Well?

>>7946401
First two.

>>7946401
A to make sure the other side is an even #.
7 to make sure the other side is not a vowel.
D to make sure the other side is not an even #.
4 to make sure the other side is a vowel.

You have to turn over all of them. No way around it.

>>7946404
How does that tell you D doesn't have 2 on the other side and 4 doesn't have F?

>>7946408
D/2 4/F doesn't break the rule

>>7946405
You don't have to flip 4 because it doesn't matter if the other side is a vowel or not. This is an if statement, not if and only if.

Flip all but 4

>>7946408
The rule says if there's a vowel on one side, there must be an even number on the other.

It doesn't say if there's an even number on one side there must be a vowel on the other.

>>7946409
The rule says "one side," not the side facing up.

>>7946405
>>7946410
come on guys, the first post got it right. stop fucking the thread

Not a biconditional statement.
So only verify the case one way

Oh neat. My discrete math professor did this one at the beginning of the semester.
It's an implication, meaning it is only false when there is a vowel on one side and an odd on the other. You don't need to turn 4 over, because the statement can still be true, no matter what's on the other side of the card.
So, A and 7.

>>7946404
A isn't a vowel.

>a speech sound that is produced by comparatively open configuration of the vocal tract, with vibration of the vocal cords but without audible friction and is a unit of the sound system of a language that forms the nucleus of a syllable.

A isn't a sound last time I checked.

A is a letter.

wait, why 7?

>>7946502
7 is odd, so if there is a vowel on the other side, the implication is false.

A and 4. What kind of babby shit is this?

>>7946513
Which of the cards would need to be turned over to determine whether the rule is true or false?

vs.

Which of the cards needs to be turned over to determine whether the rule is true or false?

Grammar.

It saves lives

A and 7.

>>7946513
Which of the cards would need to be turned over to determine whether rule is true or false?

vs.

Which of the cards needs to be turned over to determine whether rule is true or false?

Grammar.

It saves lives.

>>7946527
*the

>>7946527
Which of the cards would need to be turned over to determine whether rule is true or false?

vs.

Which of the cards needs to be turned over to determine whether rule is true or false?

Grammar.

It saves lives.

A, 7, and D you brainlets

A has anything that isn't an even number on the other side: rule is disproved

7 has a vowel on the other side: rule is disproved

D has a vowel on the other side: rule is disproved

4 obviously can't break the rule in any way so nothing is gained from flipping it

>>7946604
>D has a vowel on the other side: rule is disproved

>Each of them has a letter on one side and a number on the other.
Are you Roman?

>>7946410
No, only flip the first two because the reason we don't flip D is the same as your reason for not flipping 4.

>>7946604
The fact that there is a letter on one side and a number on the other is given and is not part of the rule being proven or disproven.

Therefore, we do not need to flip D.

Re-read the puzzle.

File: FUCKING_ANGRY.jpg (30KB, 712x662px) Image search: [Google]

30KB, 712x662px

fucking logic problems
the problem creators seem to think that normal people will interpret logic problems like logicians do when normal people can barely put their fucking pants on

let me try to make this intuitive

we are given the statement:
(If Vowel, then Even)
meaning that if we have a vowel, then the other side is an even number

so we know that if there is a Vowel, the other side MUST BE EVEN
but if we just have an Even number, there is nothing saying that there must be a vowel on the other side of an Even, only that there must be an Even on the other side of a Vowel

we're given cards A 7 D 4 and asked which cards we need to check to determine if the statement (If Vowel, then Even) is correct
The first choice is A. We flip A over, and see if it has an Even.
Now, looking at 7, we know that if there is a Vowel on the other side of the 7, then the statement is false, because every Vowel must have an Even on the other side, and because 7 is Odd, the rule does not hold
We look at D next. D is not a vowel, so it doesn't matter what's on the other side.
Then, we're left with 4. 4 is not a vowel, so it doesn't matter what's on the other side.

