Does "All I know are ..." = "I don't know any ..."?

[lewisnotmiller]

OK. Noted.

How about just a simple yes/no question. Feel free to elaborate if you wish. But I'm curious.

Is it ridiculous to say "All the Jews I know are atheists AND I don't know any Jews"?

I wouldn't say ridiculous without context. But in most cases it would be.

(As an example I could certainly see it being a decent setup for a joke, but I don't think that's what you meant)

 

[dybmh]

I wouldn't say ridiculous without context. But in most cases it would be.

(As an example I could certainly see it being a decent setup for a joke, but I don't think that's what you meant)

Thank you.

 

[dybmh]

Just a quick note to all,

This is a debate forum. So It's OK to argue with me if you want.

 

[Polymath257]

I know. The conversation advanced.



The conversation advanced.



Right. So if you don't know any that have that property, that is an obvious contradiction.



In English, we read left to right, not right to left.

"All dogs I know are brown" does not imply "I don't know any dogs"

If it is read in English, it is ridiculous. Maybe if a person reads it backwards, it isn't. Let's ignore all those things though. Let's look at the proof.

As I said, the second implies the first, so the two statements are consistent.

First question:

​

not p ---> (p ---> q) ???​

​

What's p and what's q?​

Any proposition.

In essence, a falsehood implies anything. This is a well-known, though somewhat counter-intuitive aspect of logic.

The truth table for implication is:

p | q | p-->q
T | T | T
T | F | F
F | T | T
F | F | T

In this particular case, p is the statement dog(x) and known(x) and q is the statement brown(x).

Second question:

Ax not(dog(x) and known(x))​

​

What's known is not a dog?​

​

​

No: it is never the case that something is both known and a dog.

 

[Polymath257]

To say "Of the Jews that I know" implies that I know some. Therefore it implies that I have knowledge about some Jews.

No, it does not.

I personally have met several Jews, at least one of whom I am convinced is not an atheist. Do I really know these Jews' minds? No. However despite that I cannot honestly say that of the zero Jews that I know they are all atheists. Somehow that does not work, even though I don't really know them. It would imply that I knew more than I did, which would not be true.

 

[Brickjectivity]

No, it does not.

In regular informal conversation if you said "Of the Jews that I know..." you'd be implying not only that you knew one but that you knew multiple Jews.

In a logic puzzle book perhaps it would be mistaken for a formal proposition and then would mean something else such as a mathematical proposition, which it would not be normally.

 

[Polymath257]

It's pretty esoteric and you'll never encounter it unless you're a professional programmer, philosopher or electrical engineer. But it helps you to think logical.

Or in mathematics.

 

[Polymath257]

All zero can be said to be Atheists? How?
If so, all zero can be said to theists.

So the statement is both true and false?

Every element of the empty set is an atheist. And, also, every element of the empty set is a theist.

The statement 'every x is a y' is true if there are no x's.

 

[Polymath257]

OK. Noted.

How about just a simple yes/no question. Feel free to elaborate if you wish. But I'm curious.

Is it ridiculous to say "All the Jews I know are atheists AND I don't know any Jews"?

No. And, in the case where you don't know any Jews, it is a true statement. But it would also be true that all Jews that you know are theists.

 

[dybmh]

In essence, a falsehood implies anything. This is a well-known, though somewhat counter-intuitive aspect of logic.

Anything?

Not P ---> ( P ---> Q )?
Not P ---> ( P ---> ( Not Q ))?
Not P ---> ( P ---> ( Q AND/OR Not Q ))?

"Not P" doesn't tell us anything conclusive about "Q"?

The truth table for implication is:

p | q | p-->q
T | T | T
T | F | F
F | T | T
F | F | T

OK. Confirmed.

p | q | p-->q
F | T | T
F | F | T

Q is both true and false and the implication holds.

Not P ---> ( P ---> ( Q AND/OR Not Q ))

You said "counter-intuitive". I think it goes further than that. This "truth" table does not filter out non-sequiturs. P and Q don't need to be related in anyway. Here's my question:

What's p and what's q?

Here's your answer:

Any proposition.

