$\neg(A\Rightarrow B) \iff A\land \neg B$ When considering the question:

Rewrite the following using only the symbols $A, B, \lor, \land, \neg$ :
  $$\neg(A\Rightarrow B)$$

I do not understand how to interpret this and what method to use to show the statement in a different form. If A and B are statements, should I assume that $A \Rightarrow B$ is true and then look for a false statement?
The answer is $A\land \neg B$
 A: As suggested in the comment, a truth table may be used to find these equivalences. However, there are some Logical Equivalences that you can build on. Particularly, you can use the facts that


*

*$A \rightarrow B \Leftrightarrow \neg A \lor B$

*$\neg ( A \lor B ) \Leftrightarrow (\neg A \land \neg B)$  (De Morgan's Law)


Applied to your problem, you can state that
$\neg (A \rightarrow B) \Leftrightarrow$ (according to first equaivalence)
$\neg (\neg A \lor B) \Leftrightarrow$ (according to De Morgan)
$A \land \neg B$
(although the first step would already qualify for "writing it only with $A, B, \lor, \land, \neg$")
A: So, let's look at the truth table for "⇒"
p   q   (p⇒q)
0   0     1
0   1     1
1   0     0
1   1     1

Thus, the only time that (p⇒q) ends up false is when p holds true, and q is false.  Otherwise (p⇒q) is true.  So, if it is not the case that p holds true, and q is false, then (p⇒q) is true.  Or in symbols if ¬(p$\land$$\lnot$q), then (p⇒q) holds true.  If you check the truth table for ¬(p$\land$$\lnot$q) you can also observe that if (p⇒q) holds true, then ¬(p$\land$$\lnot$q) holds true.  Consequently, in
¬(A⇒B)
we can replace (A⇒B) by $\lnot$(A$\land$$\lnot$B).  Doing such we obtain
¬$\lnot$(A$\land$$\lnot$B).
Since ¬¬p is logically equivalent to p, from the above we can obtain
(A$\land$$\lnot$B).
A: $\neg(A \implies B) \iff  \neg [ (\neg A) \vee B] \iff  [\neg(\neg A)] \wedge[\neg B]
  \iff A \wedge(\neg B)$.
As an exercise, you can show how each of these statements is logically equivalent to its successor.
