have full control over substitution in Isabelle

In Isabelle I find myself often using
apply(subst xx)
apply(rule subst)
apply(subst_tac xx)
and similar command but often it is a hit or miss. Is there any resources on how to guide the
term unification and how to precisely specify the terms that should be substituted for?
For example if there are multiple ways to perform unification, how can I disambiguate?
If I have multiple equalities among the premises, how can I tell Isabelle which one of them to use? I spend way too much time wrestling with such seemingly simple problems.
This book https://www21.in.tum.de/~nipkow/LNCS2283/ has a chapter dedicated to substitution but it's far too short, only covers erule ssubst and doesn't really answer my questions.
To give some examples, this is ssubst
lemma ssubst: "t = s ⟹ P s ⟹ P t"
by (drule sym) (erule subst)
but what about
lemma arg_cong: "x = y ⟹ f x = f y"
by (iprover intro: refl elim: subst)
How can I do erule_tac arg_cong and specify exactly the desired f, x and y? Anything I tried resulted in Failed to apply proof method which is not a particularly enlightening error message.
As I recall, a more elaborate substitution method is available, in particular for restricting substitution to certain contexts. But to answer your question properly it's essential to know what sort of assertions you are trying to prove. If you are working with short expressions (up to a couple of lines long), then it's much better to guide the series of transformations using equational reasoning, via also and finally. (See Programming and Proving in Isabelle/HOL, 4.2.2 Chains of (In)Equations.) Then at the cost of writing out these expressions, you'll often find that you can prove each step automatically, without detailed substitutions, and you can follow your reasoning.
Also note that if your expressions are long because they contain large, repeated subexpressions, you can introduce abbreviations using define. You will then not only have shorter and clearer proofs, but you will find that automation will perform much better.
The other situation is when you are working with verification conditions dozens of lines long, or even longer. In that case it is worth looking for more advanced substitution packages.

Does using a free monad in F# imply a higher startup time and limited instructions?

I am reading this excellent article by Mark Seemann.
In it, he gives a simple demonstration of using the free monad to model interactions using pure functions.
I understand it enough to be able to write such a program, and I can appreciate the virtues of such an approach. There is one bit of code though, that has me wondering about the implications.
let rec bind f = function
| Free instruction -> instruction |> mapI (bind f) |> Free
| Pure x -> f x
The function is recursive.
Given that this is not tail recursive, does that mean that the number of instructions I include is limited by stack space, because every time I add an instruction, it has to unwrap the whole thing to add it?
Similarly to the above, does this imply a higher startup time because each new instruction has to go further down to be added? It might be negligible, but that is not the point of the question; I am trying to understand the theory, not the practicality.
If the answer to the above questions is 'Yes', would an alternative implementation along the lines of (make a list of instructions in the order you want them, then process the list in reverse order such that each instruction gets added to the top of your program, rather than deep down in the bottom) be preferable?
I may (probably) have missed something fundamental here that means that the recursion never has to go very far, please let me know.
I think the answer to the first two questions is "it depends" - there is some overhead associated with free monads and you can certainly run into stack overflows, but you can always use an implementation that avoids this and if you're doing I/O, then the overhead probably is not too bad.
I think the much bigger issue with free monads is that they are just too complicated abstraction. They let you solve one problem (abstracting over how you run code with lots of iteractions), but at a cost that you are making code very complicated. Most of the time, I think you can just keep a "functional core" as a nice testable pure part and "imperative wrapper" around it that is simple enough that you do not need to test it and abstract over it.
Free monads are very universal way of modeling this and you could use a more concrete representation. For reading and writing, you could do:
type Instruction =
| WriteLine of string
| ReadLine of (string -> Instruction list)
As you can see, this is not just a simple list - ReadLine is tricky, because it takes a function that, when called with the string from the console, produces more instructions (that may, or may not, depend on this string):
let program =
[ yield WriteLine "Enter your name:"
yield ReadLine (fun name ->
[ if name <> "" then
yield WriteLine ("Hello " + name) ] ) ]
This says "Hello " but only when you enter non-empty name, so it shows how the instructions depend on the name you read. The code to run a program then looks like this:
let rec runProgram program =
match program with
| [] -> ()
| WriteLine s :: rest ->
Console.WriteLine s
runProgram rest
| ReadLine f :: rest ->
let input = Console.ReadLine()
runProgram (f input # rest)
The first two cases are nice and fully tail-recursive. The last case is tail-recursive in runProgram, but it is not tail recursive when calling f. This should be fine, because f only generates instructions and does not call runProgram. It also uses slow list append #, but you could fix that by using a better data structure (if it actually turned out to be a problem).
Free monads are basically an abstraction over this - but I personally think that they are going one step too far and if I really needed this, then something like the above - with concrete instructions - is better solution because it's simpler.

