1) K forces (for all x A(x)) –> Q

2) but K does not force exists x (A(x) –> Q)

The intuitive reason behind it is that

for all x A(x) or exists x not A(x)

is not constructive or in computability sense, it is not decidable.

If we think the implication in A –> B as a universal machine that construct a proof of B by getting a proof of A as input (BHK interpretation),

A –> B

and

not A or B

are not alike.

So, by knowing the set X, we can not construct or compute a real number ‘b’ in X such that

|b| diam(X)<= 2

Or am I missing something? ]]>

Saying “But that is clearly nonsense. If all we know is that one particular element of X has modulus at most 1, it can’t possibly imply that X has diameter at most 2.” is similar to saying that the following is clearly nonsense. If my cat is black, then 1+1=2. Important thing to note is that “1+1=2” is true regardless of my cat’s color (even existence of my cat).

In that sense, if diam(X) \le 2, then there is nothing to worry about. We can choose any x in X either with |x| \le 1 or |x| > 1, the conclusion, and hence, the implication is true. However, if diam(X) > 2, then surely we can find x with |x| > 1 which makes the implication true, since premise is false.

If there is no element in X, then premises in both statements are vacuously true and thus, the conclusions and the implications have the same values. Since both conclusions are the same, namely diam(X) \le 2, both statements have the same truth value. (Of course, the natural definition would be diam(empty set) = 0 which makes both statements true.)

]]>Then, the second to last statement would also be false, as the left part is true and the right part is false. By establishing the contrapositive, we see that indeed the second to last statement implies the last statement.

]]>I think this does not match the formula. For that sentence I would give this formula:

forall X \subset R, (\exists x \in X, |x| \le 1) \implies diam(X) \le 2 (*)

Sorry for the non-latex.

(*) differs from the last formula in your post because of the bracketing. (*) is certainly false as your counter-example shows.

It is also interesting to convert the last formula of your post to its contra-positive and then move the constant proposition out of the scope of the \exists x. That does say something interesting:

\forall X \subset \R, diam(X) > 2 \implies (\exists x \in X, |x| > 1).

]]>The original logical argument, as applied to the counter-example, appears legitimate until the violation of the “we haven’t done something foolish” prior to the promotion of the existential quantifier.

]]>Here is how to get from the last formula to the verbal description: the formula says that there is an x in X such that the following statement holds:

If all we know is that x has modulus at most 1, then that implies X has diameter at most 2.

Now this statement is nonsense for any given x. Thus the last formula is false.

———

I think the solution lies in the meaning of “implies,” as others have said.

]]>But consider how you would go about proving that second (purportedly different) claim. If you can’t exhibit the x with that bizarre property (how could you?), the best way to prove it would be to assume it is false and then find a contradiction. Then you are assuming that all the x’s are such that they are P and Q is false. But this is exactly what you would have gotten if you had tried to prove the original statement by contradiction. So the two really are equivalent, since their negations are.

]]>I think that, in a proof, a chain like “A implies B implies C” is actually a shorthand notation of “(A implies B) and (B implies C)”, similar to “a < b < c" in arithmetics. In a strict sense, such a chain is syntactically incorrect, if you think of "<" as a binary operator that takes two numbers and returns a boolean. However, everybody knows you actually mean a<b and b<c.

]]>I realised that I was using “implies” in two different senses when I was an undergraduate. I would write a proof “A implies B implies C” (often a longer chain) meaning “If I know A to be true then I can establish the truth of C by proving that B is true as an intermediate step” or something of the kind. But this is not the same kind of statement as either “(A implies B) implies C” or “A implies (B implies C)” and that always worried me a bit. The kind of implication I use in an informal way in a proof is associative. The same word, two different ideas. I think Thurston’s idea of a “verb” comes from the informal version.

]]>Thanks for that quote. Actually, in a way I disagree with Thurston that “or”, “and” and “implies” have identical formal usage, though it is hard to explain what I mean by that, when it seems obviously false. Roughly speaking, it’s fine to say “P and Q” or “P or Q” when we have two statements P and Q that contain nothing but free variables. But it’s very rare to say that P implies Q when P and Q do not involve free variables.

That’s not quite true as stated, so let me illustrate with an example in an effort to say what I mean. Suppose I say “‘n is a prime greater than 2’ implies ‘n is odd'”. Then what I’m really saying is that *for every n* that holds. So n isn’t (despite appearances) a free variable. It’s slightly complicated because I could be in the middle of the argument and I might have said at some point “Let n be an integer such that P(n)” and I might have deduced that n is a prime greater than 2. So technically n would be a free variable. But it would be the kind of “arbitrary positive integer” rather than some fixed integer.

I wouldn’t, for instance, find it natural to say “’17 is a prime greater than 2′ implies ’17 is odd'” because that suggests the possibility that 17 is not a prime greater than 2. (I would find it natural to say, “17 is a prime greater than 2. Therefore, 17 is odd.” But that’s different, and is appealing to the more general fact about all n.)

