Is there a distributive rule for the implication?

21 October 2018 in MAE 106: Discrete Mathematics with Probability Fall 2018 | No comments

We know that we have distributive rules that handle the disjunction and the conjunction, such as the following

$\begin{align*} p \and (q \orr r) &\Leftrightarrow (p \and q) \orr (p \and r)\\ p \orr (q \and r) &\Leftrightarrow (p \orr q) \and (p \orr r). \end{align*}$

But, does there exist a rule, for example, that would handle distributivity of the conjunction with the implication? In other words,

$\text{is } p \and (q \rightarrow r) \text{ logically equivalent to } (p \and q) \rightarrow (p \and r)?$

At least, the above case, the answer is no. We can see this by constructing a truth table like the one below; the truth values of $p \and (q \rightarrow r)$ are not the same as $(p \and q) \rightarrow (p \and r)$ . Hence, $p \and (q \rightarrow r)$ and $(p \and q) \rightarrow (p \and r)$ are not logically equivalent. Therefore, it’s not a rule that we can apply in our proofs.