Isabelle/HOL foundations

I have seen a lot of documentation about Isabelle's syntax and proof strategies. However, little have I found about its foundations. I have a few questions that I would be very grateful if someone could take the time to answer:
Why doesn't Isabelle/HOL admit functions that do not terminate? Many other languages such as Haskell do admit non-terminating functions.
What symbols are part of Isabelle's meta-language? I read that there are symbols in the meta-language for Universal Quantification (/\) and for implication (==>). However, these symbols have their counterpart in the object-level language (∀ and -->). I understand that --> is an object-level function of type bool => bool => bool. However, how are ∀ and ∃ defined? Are they object-level Boolean functions? If so, they are not computable (considering infinite domains). I noticed that I am able to write Boolean functions in therms of ∀ and ∃, but they are not computable. So what are ∀ and ∃? Are they part of the object-level? If so, how are they defined?
Are Isabelle theorems just Boolean expressions? Then Booleans are part of the meta-language?
As far as I know, Isabelle is a strict programming language. How can I use infinite objects? Let's say, infinite lists. Is it possible in Isabelle/HOL?
Sorry if these questions are very basic. I do not seem to find a good tutorial on Isabelle's meta-theory. I would love if someone could recommend me a good tutorial on these topics.
Thank you very much.
You can define non-terminating (i.e. partial) functions in Isabelle (cf. Function package manual (section 8)). However, partial functions are more difficult to reason about, because whenever you want to use its definition equations (the psimps rules, which replace the simps rules of a normal function), you have to show that the function terminates on that particular input first.
In general, things like non-definedness and non-termination are always problematic in a logic – consider, for instance, the function ‘definition’ f x = f x + 1. If we were to take this as an equation on ℤ (integers), we could subtract f x from both sides and get 0 = 1. In Haskell, this problem is ‘solved’ by saying that this is not an equation on ℤ, but rather on ℤ ∪ {⊥} (the integers plus bottom) and the non-terminating function f evaluates to ⊥, and ‘⊥ + 1 = ⊥’, so everything works out fine.
However, if every single expression in your logic could potentially evaluate to ⊥ instead of a ‘proper‘ value, reasoning in this logic will become very tedious. This is why Isabelle/HOL chooses to restrict itself to total functions; things like partiality have to be emulated with things like undefined (which is an arbitrary value that you know nothing about) or option types.
I'm not an expert on Isabelle/Pure (the meta logic), but the most important symbols are definitely
⋀ (the universal meta quantifier)
⟹ (meta implication)
≡ (meta equality)
&&& (meta conjunction, defined in terms of ⟹)
Pure.term, Pure.prop, Pure.type, Pure.dummy_pattern, Pure.sort_constraint, which fulfil certain internal functions that I don't know much about.
You can find some information on this in the Isabelle/Isar Reference Manual in section 2.1, and probably more elsewhere in the manual.
Everything else (that includes ∀ and ∃, which indeed operate on boolean expressions) is defined in the object logic (HOL, usually). You can find the definitions, of rather the axiomatisations, in ~~/src/HOL/HOL.thy (where ~~ denotes the Isabelle root directory):
All_def: "All P ≡ (P = (λx. True))"
Ex_def: "Ex P ≡ ∀Q. (∀x. P x ⟶ Q) ⟶ Q"
Also note that many, if not most Isabelle functions are typically not computable. Isabelle is not a programming language, although it does have a code generator that allows exporting Isabelle functions as code to programming languages as long as you can give code equations for all the functions involved.
3)
Isabelle theorems are a complex datatype (cf. ~~/src/Pure/thm.ML) containing a lot of information, but the most important part, of course, is the proposition. A proposition is something from Isabelle/Pure, which in fact only has propositions and functions. (and itself and dummy, but you can ignore those).
Propositions are not booleans – in fact, there isn't even a way to state that a proposition does not hold in Isabelle/Pure.
HOL then defines (or rather axiomatises) booleans and also axiomatises a coercion from booleans to propositions: Trueprop :: bool ⇒ prop
Isabelle is not a programming language, and apart from that, totality does not mean you have to restrict yourself to finite structures. Even in a total programming language, you can have infinite lists. (cf. Idris's codata)
Isabelle is a theorem prover, and logically, infinite objects can be treated by axiomatising them and then reasoning about them using the axioms and rules that you have.
