Proof through contradiction My textbook says: To prove $A\Rightarrow B$ we have to lead $A \wedge \neg B$ to a contradiction. Does it imply, that $B\Rightarrow A$ would also be true? As far as I know $\wedge$ is commutative.
 A: The fact that $\wedge$ is commutative means that $A\wedge(\neg B)$ is logically equivalent to $(\neg B)\wedge A$.  But to prove $B\Rightarrow A$ by contradiction you would start by assuming $B\wedge(\neg A)$, which is not the same.
A: The conditional statement $A \Rightarrow B$ is true when it is not the case that $A$ is true and $B$ is false. That is, the conditional is true when we have $\lnot (A \land \lnot B)$. That is why the textbook says

To prove $A \Rightarrow B$ we have to lead $A \land \lnot B$ to a contradiction.

Here is the question:

Does it imply, that $B \Rightarrow A$ would also be true? As far as I know $\land$ is commutative.

To see how the converse, $B \Rightarrow A$, of the original conditional fits into this consider the following.
Using De Morgan's laws, $\lnot (A \land \lnot B)$ is equivalent to $\lnot A \lor B$ which gives us the material implication rule of inference:
$$A \Rightarrow B \iff \lnot A \lor B$$
That the disjunction is commutative gives us the following:
$$\lnot A \lor B \iff B \lor \lnot A$$
That corresponds to the equivalence between the original conditional and its contrapositive:
$$A \Rightarrow B \iff \lnot B \Rightarrow \lnot A$$
However, $B \Rightarrow A$ is the converse, not the contrapositive, of the conditional $A \Rightarrow B$. If both the conditional and its converse are true then the biconditional, $A \iff B$, is true.  The contrapositive of the converse is called the inverse of the original conditional:
$$B \Rightarrow A \iff \lnot A \Rightarrow \lnot B$$
