A Logic Puzzle

Use of non sequitur noted.

Perhaps I am missing something, though?

You have two options for basket one:
a) This basket contains nothing.
b) The basket with the false note contains nothing.

You have two options for basket two:
a) This basket contains $1million and the basket with the false note contains nothing.
b) This box contains nothing and the box with the true note contains $1million.

Further, you specify “One note is true, and one note is false. One basket contains $1million, and one contains nothing.” We’ll designate this “R”.

So if 1a is true, R dictates that 1b is false.
1b is true by default.
If 2a is true, note one is false and R is satisfied.
If 2b is true (thus 1a is true), R is not satisfied.

false

Okay, not by default. 2a being true, 1b happens to be true as well.

there are more things wrong with your logic as well. i’d point them out but i don’t want to end up giving the answer/logic away by helping too much.

B1 note is true, money is in B2

Assume B1 is true:
Note 1 says either B1 is empty and B2 is $ or B2 is empty and B1 is $ - logically ok
Test if B2 being false makes sense:
Note 2 says either B2 is $ and B2 is empty - always false which is ok
or B2 is empty and B1 is $ - that is only false if the money is in B2

Could B2 be true instead?
Note 2 says B2 is $ and B1 is empty or B2 is empty and B2 is $ - already inconstant on second part
Test if B1 being false makes sense:
Note 1 says B1 is empty - only false if money is in B1
or B1 is empty - only false if money is in B1
They can’t both be the same statement so B2 being true and B1 being false doesn’t make sense.

How did I do?

Oh I see. Basket one could contain $1 million dollars if note two is false, satisfying both 1b and r. So it’s in basket one.

Some of your wording is hard to follow, especially in the second paragraph, so i can’t really tell you Phyllo

Wait, isn’t it a paradox? Both notes offer options that satisfy both situations. That can’t be the case unless R is false.

nope, not a paradox. i promise, there’s one answer and it’s provable.

Then it is basket two.
Note one is true and note two is false.
2a contradicts 1a. (If 2a is true, 1a is true and 1b agrees with 2a, therefor note 1 is not false)
2b contradicts itself. (If basket 2 contains nothing, 2b is true and therefor should contain the money)
1b contradicts itself. (If 1b is true, note 1 must be true and therefor should contain the money, making 2b true)
1a is true. (If 1a is true, neither 1b nor note 2 can be true, so by R, the money must be in basket 2)
1a is true and the money is in the basket with the false note.

I’ve only read the opening post of this thread. I say the note on basket 1 is the false note. What’s false about it is that that basket is the basket with the false note, so that basket and the basket with the false note do both contain nothing. The note written on basket 2, then, is true: for though it does not contain nothing, it does contain $1 million and the basket with the false note contains nothing.

wrong

if note1 is false and basket1 doesn’t have anything, then note1 is true, because note1 says …“or the basket with the false note contains nothing.”

Basket 1 has the false note and contains nothing. It contains nothing and is has the false note, making both parts of the disjunction true (which its note says aren’t both true).

Basket 2 has the true note and contains $1 million. The note is a simple disjunction, and when one part of a disjunction is true (‘x’ or ‘y’) the whole statement is true. ‘X’ = ‘this basket contains $1 million and the basket with the false note contains nothing’ and is true, while ‘Y’ = ‘this box contains nothing and the box with the true note contains $1 million’ and is false because this box does not contain nothing. Since at least ‘X’ is true, though, the disjunction is true, thus the note is true.

Sauwelios and I are right. Note 1 says both parts of its disjunction cannot be true, but the both are: the basket contains nothing and it has the false note.

Let me try to be more clear. With the conditions of the puzzle in mind (one note must be false and one must be true, one basket must contain $1 million and the other nothing), basket 1’s note is self-contradictory. For its note to be true, one part of the disjunction has to be true. If either part is true, though, then both parts are true. But it says both can’t be true, from which it follows that neither can be true. So note 1 is false, which means note 2 has to be true.

Fuck, I got it wrong.

Basket 1 contains nothing, but has the true note.

Basket 2 contains $1 million, but has the false note.

If basket 1 contains nothing, then the first part of it’s disjunction is true, but the second might not be (the false note basket might contain $1 million). And since both aren’t true the whole statement would be true.

If basket 1 has the true note, then basket 2 has the false note, so it can’t be true that the false note contains nothing and the true note contains $1 million. Thus note 2 is false, but the basket contains $1 million.

I’m not quite sure about your logic, but since there have been 2 right answers so far with logic that i didn’t fully understand, I’ll give my logic:

there are only 4 possibilities given the initial criteria: basket 1 has money and is true, basket 1 has nothing and is true, basket 1 has money and is not true, and basket 1 has nothing and is not true. i will go through each one one-by-one to show that only one answer is logically consistent.

Assume Basket 1 is true and basket 1 has nothing. If Basket 1 has nothing and is true, basket 2 must be false and have the money. Both parts of the disjunction of b2 must be false. The first part of the disjunction of B2, “this basket contains $1million and the basket with the false note contains nothing,” is false, because the basket with the false note contains the money. The second part of B2, “this box contains nothing and the box with the true note contains $1million,” is also false because “this box” does contain the money, and the true one has nothing.

So, b1 = true and having nothing is consistent logically

Assume basket 1 is true and has something. basket 2 must be false and have nothing. once again, both parts of the disjunction must be false for b2 to be false. “this basket contains $1million and the basket with the false note contains nothing” ← this is clearly false, as “this basket” does not contain the money. “this box contains nothing and the box with the true note contains $1million.” ← this one, however, is true, making b2 true.

b1 = true and having something is logically inconsistent, as it produces a result of b2 = true and b2 = false.

Assume basket 1 is false and has nothing. the first part of basket 1 states, “this basket contains nothing,” and it does in this case. the second part of b1 states “the one with the false note contains nothing” ← that is also true. b1 ends with “but not both,” so since they’re both true, b1 = false, so far consistent. now to test b2 (which needs to be true if b1 is false): the first part of the disjunction is true, “this basket contains $1million and the basket with the false note contains nothing”, so b2 is true.

b1 = false and having nothing is consistent logically

Assume b1 is false and has something. It’s consistent with both of b1’s standards – they’re both false. b2 has to be true. first part of the disjunction is false, because “this basket” does not contain the money, b1 does. second part is false because the true one does not contain 1million. b2 is false

b1 = false and having something is logically inconsistent, as it produces a result of b2 = true and b2 = false.

There are only 2 logically consistent answers in this (I actually started this thread with the belief that there was only 1, because i hadn’t fully considered the ‘not both’ addendum, but that’s ok), and both of the consistent answers have the money in box 2. So you can safely choose box 2 and expect to be a millionaire. Unless you’re dealing with a liar.

Now, for my promised pic:

Way to go, fuse. Still, I think your former solution (and mine) is also correct: it is possible for basket 1 to contain nothing and have the false note attached to it, because then both “this basket contains nothing” and “the basket with the false note contains nothing” are true, which means the “but not both” is false and thereby the note as a whole is false.

This would be better expressed in propositional calculus.

AvB
~(A[and symbol]B)

both given?

Yeah, but this is ILP. I just assume most people here aren’t going to understand symbolic logic and I don’t feel like giving a course at the moment.

Basket 1: (A v B) * ~(A * B)

v → “or”

• → “and/but”
  ~ → “not”

A = “this basket contains nothing”
B = “the basket with the false note contains nothing”

Basket 2: (C * B) v (A * D)

A & B are the same as above.
C = “this basket contains $1 million”
D = “the box with the true note contains $1 million”

Then it’s just a matter of plugging in truth values.