For instance, HOL assumes the existence of an infinite type and defines the natural numbers on that. That already gives you access to functions nat ⇒ 'a, which are essentially infinite lists.
You can also define infinite lists and other infinite data structures as codatatypes with the (co-)datatype package, which is based on bounded natural functors.
Let me add some points to two of your questions.
1) Why doesn't Isabelle/HOL admit functions that do not terminate? Many other languages such as Haskell do admit non-terminating functions.
In short: Isabelle/HOL does not require termination, but totality (i.e., there is a specific result for each input to the function) of functions. Totality does not mean that a function is actually terminating when transcribed to a (functional) programming language or even that it is computable at all.
Therefore, talking about termination is somewhat misleading, even though it is encouraged by the fact that Isabelle/HOL's function package uses the keyword termination for proving some property P about which I will have to say a little more below.
On the one hand the term "termination" might sound more intuitive to a wider audience. On the other hand, a more precise description of P would be well-foundedness of the function's call graph.
Don't get me wrong, termination is not really a bad name for the property P, it is even justified by the fact that many techniques that are implemented in the function package are very close to termination techniques from term rewriting or functional programming (like the size-change principle, dependency pairs, lexicographic orders, etc.).
I'm just saying that it can be misleading. The answer to why that is the case also touches on question 4 of the OP.
4) As far as I know Isabelle is a strict programming language. How can I use infinite objects? Let's say, infinite lists. Is it possible in Isabelle/HOL?
Isabelle/HOL is not a programming language and it specifically does not have any evaluation strategy (we could alternatively say: it has any evaluation strategy you like).
And here is why the word termination is misleading (drum roll): if there is no evaluation strategy and we have termination of a function f, people might expect f to terminate independent of the used strategy. But this is not the case. A termination proof of a function rather ensures that f is well-defined. Even if f is computable a proof of P merely ensures that there is an evaluation strategy for which f terminates.
(As an aside: what I call "strategy" here, is typically influenced by so called cong-rules (i.e., congruence rules) in Isabelle/HOL.)
As an example, it is trivial to prove that the function (see Section 10.1 Congruence rules and evaluation order in the documentation of the function package):
fun f' :: "nat ⇒ bool"
where
"f' n ⟷ f' (n - 1) ∨ n = 0"
terminates (in the sense defined by termination) after adding the cong-rule:
lemma [fundef_cong]:
"Q = Q' ⟹ (¬ Q' ⟹ P = P') ⟹ (P ∨ Q) = (P' ∨ Q')"
by auto
Which essentially states that logical-or should be "evaluated" from right to left. However, if you write the same function e.g. in OCaml it causes a stack overflow ...
EDIT: this answer is not really correct, check out Lars' comment below.
Unfortunately I don't have enough reputation to post this as a comment, so here is my go at an answer (please bear in mind I am no expert in Isabelle, but I also had similar questions once):
1) The idea is to prove statements about the defined functions. I am not sure how familiar you are with Computability Theory, but think about the Halting Problem and the fact most undeciability problems stem from it (such as Acceptance Problem). Imagine defining a function which you can't prove it terminates. How could you then still prove it returns the number 42 when given input "ABC" and it doesn't go in an infinite loop?
If instead you limit yourself to terminating functions, you can prove much more about them, essentially making a trade-off (or at least this is how I see it).
These ideas stem from Constructivism and Intuitionism and I recommend you check out Robert Harper's very interesting lecture series: https://www.youtube.com/watch?v=9SnefrwBIDc&list=PLGCr8P_YncjXRzdGq2SjKv5F2J8HUFeqN on Type Theory
You should check out especially the part about the absence of the Law of Excluded middle: http://youtu.be/3JHTb6b1to8?t=15m34s
2) See Manuel's answer.
3,4) Again see Manuel's answer keeping in mind Intuitionistic logic: "the fundamental entity is not the boolean, but rather the proof that something is true".
For me it took a long time to get adjusted to this way of thinking and I'm still not sure I understand it. I think the key though is to understand it is a more-or-less completely different way of thinking.

