How to show a statement of the form $p\Leftrightarrow(q\wedge r)$?

I am trying to prove a statement of the form: \begin{gather} p\Leftrightarrow(q\wedge r) \end{gather} Therefore, I need to show the following two statements: \begin{gather} \text{(a) }\;p\Rightarrow(q\wedge r)\\ \text{(b) }\;(q\wedge r)\Rightarrow p \end{gather} Given the nature of the statement, my approach is to show the equivalent statements: \begin{gather} \text{(a’) }\;\neg(q\wedge r)\Rightarrow\neg p\\ \text{(b’) }\;\neg p\Rightarrow\neg(q\wedge r) \end{gather} What I am currently doing is this:

1. To show (a’), I just show that $\neg q\Rightarrow\neg p$ and $\neg r\Rightarrow\neg p$;
2. To show (b’), I just show that $\neg p\Rightarrow(\neg q\vee\neg r)$.

Unfortunately, I have two doubts regarding my approach:

1. When showing (a’), is it enough with what I am doing or do I need to also show $(\neg q\wedge\neg r)\Rightarrow\neg p$?
2. When showing (b’), is it enough with what I am doing or do I need to also show $\neg p\Rightarrow(\neg q\wedge\neg r)$?

Thank you all very much for your time.

Solution 1:

\begin{gather} \text{(a’) }\;\neg(q\wedge r)\Rightarrow\neg p\\ \text{(b’) }\;\neg p\Rightarrow\neg(q\wedge r) \end{gather} What I am currently doing is this:

1. To show (a’), I just show that $\neg q\Rightarrow\neg p$ and $\neg r\Rightarrow\neg p$;

This is correct, since \begin{align}&(¬q→¬p )∧ (¬r→¬p)\\\equiv &(p→q)∧ (p→r)\\\equiv &p→(q∧r)\\\equiv &¬(q∧r)→¬p.\end{align}

(It might be worth noting that $(¬q→¬p ) ∨ (¬r→¬p)\;\not\models\; ¬(q∧r)→¬p.$)

2. To show (b’), I just show that $\neg p\Rightarrow(\neg q\vee\neg r)$.

This too is correct, since \begin{align}&¬p→(¬q∨¬r)\\\equiv &p ∨(¬q∨¬r)\\\equiv &p ∨¬(q∧r) \\\equiv &¬p→¬(q∧r).\end{align}