So, let's work through the "truth" table with some examples. Maybe you could tell me if I made an errors?

p = the car starts
q = it's noon

p | q | p-->q
T | T | T
T | F | F
F | T | T
F | F | T

"the car starts and it's noon" is true. "IF the car starts THEN it's noon" is true?
"the car starts and it's not noon" is true. "IF the car starts THEN it's noon" is false.
"the car doesn't start and it's noon" is true. "IF the car starts THEN it's noon." is true.
"the car doesn't start and it's not noon" is true. "IF the car starts THEN it's noon." is true.

So the first problem is, my car starting has nothing to do with time of day.
The second problem is that if the car doesn't start, no conclusion can be made at all about when it starts.

If I haven't made any mistakes, this "truth" table is not useful at all. It's not just counter-intuitive, the last two rows do not match reality.

In this particular case, p is the statement dog(x) and known(x) and q is the statement brown(x).

Oh? This is the original statement:

"All dogs I know are brown"

An implication is IF ... THEN. Wasn't one of your comments that I had changed the wording? "All dogs I know are brown" is not an IF ... THEN.

No: it is never the case that something is both known and a dog.

Well, that's what I said. Even if you disagree, that's what I meant. Either way, I'm happy to use your wording.

"All dogs I know are brown" AND " it is never the case that something is both known and a dog"

"All dogs I know are brown" CANNOT be talking about dogs?

 

[dybmh]

Every element of the empty set is an atheist. And, also, every element of the empty set is a theist.

The statement 'every x is a y' is true if there are no x's.

So nothing conclusive can be said about an empty-set since nothing and no one can be both a theist and an atheist simultaneously. Nothing conclusive except that it's empty, of course.

No. And, in the case where you don't know any Jews, it is a true statement. But it would also be true that all Jews that you know are theists.

Is there any person anywhere that can be both theist and an atheist simultaneously? Jews are people. So.... the statement wasn't talking about Jews.

It's NOT ridiculous to make assertive claims about things that cannot exist?

Maybe ridiculous isn't the right word.

It's absurd. It's a contradiction.

 

[Polymath257]

Anything?

Not P ---> ( P ---> Q )?
Not P ---> ( P ---> ( Not Q ))?
Not P ---> ( P ---> ( Q AND/OR Not Q ))?

"Not P" doesn't tell us anything conclusive about "Q"?

Correct, but p being false *does* tell us that p-->q is true. Hence, not p --> (p-->q)

OK. Confirmed.

p | q | p-->q
F | T | T
F | F | T

Q is both true and false and the implication holds.

Yes.

Not P ---> ( P ---> ( Q AND/OR Not Q ))

You said "counter-intuitive". I think it goes further than that. This "truth" table does not filter out non-sequiturs. P and Q don't need to be related in anyway. Here's my question:

Correct. To filter out non-sequiters uses entailment, not implication. If you want to only base on truth values, the truth table I give is the only one that works.

Here's your answer:

So, let's work through the "truth" table with some examples. Maybe you could tell me if I made an errors?

p = the car starts
q = it's noon

p | q | p-->q
T | T | T
T | F | F
F | T | T
F | F | T

"the car starts and it's noon" is true. "IF the car starts THEN it's noon" is true?
"the car starts and it's not noon" is true. "IF the car starts THEN it's noon" is false.
"the car doesn't start and it's noon" is true. "IF the car starts THEN it's noon." is true.
"the car doesn't start and it's not noon" is true. "IF the car starts THEN it's noon." is true.

So the first problem is, my car starting has nothing to do with time of day.
The second problem is that if the car doesn't start, no conclusion can be made at all about when it starts.

And again, if the car doesn't start, the implication (if the car starts, then it is noon) is true. It doesn't say what happens to q, only to p-->q.

If I haven't made any mistakes, this "truth" table is not useful at all. It's not just counter-intuitive, the last two rows do not match reality.

Actually, it is because it filters out irrelevancies. Think about it like this.

When *should* the implication p-->q be false? Well, it should be false if the hypothesis (p) is true and the conclusion (q0 is false. That would be enough to show that p-->q is false. But, if p is false, there is no way to show that p-->q is false because the hypothesis doesn't apply. ANYTHING could happen and it would be consistent with the claim that p-->q.