A note about logic that seems to be confusing people:
We are given the logical proposition [math] \text{Vowel} \rightarrow \text{Even} [/math]
this is logic speak for [math] \text{If Vowel, then Even} [/math]
the logical converse of this proposition is [math] \text{Even} \rightarrow \text{Vowel} [/math]
this is logic speak for [math] \text{If Even, then Vowel} [/math]

the reason we don't need to check D and 4 is because the converse of a proposition is not always true
consider this situation:
if you shitpost, you are a fucking idiot
which is obviously a true proposition

now look at the converse
if you are a fucking idiot, you shitpost
this is not always true, i can think of a bunch of fucking idiots who are too stupid to use a computer, much less shitpost

i hope a fucking idiot reads this and becomes a little less stupid

>>7946412
the rule does not say anything about the four cards that we have

>>7946401

the rule is VOWEL=>EVEN and also ODD=>CONSONANT in classical logic.

so in classical logic, to prove the VALIDITY of the rule, you must find that
-in each instance of a vowel, you get even
OR
-in each instance of a odd, you get a consonant


in classical logic, to prove the INVALIDITY of the rule, you must find that
-in ONE instance of a vowel, you get ODD
OR
-in ONE instance of a odd, you get a vowel


then what do you do ?
you try to turn over all the cards until you reach ONE card that invalidates the rule, since there is no guarantee that, if some of the cards follow the rules, other cards will follow the rule. there are no link between the cards. [all you know is that the cards have a number and a letter]

File: 200_s.gif (35KB, 310x200px) Image search: [Google]

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>sci makes my homework again

>>7946478
How is it not a sound?

4 cards, each letter <-> number
rule to prove=if vowel then <-> even number
1. A is vowel so turn to check for even
2. 7 is number so turn to check for vowel
3. D isnt a vowel so dont turn
4. 4 is even but we dont have to turn because the rule is true either way vowel or not

>>7946703
>hurf durf you can't assume that the cards follow the rules
Why do people think this is a solution?

>>7948830
because it is not said in the problem that the cards are linked somehow.

>>7946401
the only rule is
vowel --> even_number,
with the "-->" being a mathematical implication, i.e. it's only false if true implies false

We can conclude:
>if it's not a vowel, it is always true
>if it's an even number, it is always true

that means we need to check the cards with vowels, and the cards with odd numbers. These are card 1 and 2.

>>7946703
you assume that the opposite rule also holds, which is a fallacy.
Nowhere does it state that "ODD=>CONSONANT", which is an easy pitfall many logic beginners fall into. You literally pulled that out of your ass because you thought it's "make sense"

>>7949044
P => Q is equivalent to ~Q => ~P you brainless fucktard

>>7949049
prove it.

>>7949055

Both are easily equivalent to (~P or Q).

(http://math.stackexchange.com/questions/1002811/proof-that-p-implies-q-entails-not-p-or-q)

>>7949062
correct.

>>7949044
>>7949055
>>7949068
stop shitposting, faggot

>>7949070
give me a task, then.

>>7949072
Go to /sqt/. There are many questions in need of answers there.

>>7946604

Indisputable proof that people who use the word "brainlet" are retarded highschoolers

>>7946401
Retards.

vowel on one side -> even on other

Says NOTHING about the following:
even on one side -> vowel on other
odd on one -> consonant on other
etc.

First card only, as that's the only one it made a prediction about.

>>7949109

Not quite, we need to check 7 as well since a wovel on the other side would break the rule. Think of it this way, if you have P → Q, then card 1 is evidence of P = 1, which means Q = 1, while card 2 is evidence of Q = 0, which means that if P = 1 the rule is broken since it entails 1 → 0

A
not 4 because opposite could be either
not D for same reason
not 7 for same reason
Rule clearly says if there is vowel, opposite must be even, so we have to check.

>>7949044
>>you assume that the opposite rule also holds, which is a fallacy.
I explicitly say that it holds in classical logic, which is called classical for a reason.

>>7949121
Rule is wrong if the side opposite 7 is a vowel.

>>7949124
yep you're right

and for people who try to be rigorous, even though ANY little problem like this is ALWAYS poorly formulated, there is no mention of any choice of formal logic.
So for all we know, the rule ''if then'' is not even a good formula in the logic that the guy thought about while writing the problem (when he talks about truths).

