How to prove that $p\to(q\wedge r)$ and $(p\to q)\wedge(p\to r)$ are logically equivalent

Question

Is it true that

1. p->(r ∧ q) is logically equivalent to (p->r) ∧ (p->q),

2. p->(r ∨ q) is logically equivalent to (p->r) ∨ (p->q)?

I am solving problems on logical arguments (check whether they are valid) and it would be easier if I could break down the statements above and check them seperately. I wrote the truth tables, which turned out to match, so I think they should be equivalent, but I am not sure.

Below are the truth tables that I used

Answer 1

You can also apply the properties of the logical operators involved: \begin{align*} (p\to(q\wedge r)) & \Longleftrightarrow (\neg p\vee(q\wedge r))\\\\ & \Longleftrightarrow (\neg p\vee q)\wedge(\neg p\vee r)\\\\ & \Longleftrightarrow (p\to q)\wedge(p\to r) \end{align*}

Similarly, we do also have that \begin{align*} (p\to(q\vee r)) & \Longleftrightarrow (\neg p\vee (q\vee r))\\\\ & \Longleftrightarrow ((\neg p\vee\neg p)\vee(q\vee r))\\\\ & \Longleftrightarrow (\neg p\vee q)\vee(\neg p\vee r)\\\\ & \Longleftrightarrow (p\to q)\vee(p\to r) \end{align*} and we are done.