Logic Problem - wrong?

Tangram · 2015-08-08 23:25:27

Alice, Bill, and Carol are considering going to a party. Dave says he will go only if Carol does and Bill does not. Ed says he will go if either Dave or Alice or both go. You learn that Ed and Bill did not go. Who did go?

According to the source that this problem came from, the answer is that no-one went. But according to my reckoning, this is wrong.

Let A = "Alice goes", B = "Bill goes", etc.

I translated the premises as follows-

1. If D, then C and not-B.
2. if A or D, then E.
3. not-E
4. not-B

From these I generated the following-

5. not-(A or D) from 2 and 3.
6. not-A and not-D from 5.

I don't think anything else can be concluded, but it's not the case that no-one went, because whether C (Carol) went is indeterminate.

Any help appreciated!

Last edited by Tangram (2015-08-08 23:26:09)

Bob · 2015-08-09 00:13:27

hi Tangram,

I agree with your deductions but .......

I made a truth table:

The only line containing 'not' for both Bill and Ed is the bottom line => no one goes.

Your version is essentially the same except you haven't got a truth value (yet) for Carol. I think it must be hidden in there somewhere so I'll work on that.

LATER EDIT: I have it.

Suppose that Carol goes.

Then we have Carol goes and Bill does not => Dave goes. =><= contradicts Dave doesn't go => Carol doesn't go.

Bob

Last edited by Bob (2015-08-09 00:19:53)

Tangram · 2015-08-09 19:38:46

Hi Bob,

Thanks for that. It seems to depend on how you interpret "Dave says he will go only if Carol does and Bill does not". Specifically, "if" is not the same as "only if".

I interpreted the statement to mean "D => C.¬B". If you read it as meaning "C.¬B => D" then the argument does indeed come out as valid, but not when using
"D => C.¬B".

I checked the argument both ways using an online truth table generator:

http://turner.faculty.swau.edu/mathemat … ary/truth/

Copy and pasted the following one at a time:

(~e & ~b & ((a + d) > e) & (c & ~b > d)) > (~a & ~b & ~c & ~d & ~e) This comes out as valid (all values TRUE)
(~e & ~b & ((a + d) > e) & (d > (c & ~b))) > (~a & ~b & ~c & ~d & ~e) This gives one FALSE value in the last-but- one row.

Bob · 2015-08-09 21:02:39

hi Tangram,

Actually, I was interpreting it to mean Dave goes "is equivalent to" (Carol goes AND Bill does not)

This is because Dave only makes his decision after Alice, Bill and Carol have made theirs. That's why I wrote the truth table with go and not rather than the usual TRUE and FALSE. Let's say (Carol goes and Bill does not) is statement A . Statement A could be TRUE or FALSE.

Then the sentence becomes Dave goes only if statement A is TRUE. If statement A is TRUE then surely that means that he goes. And if statement A is FALSE, surely that means he doesn't go.

Under your interpretation it is TRUE that Dave doesn't go when statement A is TRUE.

So (Dave goes) = (Carol goes AND Bill does not) Here I'm using = to mean "is equivalent to" because I don't want to have to resort to Latex.

Bob

Tangram · 2015-08-09 23:01:30

Bob,

Yes, that's how the author of the book where I got the problem interpreted the statement - as an equivalence. But in that case it seems to me it should have been worded as "if and only if", and not just "only if". But maybe I'm nit-picking... ;-)

Last edited by Tangram (2015-08-09 23:02:49)

Bob · 2015-08-10 04:58:19

I agree that "if" and "only if" mean different things and so the wording is loose. But you should also take account of the problem in context. Because Dave only decides after Carol and Bill have declared this means that certain logical possibilities cannot occur as they would if time was not a factor.

Bob

Math Is Fun Forum

#1 2015-08-08 23:25:27

Logic Problem - wrong?

#2 2015-08-09 00:13:27

Re: Logic Problem - wrong?

#3 2015-08-09 19:38:46

Re: Logic Problem - wrong?

#4 2015-08-09 21:02:39

Re: Logic Problem - wrong?

#5 2015-08-09 23:01:30

Re: Logic Problem - wrong?

#6 2015-08-10 04:58:19

Re: Logic Problem - wrong?

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