Oh? This is the original statement:

"All dogs I know are brown"

An implication is IF ... THEN. Wasn't one of your comments that I had changed the wording? "All dogs I know are brown" is not an IF ... THEN.

To say 'all dogs are known' is equivalent to 'for all x, if x is a dog, then x is known'. We are doing quantifier logic, not just propositional logic. So, yes, there *is* an implication.

Well, that's what I said. Even if you disagree, that's what I meant. Either way, I'm happy to use your wording.

"All dogs I know are brown" AND " it is never the case that something is both known and a dog"

"All dogs I know are brown" CANNOT be talking about dogs?

Correct, the hypothesis is false, so the *implication* is true. Basic logic.

 

[Polymath257]

So nothing conclusive can be said about an empty-set since nothing and no one can be both a theist and an atheist simultaneously. Nothing conclusive except that it's empty, of course.

Incorrect. many things can be said about the empty set. For example, it is a subset of any set. If two sets do not overlap, their intersection is the empty set. The union of the empty set with any set is that other set. Etc.

Is there any person anywhere that can be both theist and an atheist simultaneously? Jews are people. So.... the statement wasn't talking about Jews.

No, and that is the point. Anyone that is both is an element of the empty set (i.e, does not exist).

It's NOT ridiculous to make assertive claims about things that cannot exist?

Not quite the same thing. Once again, we are looking at an implication. And in an implication, a false implies anything.

Maybe ridiculous isn't the right word.

View attachment 77195

It's absurd. It's a contradiction.

Correct. And this says
((not p--> q) and not q)--> not p
In other words, if the negation of p implies something that is known to be false, then p must be false. This should hold true for any combination of truth values for both p and q. If you look at what happens when q is true, you will find that the left hand side is false, but we need the implication to be true no matter what p is.

So, to be a member of the empty set is a contradiction. It is always false. But then, for the empty set to be a subset of any set requires a false to imply anything.

Any way, it might be helpful to read this:

Truth table - Wikipedia

en.wikipedia.org

or this:

Implication | Truth Tables, Propositional Calculus, Deductive Reasoning

Implication, in logic, a relationship between two propositions in which the second is a logical consequence of the first. In most systems of formal logic, a broader relationship called material implication is employed, which is read “If A, then B,” and is denoted by A ⊃ B or A → B. The truth or

www.britannica.com

In particular:
"implication, in logic, a relationship between two propositions in which the second is a logical consequence of the first. In most systems of formal logic, a broader relationship called material implication is employed, which is read “If A, then B,” and is denoted by A ⊃ B or A → B. The truth or falsity of the compound proposition A ⊃ B depends not on any relationship between the meanings of the propositions but only on the truth-values of A and B; A ⊃ B is false when A is true and B is false, and it is true in all other cases. Equivalently, A ⊃ B is often defined as ∼(A·∼B) or as ∼A∨B (in which ∼ means “not,” · means “and,” and ∨ means “or”).

This way of interpreting ⊃ leads to the so-called paradoxes of material implication: “grass is red ⊃ ice is cold” is a true proposition according to this definition of ⊃."

 

[Polymath257]

So, let's work through the "truth" table with some examples. Maybe you could tell me if I made an errors?

p = the car starts
q = it's noon

p | q | p-->q
T | T | T
T | F | F
F | T | T
F | F | T

"the car starts and it's noon" is true. "IF the car starts THEN it's noon" is true?
"the car starts and it's not noon" is true. "IF the car starts THEN it's noon" is false.
"the car doesn't start and it's noon" is true. "IF the car starts THEN it's noon." is true.
"the car doesn't start and it's not noon" is true. "IF the car starts THEN it's noon." is true.

So the first problem is, my car starting has nothing to do with time of day.
The second problem is that if the car doesn't start, no conclusion can be made at all about when it starts.

If I haven't made any mistakes, this "truth" table is not useful at all. It's not just counter-intuitive, the last two rows do not match reality.

