How to Prove Something is a Contradiction only by Logical Equivalencies

By the distributive rule, we have:

$$\begin{align}(p \lor q) \land (p \lor \lnot q) \land (\lnot p \lor q) \land (\lnot p \lor \lnot q)&\equiv [p \lor (q\land \lnot q)] \land [\lnot p \lor (q \land \lnot q)]\\ \\ &\equiv (p \lor \bot) \land (\lnot p \lor \bot)\\ \\ &\equiv (p \land \lnot p)\\ \\ &\equiv \bot\end{align}$$

Note that $\bot$ means "logically false," aka a "contradiction".

We are going to prove that

$(\lnot p \rightarrow q) \land (\lnot p \rightarrow \lnot q)\equiv p\;.$

Indeed,

$(\lnot p \rightarrow q) \land (\lnot p \rightarrow \lnot q)\equiv (p\lor q)\land(p\lor\lnot q)\equiv\\\equiv p\lor(q\land\lnot q)\equiv p\lor\text{False}\equiv p\;.$

Moreover it results that

$(p \rightarrow q) \land (p \rightarrow \lnot q)\equiv \lnot p\;.$

Indeed,

$(p \rightarrow q) \land (p \rightarrow \lnot q)\equiv (\lnot p\lor q)\land(\lnot p\lor\lnot q)\equiv\\\equiv \lnot p\lor(q\land\lnot q)\equiv \lnot p\lor\text{False}\equiv \lnot p\;.$

Hence,

$(\lnot p \rightarrow q) \land (\lnot p \rightarrow \lnot q)\land(p \rightarrow q) \land (p \rightarrow \lnot q)\equiv\\\equiv p\land\lnot p\equiv\text{False}\;.$