Understanding Recursive Rule and Unification in Prolog

I'm a beginning Prolog student following the "LearnPrologNow!" set of tutorials. I'm doing my best to get a grip on the concepts and vocabulary. I've been able to understand everything up until Chapter 3 on Recursive Definitions when presented with this problem:
numeral(0).
numeral(succ(X)) :- numeral(X).
given the query
numeral(X).
Now, I understand that the idea of the program is that Prolog will begin counting numbers in this system in a sequence such as
X=0
X=succ(0)
X=succ(succ(0))
But I do not understand what causes it to "scale back" and ascend each time. I understand the principle of unification in that the program is trying to unify the query of X, but should it just follow the recursive rule once, and then return zero? What allows it to add a succ() around the query? Is that not traversing the recursive rule in the opposite direction?
Please think declaratively:
The rule
numeral(succ(X)) :- numeral(X).
means:
If X is a numeral, then succ(X) is a numeral.
:- is like an arrow used in logical implication (it looks similar to <==).
Seeing that you successfully derived that 0 is a numeral (first answer), it is thus no surprise that succ(0) is another solution.
I recommend you think in terms of such relations, instead of trying to trace the actual control flow.
Notice that succ/1 is not added "around the query", but is a part of the actual answer. The term succ(0) is just a normal Prolog term, with functor succ and argument 0.
Already given answer being good, i'll add some more:
Prolog uses denotational syntax (or declarative syntax) to define logical relations/"equations" between terms
A term is an object comprised of variables/functions/placeholders etc..
Unification is the process to check if two expressions (or two terms) can be equal with respect to the given relations/equations.
numeral(succ(X)) :- numeral(X)
Is such a relation/equation which says that the fact that variable-term X is of numeral type (or class), implies the successor functional succ is also of same type. So Prolog here can unify the expression (in other words solve the equation) and replace X with succ(X) and so on, untill the domain of X is covered. So this unification implies X replaced by succ(X) and then unification can be re-applied.
Just to add a proof tree to this answers, which may make things more clear for others:
Base call: numeral(succ(succ(0))).
: ^
rule1 : : {X/0}
: :
V :
numeral(succ(X)).
: ^
rule1 : : {X/0}
: :
V :
numeral(X).
: ^
fact1 : : {X/0}
: :
V :
Found proof [.]
You start with the downwards arrows and move then back up to the previous calls with the new found unificator in the last step. Please note that Prolog declares each variable in each step as a new variable, which I omitted in this scheme.

Fixed-Point Combinators

I am new to the world of fixed-point combinators and I guess they are used to recurse on anonymous lambdas, but I haven't really got to use them, or even been able to wrap my head around them completely.
I have seen the example in Javascript for a Y-combinator but haven't been able to successfully run it.
The question here is, can some one give an intuitive answer to:
What are Fixed-point combinators, (not just theoretically, but in context of some example, to reveal what exactly is the fixed-point in that context)?
What are the other kinds of fixed-point combinators, apart from the Y-combinator?
Bonus Points: If the example is not just in one language, preferably in Clojure as well.
UPDATE:
I have been able to find a simple example in Clojure, but still find it difficult to understand the Y-Combinator itself:
(defn Y [r]
((fn [f] (f f))
(fn [f]
(r (fn [x] ((f f) x))))))
Though the example is concise, I find it difficult to understand what is happening within the function. Any help provided would be useful.
Suppose you wanted to write the factorial function. Normally, you would write it as something like
function fact(n) = if n=0 then 1 else n * fact(n-1)
But that uses explicit recursion. If you wanted to use the Y-combinator instead, you could first abstract fact as something like
function factMaker(myFact) = lamba n. if n=0 then 1 else n * myFact(n-1)
This takes an argument (myFact) which it calls were the "true" fact would have called itself. I call this style of function "Y-ready", meaning it's ready to be fed to the Y-combinator.
The Y-combinator uses factMaker to build something equivalent to the "true" fact.
newFact = Y(factMaker)
Why bother? Two reasons. The first is theoretical: we don't really need recursion if we can "simulate" it using the Y-combinator.
The second is more pragmatic. Sometimes we want to wrap each function call with some extra code to do logging or profiling or memoization or a host of other things. If we try to do this to the "true" fact, the extra code will only be called for the original call to fact, not all the recursive calls. But if we want to do this for every call, including all the recursive call, we can do something like
loggingFact = LoggingY(factMaker)
where LoggingY is a modified version of the Y combinator that introduces logging. Notice that we did not need to change factMaker at all!
All this is more motivation why the Y-combinator matters than a detailed explanation from how that particular implementation of Y works (because there are many different ways to implement Y).
To answer your second question about fix-point combinators other than Y. There are countably infinitely many standard fix-point combinators, that is, combinators fix that satisfy the equation
fix f = f (fix f)
There are also contably many non-standard fix-point combinators, which satisfy the equation
fix f = f (f (fix f))
etc. Standard fix-point combinators are recursively enumerable, but non-standard are not. Please see the following web page for examples, references and discussion.
http://okmij.org/ftp/Computation/fixed-point-combinators.html#many-fixes

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