In the last two, think about whether the situation could be used to show that the implication p-->q is false. And, you will find that in both, the falsity of p means that it is irrelevant to the statement that p-->q. So it cannot be used to show it false. hence, it must be true.

The *only* case where p-->q is false is when p is true and q is false.

Maybe I should try to state a thread on truth values and logic?

 

[PureX]

I know. In another thread, someone is claiming that no "thinking person" would object to it.

Here's what they are saying:

​

It is not ridiculous to say "All the Jews I know are Atheists AND I don't know any Jews"​

​

And they're trying desperately to show the so-called logic behind it. I have been arguing against it, naturally. The only argument they have brought is:

"It's obvously absolutely true. It's ridiculous to say it's false. Every thinking person agrees with me."​

It's deliberately misleading gibberish. The actual content being, "I don't know any Jewish atheists".

 

[dybmh]

It's deliberately misleading gibberish. The actual content being, "I don't know any Jewish atheists".

Is the gibberish true or false?

Is gibberish a statement?

 

[Stevicus]

Hopefully this is a simple question. Please answer the poll.

Yes or No? Does "All I know are ..." = "I don't know any ..."

Examples:

"All I know are dogs" = "I don't know any dogs" ??
"All I know are cats" = "I don't know any cats" ??
"All I know are green-martians" = "I don't know any green-martians" ??

Thank you,

This reminds me of the song "Both Sides Now" by Joni Mitchell. "I really don't know clouds at all."

I guess I could say the same about cats or dogs. I've had cats, I've known cats, but when you really come down to it, I really don't know cats at all.

 

[PureX]

Is the gibberish true or false?

Is gibberish a statement?

The truth is it's a nonsensical statement.

 

[dybmh]

Correct, but p being false *does* tell us that p-->q is true. Hence, not p --> (p-->q)

That is incomplete. Not P --> ( P --> Q ) is incomplete. Not P --> ( P --> ( Q AND/OR Not Q )) is complete.

Correct. To filter out non-sequiters uses entailment, not implication. If you want to only base on truth values, the truth table I give is the only one that works.

Then evaluating "true/false" using "only" the truth table is a poor method.

And again, if the car doesn't start, the implication (if the car starts, then it is noon) is true. It doesn't say what happens to q, only to p-->q.

Not if it's noon.

If the car DOESN'T start and it IS noon, then how is the implication true?

p | q | p-->q
T | T | T
T | F | F
F | T | T
F | F | T

"the car doesn't start and it's noon" is true. "IF the car starts THEN it's noon." is true.

Actually, it is because it filters out irrelevancies. Think about it like this.

When *should* the implication p-->q be false? Well, it should be false if the hypothesis (p) is true and the conclusion (q0 is false. That would be enough to show that p-->q is false.

Agreed. That's the 2nd row of the truth table.

But, if p is false, there is no way to show that p-->q is false because the hypothesis doesn't apply. ANYTHING could happen and it would be consistent with the claim that p-->q.

This only works with the fourth row where both p is false and q is false. If p is false and q is true, that is a valid counter-example.

"In logic a counterexample disproves the generalization, and does so rigorously in the fields of mathematics and philosophy."

Counterexample - Wikipedia

en.m.wikipedia.org

To say 'all dogs are known' is equivalent to 'for all x, if x is a dog, then x is known'. We are doing quantifier logic, not just propositional logic. So, yes, there *is* an implication.

Would you please reconstruct the following statement into an implication?

"All the Jews I know are atheists." Please try to make into something that is not too awkward. I think when you do that, no one will object to it, because when the statement is CHANGED into an "If ... Then" statement, it accurately communicates ignorance and becomes less absurd.

If it's OK to make changes to the statement to reduce it's absurdity, then it should be OK to make similar ( yet opposite ) changes to the statement to increase its absurdity.

"All I know are ... " =/= "I don't know any ..." makes a similar change, in scope, to the statement as the addition of "IF ... THEN".

Correct, the hypothesis is false, so the *implication* is true. Basic logic.

I disagree that if the hypothesis is false that it implies anything, unless everything is assumed to be true unless it is proven false. That's not an implication, that is extreme optimism. I am an optimist, I know optimism when I see it. Optimism has a place and purpose and is useful. Evaluating true/false is not one of them.

