drinking in the bar

[BMAD]

Two men walk by a bar. The man closest to the window says, "I am impressed, everyone is drinking in there". Shocked by his friend's abillity to be so observant, the friend asked how he knew. The first man simply said, I saw the first man. Since he was drinking, all people in there were drinking. "What?" Replied the friend.

Smiling the man said, "it is an axiom: if someone in the bar, is drinking, then everyone is drinking".

Disbelieving his friend, the friend marched back to the bar to have a look inside.

Did he confirm or deny the axiom?

[DeGe]

He confirmed that everyone was drinking... The first friend's theory is also supported by "induction"

[DeGe]

While in school our math teacher gave us a funny and a practical example of mathematical induction. At the start of the class, she said "before the end of this month, you will have a surprise quiz. Can anyone guess when it will be?" We all gave some random days and she ultimately handed us the quiz in the next 10 mins!
Thinking back, here's how I reason it...
Let's say the month is of 31 days. If by the 30th, there is no test done, it will be done on the 31st and is no longer a "surprise test". The test must therefore be done before the 31st.. Now, on 30th, if there has been no test by the 29th, the test must be done before otherwise it will not be a surprise... working backwards by induction, the test must be today!

[BMAD]

Now what if the friend walks in the bar and he sees the first person drinking but the next person isn't drinking, does this contradict the axiom?

[witzar]

working backwards by induction, the test must be today!

If you can prove by induction that the test is today, how can it's date be a surprise?

[DeGe]

working backwards by induction, the test must be today!

If you can prove by induction that the test is today, how can it's date be a surprise?

Because the announcement that there will be a surprise test can be made on any random day, which is a surprise.

[DeGe]

Now what if the friend walks in the bar and he sees the first person drinking but the next person isn't drinking, does this contradict the axiom?

No, it doesn't contradict the axiom. In truth, the actual probability that a person is drinking or not is 0<= p <= 1. However, as there is no physics that determines what the exact probability of everyone or any one person drinking would be, one must rely on inference and induction.

[bonanova]

While in school our math teacher gave us a funny and a practical example of mathematical induction. At the start of the class, she said "before the end of this month, you will have a surprise quiz. Can anyone guess when it will be?" We all gave some random days and she ultimately handed us the quiz in the next 10 mins!

Thinking back, here's how I reason it...

Let's say the month is of 31 days. If by the 30th, there is no test done, it will be done on the 31st and is no longer a "surprise test". The test must therefore be done before the 31st.. Now, on 30th, if there has been no test by the 29th, the test must be done before otherwise it will not be a surprise... working backwards by induction, the test must be today!

This is one version of Martin Gardner's "Unexpected Hanging" paradox.

The resolution turns on the truthfulness of the person saying the test (hanging) will happen at a time that cannot be anticipated.

[BMAD]

Yes, but this op actually refers to a different paradox.

[gavinksong]

If this is a paradox, it should probably go in the paradox section.

I'm kinda curious as to what this one is about, because I am pretty confused right now.

[BMAD]

What would it take to contradict the axiom?

[Rainman]

The statement "if someone is drinking, then everyone is drinking" is false if and only if someone is drinking but not everyone is. So it takes at least one person who is drinking, and at least one who isn't.

However, I wonder if you actually mean axiom rather than statement? An axiom can't be contradicted simply by finding a model where it isn't true. An axiom can only be contradicted if it is contradictory by itself, or if we have another set of axioms to contradict it. Since the axiom itself isn't contradictory, we would need more axioms to contradict it.

[BMAD]

You can only consider two possible situations based on the op:

(1) everyone is drinking

(2) at least 1 person isn't drinking

"it is an axiom: if someone in the bar, is drinking, then everyone is drinking".

This statement does not allow for rainman's if and only if statement.

Also, I was using axiom socially, let's not get too caught with its precise meaning. I realize now that it may have been a poor choice in words.

[Rainman]

1) Everyone is drinking. We don't have a problem, the statement is true.

2) There is at least one person who is not drinking. Here the statement is false, because the OP states that someone is drinking. So someone is drinking, but everyone is not drinking.

[BMAD]

Think about which in the P&Q (if p then q) is false with the second scenario.

[Rainman]

P (someone is drinking) is true. Q (everyone is drinking) is false. So "if P then Q" is false.

[BMAD]

So if someone is not drinking then p is false, so...

[Rainman]

I see what you mean now, but no. The negation of P is "no one is drinking", not "someone is not drinking".

Someone is drinking = at least one person is drinking.

Someone is not drinking = at least one person is not drinking.

These statements are not mutually exclusive. If you are drinking and I am not drinking, then both are true.

[BMAD]

From wikipedia

Suppose that at least one person is not drinking. For any particular nondrinking person, it still cannot be wrong to say that if that particular person is drinking, then everyone in the pub is drinking because that person is, in fact, not drinking. In this case the condition is false, so the statement is vacuously true due to the nature of material implication in formal logic, which states that "If P, then Q" is always true if P (the condition or antecedent) is false. [1][2] Either way, there is someone in the pub such that, if he is drinking, everyone in the pub is drinking. A slightly more formal way of expressing the above is to say that if everybody drinks then anyone can be the witness for the validity of the theorem. And if someone doesn't drink, then that particular non-drinking individual can be the witness to the theorem's validity. [3] The proof above is essentially model-theoretic (can be formalized as such).

http://en.m.wikipedia.org/wiki/Drinker_paradox

Edited August 3, 2013 by BMAD

[Rainman]

That's not the same statement. "There is someone in the pub, such that if he is drinking, everyone is drinking" is not the same as "if someone in the pub is drinking, then everyone is drinking". The first statement checks each person individually for the "if-then" statement, and asserts that at least one of those is true. The second statement is just another way of saying "no one is drinking or everyone is drinking", which is often false.

To illustrate the difference, suppose persons A and B are in the bar. A is drinking, B is not drinking. So someone in the bar is drinking. But everyone is not drinking.

The statement in the OP, "if someone is drinking, then everyone is drinking", is false (P is true, Q is false).

The statement on wikipedia, "there is someone in the pub, such that if he is drinking, then everyone is drinking", is in fact true. Let us check these two statements:

• If A is drinking, then everyone is drinking. A is drinking, but everyone is not drinking. So this one is false.
• If B is drinking, then everyone is drinking. Since B is not drinking, this statement is true by default.

So there is indeed such a person in the pub that the conditional statement is true, and that person is B. Hence there is someone in the pub, such that if he is drinking, then everyone is drinking.

In the OP, the man said something to the effect of "I noticed the first man was drinking, and if he is drinking then everyone is drinking". This is false unless everyone is in fact drinking. But, if he had seen someone who was not drinking, he could have safely pointed to the non-drinker and said "if that man is drinking, then everyone is drinking". That statement is true because the "if" part is false.

[witzar]

I agree with Rainman.

The proper version comes from "What is the Name of this Book" by Raymond Smullyan and is called 'The Drinking Principle'.

Smullyan also states a dual paradox:

There is at least one person such that if anybody drinks, then he does,

and more dramatic version of TDP:
There is a woman on earth such that if she becomes sterile, the whole human race will die out.

[OmegaScales]

1) By walking into the bar, they have disproven the axiom as they are not drinking, merely observing, even if everyone else is.

If that wasn't simple enough

2) For a bartender to drink whilst on the job is extremely inadvisable, even illegal in some places.

[kestrelknight]

You Logic/Math people are amazing. Honestly, I have no idea what most of you are talking about, but I feel like I would just say that:

P'raps theres one person in the bar (likely the bartender), who is drinking. He could be drinking water, or some other non-alcoholic beverage.