# 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.

• Your approach is fine. (a') implies (a) and (b') implies (b), so if you show (a') and (b') are correct then it follows that (a) and (b) are as well. If this is an exercise in a mathematical logic course you might want to add an explanation for why (a') and (b') imply (a) and (b).
– Snaw
Commented Jan 5, 2022 at 13:08
• Thank you for your comment. Unfortunately, my question is not about whether (a’) implies (a) or whether (b’) implies (b) (which I know is true); but rather whether my approach to showing (a’) and (b’) is correct and complete. Commented Jan 5, 2022 at 13:15
• Right. I meant to write that (a'') implies (a') and (b'') implies (b') where (a''),(b'') are the items you wrote in 1, 2. For instance (b'') is equivalent to (b') by De Morgan's laws en.wikipedia.org/wiki/De_Morgan%27s_laws
– Snaw
Commented Jan 5, 2022 at 13:20

$$\begin{gather} \text{(a’) }\;\neg(q\wedge r)\Rightarrow\neg p\\ \text{(b’) }\;\neg p\Rightarrow\neg(q\wedge r) \end{gather}$$ 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 $$(¬q→¬p ) ∨ (¬r→¬p)\;\not\models\; ¬(q∧r)→¬p.$$)
1. 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}