You have agreed that the statement: "All the dogs I know are brown" is intentionally not talking about "dogs" if the speaker correctly states they do not know any "dogs".

It actually goes further than this, doesn't it?

The statement: "I don't know any dogs" was translated into: Ax not(dog(x) and known(x)).

not ( dog and known ) = ( not dog ) AND/OR ( not known ).

This is the negation of a conjunction. I won't be rude and call it "Basic Logic". But it is demonstrably true.

So, when a person says "All the dogs I know are brown" AND "I don't know any dogs". They are not talking about dogs, or they are not talking about knowledge, or they are not talking about dogs AND knowledge.

This produces 3 and ONLY 3 possible meanings:

All the dogs I know are brown.
All the dogs I know are brown.
All the dogs I know are brown.

How are any of these true?

 

[dybmh]

Incorrect. many things can be said about the empty set. For example, it is a subset of any set. If two sets do not overlap, their intersection is the empty set. The union of the empty set with any set is that other set. Etc.

I knew you were going to say this.

None of those are conclusive. They are definitions. A definition is NOT a conclusion.

Can you prove that the "empty-set" is a subset of every set without ignoring any counter-examples? Is "empty-set is a subset of every set" a conclusion?

No, and that is the point. Anyone that is both is an element of the empty set (i.e, does not exist).

Good, so the statement is intentionally making a claim about things that cannot exist. This is not a maybe/maybe-not.

Not quite the same thing. Once again, we are looking at an implication. And in an implication, a false implies anything.

Only if every statement is assumed to be true unless it is proven false. Guilty until proven innocent?

Correct. And this says
((not p--> q) and not q)--> not p
In other words, if the negation of p implies something that is known to be false, then p must be false. This should hold true for any combination of truth values for both p and q. If you look at what happens when q is true, you will find that the left hand side is false, but we need the implication to be true no matter what p is.

Good, you have acknowledged that the statement is absurd, and is a contradiction. Why is an absurd contradiction being considered true?

So, to be a member of the empty set is a contradiction. It is always false. But then, for the empty set to be a subset of any set requires a false to imply anything.

OH! The definition of an empty-set as a subset is a contradiction? I have been saying this for weeks. In fact, "empty-set" IS "contradiction".

Any way, it might be helpful to read this:

Truth table - Wikipedia

en.wikipedia.org

or this:

Implication | Truth Tables, Propositional Calculus, Deductive Reasoning

Implication, in logic, a relationship between two propositions in which the second is a logical consequence of the first. In most systems of formal logic, a broader relationship called material implication is employed, which is read “If A, then B,” and is denoted by A ⊃ B or A → B. The truth or

www.britannica.com

I'm familiar, but I'll read it again, thanks.

In particular:
"implication, in logic, a relationship between two propositions in which the second is a logical consequence of the first.

Hmmmm.... the second is a logical consequence of the first.

"All the Jews I know are atheists" AND "I don't know any Jews".

Is "I don't know any Jews" a logical consequence of "All the Jews I know are atheists"? When is that the case?

Oh! I know, it's a logical consequence as soon as they meet a Jew who is not an atheist! But, in that case, they know a Jew, so the relationship is broken. No matter what, it's not a logical implication.

In most systems of formal logic, a broader relationship called material implication is employed, which is read “If A, then B,” and is denoted by A ⊃ B or A → B. The truth or falsity of the compound proposition A ⊃ B depends not on any relationship between the meanings of the propositions but only on the truth-values of A and B; A ⊃ B is false when A is true and B is false, and it is true in all other cases. Equivalently, A ⊃ B is often defined as ∼(A·∼B) or as ∼A∨B (in which ∼ means “not,” · means “and,” and ∨ means “or”).

This way of interpreting ⊃ leads to the so-called paradoxes of material implication: “grass is red ⊃ ice is cold” is a true proposition according to this definition of ⊃."

There it is! It is always true unless it is proven false BY DEFINITION, not by any actual logic.

So, in your opinion, is it a good method to assume everything is true unless it is proven false?