So I would say that “implies” is a verb because if we say “P(x) implies Q(x)” we are talking about a certain relationship between a general instance of P and a general instance of Q. But if we say “P(x) and Q(x)” we are talking about a specific x and are asserting that both P(x) and Q(x) hold.

]]>The result

∃x∈X (P(x)⇒Q)

becomes

(A ⇒ (A⋀B)) ∨ (B ⇒ (A⋀B))

Using (A⋀B)⇒A and (A⋀B)⇒B, we have.

(A ⇒ B) ∨ (B ⇒ A)

This proves what I’ll call “Florian’s all-connectedness theorem”:

∀A, B : (A ⇒ B) ∨ (B ⇒ A)

It proves the extraordinary fact that everything is connected. For any 2 facts A and B, if B doesn’t follow from A, then A follows from B.

]]>But then the text goes on:

“But that is clearly nonsense. If all we know is that one particular element of X has modulus at most 1, it can’t possibly imply that X has diameter at most 2.”

However, the second sentence does not describe the formula correctly. What it describes is this:

And yes, that would indeed be nonsense.

]]>∀x∈X P(x)

So this should be rewritten first, I suppose:

∃X ( ∀x∈X P(x) )

Now we could also notice that the fragment:

∀x∈X P(x)

is actually an abbreviation of:

∀x (x∈X → P(x)

which introduces an important shift in the conclusion.

Indeed,

(∀x (x∈X → P(x)) → Q

leads to:

(∃x)( (x∉X→Q) & (P(x)→Q) )

]]>It is equally true that the Riemann hypothesis implies Fermat’s last theorem (Andrew Wiles proved that) or that that the existence of a pink elephant implies FLT (idem). It just breaks the implicit promise that the premise is somehow relevant to the conclusion.

]]>with the restriction that the variable does not appear in .

The paradox is created because there is no rule

regardless if has as free variable .

More strictly, if we assume that the is a single predicate of ie then the propositions

and are not equivalent .

In the language of predicate logic the proposition

is written as and so if we apply the rule of movement of quantifiers we have

While the proposition is written as

, hence

Concerning the example you mentioned – which is a proof of why these two propositions are not equivalent – we have that the proposition

is equivalent to the proposition

which creates semantics problems.But if we look the equivalent proposition which is

which translates “for all X there exists x so that if the diameter of X is greater than 2 then x belongs to X and the distance of x from 0 is greater than 1 ” the semantics problems which the reader may face are less than before.

If we apply the rule for quantifiers movement

if has no free occurrences of then we have the equivalent proposition

then the English translation is

” For each set X if the diameter of X is greater than 2 then there is an x such that x belongs to X and the distance from 0 is greater than 1 ” and the meaning is obvious.

** We used the symbol for the logical operator implies and the symbol for tautologically equivalent propositions .

]]>Your first statement is obviously true since it is just the tautology “if the Goldbach conjecture is true then the Goldbach conjecture is true”. The final statement is more subtle, but it is still true. Indeed, either the Goldbach conjecture is true or it isn’t.

If it is, then the statement is true since any premise (true or not) implies a true conclusion. If the Goldbach conjecture happens to be wrong, then the statement is still true since you will be able to find an even number $x$ which is a counterexample, so that $P(x)$ is false and a false premise implies any conclusion.

]]>Whoops, I missed a bracket in the second bit of LaTeX error (Tim, please feel free to correct it and delate this comment). I meant to say:

to

]]>The transformation of the task upwards through the chain of reasoning works. If there exists an element such that its being P implies Q, then we can show that Q by showing that all elements are P. (The existence of an element with the property that its being P would imply Q, conveniently tells us that X is not empty.)

The transformation downwards does not work, as the counter-example shows. The dodgy step is the move from

to

This is a move from

“either there is an x that isn’t P, or Q”

to

“there is an x such that either it isn’t P, or Q”.

The former tells you to check all the members of X (in the example, to check that none has a modulus exceeding 1). The latter tells you that there is some particular member of X which, if it turns out to be P, has magical powers of implication.

]]>What has gone wrong here? If you can give a satisfactory answer, then you will have a good grasp of what mathematicians mean by “implies”.

Well, I’m a mathematician and what I mean by implies (A=>B) is that because A is true, it follows that B is true. There’s a finiky three dot symbol for this type of implies or “it follows that”, but I prefer the arrow and I suspect many others do as well.

Like a lot of mathematical symbols, meaning depends on context. the “=>” symbol in logic has a much more restricted meaning that the “=>” or “implies” used in other branches, where it is treated a more of less a one way “”. In particular “=>” in logic includes the vaccuous case which I don’t even enters in even the slightest way into the vast majority of proofs in mathematics outside of formal logic.

]]>