Propositional Calculus, Can someone answer the following?

Question

Can somebody please solve the following equations: \begin{align} 1. \quad (A \rightarrow B)\land (A\rightarrow \neg B)=\lnot A \quad \quad \\ \end{align} What I have got for it so far is

$$(¬A\lor B)\land (¬A\lor ¬B)\\ (¬A\lor B)∧¬A \lor (¬A\lor B)∧¬B\\ (¬A\lor ¬A)\lor (B∧¬A) \lor (¬A∧¬B) \lor(B∧¬B)$$

After this I'm not sure.

Thanks in advance.

Answer 1

$$ \begin{align} (A \rightarrow B)\wedge (A\rightarrow \lnot B) & \equiv (\lnot A\lor B)\land (\lnot A \lor \lnot B)\tag{Implication}\\ &\equiv \lnot A \lor (B \land \lnot B)\tag{Distribution}\\ & \equiv\lnot A \lor 0\\ &\equiv \lnot A \end{align}$$

$$\begin{align} A\rightarrow (B\rightarrow C) &\equiv \lnot A \lor (\lnot B \lor C) \tag{Implication $\times 2$}\\ &\equiv (\lnot A \lor \lnot B)\lor C \tag{Associativity of $\lor$}\\ &\equiv \lnot(A\land B) \lor C\tag{DeMorgan's}\\ &\equiv (A\land B)\rightarrow C\tag{Implication} \end{align}$$

Now go back and try working out the third exercise, using any or all of the above identities. If you'd like to check out what you get for $(3)$, feel free to comment below this post.

Answer 2

Another way to look at it: Suppose $A$ is true. Then we can obtain the contradiction $B\land\neg B$ from the two implications. So, $A$ must be false, i.e. we have $\neg A$.