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i had a puzzle and used a logical argument to show a point but some says that my argument is wrong but i can't understand the reason they provide !

the puzzles says , Given four cards laid out on a table as: $D , 3 , F , 7$ , where each card has a letter on one side and a number on the other. then Which cards should you flip over to determine if every card with a $D$ on one side has a $7$ on the other side?

i solved so , my question is not to solve the puzzle i claimed that there is no need to flip $7$ over . and my argument as follows

let $P$ = the card has $D$ on one side $Q$ = the card has $7$ on the other side

let , $A$ $=$ $P$ $\rightarrow$ $Q$

$B$ $=$ $¬Q$ $\rightarrow$ $¬P$

from the truth table we know that any wff of those A and B tautolofically implies the other so they are equvlant and we can use any one of then instead of the other

so , we want to show that , if the card has $D$ on one side then it has $7$ on the other side

so we can use the equivlant wff which says , if the card doesn't have $7$ then it hasn't $D$ on the other side

and we know that the fourth card has $7$ , so $¬Q$ is wrong so $B$ so true so $A$ is true

so we don't need to flip card 7

is this argument right ?

they say that i have to show that $ A$ is true before using the equivlant between $A$ and $B $

what is right and why ?

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It's hard to parse your question, but yes, we don't need to flip the card showing $7$. We need to flip the card showing $D$ to see if condition $A$ is satisfied. Then we need to flip the card showing $3$ as well, because we need to prove the equivalent condition $B$ for that one. Cards $7$ and $F$ represent the converse case of conditions $A$ and $B$ respectively, hence do not need to be true. – Ian Coley Apr 7 '13 at 0:57
i know that we don't need to flip 7 , my question is about my argument , is it true or no . ? – Maths Lover Apr 7 '13 at 0:59
I will defer to the answer below. – Ian Coley Apr 7 '13 at 1:00
This is called the Wason selection task; it was devised in 1966 by Peter Wason. Wason observed that although the people get the correct answer unreliably and with difficulty when the question is presented with numbers on cards, they have no trouble with it when it is translated into an equivalent question about social contract violations. For example, instead of cards, you see four people in a bar; which ones might be breaking the rule about underage drinking. – MJD Apr 7 '13 at 5:29
up vote 4 down vote accepted

You do need to flip D to check and confirm that $7$ is on the other side: to establish that indeed, $P \rightarrow Q$.

But you need to also flip $3$, to confirm that $\lnot Q \rightarrow \lnot P$: that if a number is not $7$, then you need to know whether the letter is not $D$.

Without checking this latter card, you might very well have one example in which you confirmed the statement. But if if the other side of the card with $3$ is $D$, you would have a counterexample to the statement: You would have that $P is true (card has D on it), Q is false ("the other side is 7" is false), hence an invalid implication.

Clarification: NO OTHER CARD needs to be flipped. Just those two cards $D$, $3$ need to be flipped. So there is no need to flip $7$, no need to flip $F$.

Those who argue that you would need to flip $7$ are not understanding that the assertion in question is a material implication ($\rightarrow$). They are mistken in interpreting the assertion to be biconditional ($\Longleftrightarrow)$. You would only need to check the other two cards if you were trying to determine whether $P \Longleftrightarrow Q$. But that is not what is being claimed.

Bottom line: you're right!

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my question is not about the puzzle :) i know i have to flip 3 , i ask about the argument provided not any thing else . i mean , my qestion is about card 4 which has 7 and the argument , is this argument to to show that we don't have to flip 7 or not :) nice to see your post :) – Maths Lover Apr 7 '13 at 1:05
Nice to see you too, Maths Lover! I see you've changed back to your former user name? – amWhy Apr 7 '13 at 1:19
Are you clear why you do not need to flip the card with $F$ on it? We would have for $A = P \rightarrow Q$, P is false, and $A$ evaluates to true no matter what the case is for $Q$ (truth-table: $F\rightarrow T$, $F\rightarrow F$ both evaluate to true) – amWhy Apr 7 '13 at 1:25
yup :) , they agree that we don't need to flip 7 , they refuse my argument , the question is not about the puzzle at all ! it's about the argument provided to show that we don't need to flip 7 , so , i mean , in the equivlant between $A$ = $P$ $\rightarrow$ $Q$ and $ B$= $¬Q$ $\leftrightarrow$ $¬P$ , if we have $A$ , can we always use $B$ instead of $A $ ? or only when $A$ is true ? – Maths Lover Apr 7 '13 at 1:30
If we have A, then then B is true, if we have B then A is true. If one holds, then they both hold. Yes, you can use B instead of A, but since they are logically equivalent, this essentially means that using one is using the other (in terms that they are equivalent.) Either can be used to show most directly an assertion. And your application of $B$ in your argument for $7$ not needing to be flipped is the most direct. – amWhy Apr 7 '13 at 1:33

I think you are right and they are wrong.

Logically $A$ and $B$ are equivalent. You don't have to show $A$ is true to know that.

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