Reminder that that there are logics without the material implications.


reminder that no logic talks about truths, but at best, about ''truth values''.

>>7949134

I want a big mac, fries and a sprite.

>>7949134
At least it's not as bad as the time where the rule "each card has a letter on one side and a number on the other" was left out, and the answer was you had to turn over every card to see if they had a vowel on the other side.

>>7946655
tldr

>>7946401

Just A

How many times do we have to have this thread?

>>7950957
As many as is needed for you to understand that if the 7 card has a vowel behind it, the rule fails to hold.

>>7946401
All of them. If it's a rule they should all apply. But if they all apply to the rule, there's no definitive way to prove every other card following this trend does as well. Only these four can be known

>>7946655
it's not like anyone gives a shit about these autism satiaters

some of us have real things to do, like physics

>>7946401
all of them? that's the only way you could know the rules are valid for all cards

>>7946401
there is only 1 card with a vowel on it so to determine if the rule is true it is A

but if 7 has a vowel on it it would prove the rule false

if there is a A and 7 for a answer it should be the actual answer

>>7946401
Cards A, 7, and 4.
we must know if card A has an even number on the other side
we must know if card 7 does NOT have a vowel on the other side
we must know if card 4 DOES have a vowel on the other side.

for card D, in the initial conditions, we know there is a number on the other side. whether this number is even or odd is inconsequential, so it cannot contradict the logical statement we have.

I can't believe /sci/ is actually this retarded.

The only card you don't need to turn over is D.
A needs to be checked to ensure that the number on the other side is even.
7 needs to be checked to ensure that the letter on the other side is not a vowel.
4 needs to be checked to ensure that the letter is a vowel.
D does not need to be checked because D is not a vowel and the rule makes no mention of consonants.

>>7946401
A and 4. First you need to flip a in order to check if there's an even number in the back because it's a vowel and the rule says that if it's a vowel then in the back there's an even number, in 7 you don't need to flip it because it's not an even number and the rule doesn't apply here, the same with d , d isn't a vowel so it doesn't matter what's on the back, 4 is an even so again you have to check that there's a vowel on the back of the card. As simple as that too many retards here.

>>7946401
map the problem out like this:

if the BACKSIDE of a card is even/vowel, label it as "0"
if the BACKSIDE of a card is odd/consonant, label it as "1".
label an arbitrary state as "#"

There are four cards, so we have four slots:

FRONTSIDE: A 7 D 4
BACKSIDE: _ _ _ _

note also that a consonant with an even number is acceptable per the problem.

Now, consider all the possible states of the backside (you can write out all 16 states if you want, but that will take 5 minutes and doesn't help that much).

To be true, we need the state: 0 1 # #

Since:

A MUST have an even backside
7 MUST NOT have a vowel backside
D can have any backside -- it doesn't even matter
4 can have a vowel backside (which makes statement true) or consonant backside (which is irrelevant data)

To make sure you have state 0 1 # #, you flip over A and 7. Any state other than 0 1 # # (such as 1 1 0 0) results in falsehood.

>>7946478
Try saying "Ant" and tell me how you feel

>>7956638
small

>>7953888
>>7953838
How much of an idiot can you guys be? why would you have to turn 4? if 4 has a consonant on the other side, what rule does that break?

>>7953888
If 4 does not have a vowel then the rule still holds, dummy.

>>7954482
While you're correct that the rule makes no claim about D, it does make a claim about 7 - that its backside can not be a vowel. This must be checked, along with A and 4.

>>7954968
Agree with this. Good job senpai.

A would have to be turned over. A is a vowel so we need to check to make sure the other side has an even number.

7 doesn't need to be turned over. No information was given on evens.

D doesn't need to be turned over. See above. No information/condition given on consonants.

4 doesn't need to be turned over. Yes it's an even number but no condition was given for testing. The question didn't exclude consonants from having even numbers.

Therefore A being a vowel and requiring an even number on the back, it must be checked for validation.

>>7958641
If 7 has a vowel behind it, the statement is false. You need to turn over 7 as well as A.

See >>7954968