Puzzle on Reasoning The following puzzle was introduced by the psychologist Peter Wason in 1966, and is one of the
most famous subject tests in the psychology of reasoning.

Four cards are placed on the table in front of you. You are told
(truthfully) that each has a letter printed on one side and a digit on
the other, but of course you can only see one face of each. What you
see is:
                                                     B E 4 7

You are now told that the cards you are looking at were chosen to
follow the rule \If there is a vowel on one side, then there is an odd
number on the other side." What is the least number of cards you have
to turn over to verify this rule, and which cards do you in fact have
to turn over?

My solution:

Assuming A: Vowel on one side   B: Odd number on one side
by the data given to us, A(conditional)B  {A implies B}. We have to
verify this rule. We can do this by proving that the contrapositive of
the above given statement is true, which says (NO ODD IMPLIES NO
VOWEL). By doing this, We'll have to turn the "E" and the "4" card.

The correct answer is : "E" and "7" card.
Working:

You need to turn over 2 cards (the E and the 7). The rule can be
(simply) equivalently stated like so: There isn't a vowel on one side
OR there is an odd number on one side. Thus, if you see a letter and
it isn't a vowel, you don't need to verify the rule; likewise, if you
see a number and it's odd, you don't need to verify the rule.

I have two concerns, 1. How is my reasoning wrong. and 2. How is the statement -"There isn't a vowel on one side OR there is an odd number on one side" equivalent to - "If there is a vowel on one side, then there is an odd number on the other side."
 A: Your answer (not the book's) is correct. You are given a proposition that is a unidirectional implication $vowel \implies odd$. You need to test this by turning over any vowel card(s) you see to confirm the reverse side(s) is(are) odd(s).
But every implication also implies the contrapositive. Which means $not \ odd \implies not \ vowel$ which is equivalent to $even \implies consonant$. So you need to test this by turning over any even card(s) to confirm the reverse side(s) is(are) consonant(s).
In your set, that means only $E$ and $4$ need be turned over. The rest are irrelevant to the proposition.
A: The statements $(1)~V\implies O$ and $(2)~V'\cup O$ are equivalent. See the truth table:$$\begin{matrix}V&O&(1)&(2)\\0&0&1&1\\0&1&1&1\\1&0&0&0\\1&1&1&1\end{matrix}$$But you arrived at the correct answer. The book's explanation is also correct: since $B$ is not a vowel and $7$ is odd, these two cards already satisfy the rule. The ones you will need to check are, as you found out, $E,4$, to check whether the other side is having an odd number and non-vowel respectively. There seems to be a misprint.
