Implication, conjunction and disjunction distributivity problems I have proven using theorems that implication is left distributive over conjunction:
$ x \rightarrow (y \land z) \equiv ( x \rightarrow y ) \land ( x \rightarrow z ) $
Proof:
$ x \rightarrow (y \land z)
\\
\equiv
\\
\neg x \lor ( y \land z )
\\
\equiv
\\
( \neg x \lor y ) \land ( \neg x \lor z )
\\
\equiv
\\
( x \rightarrow y ) \land ( x \rightarrow z ) $
I have also proved that implication is left distributive over disjunction in a similar way. However, I am now struggling to prove:


*

*whether or not conjunction and disjunction are left distributive over implication
$ x \land ( y \rightarrow z ) \equiv ( x \land y ) \rightarrow ( x \land z ) $
$ x \lor ( y \rightarrow z ) \equiv ( x \lor y ) \rightarrow ( x \lor z ) $

*whether or not implication is right distributive over conjunction and disjunction
$ ( y \land z ) \rightarrow x \equiv ( y \rightarrow x ) \land ( z \rightarrow x ) $  (EDIT: I think I have proven this one)
$ ( y \lor z ) \rightarrow x \equiv ( y \rightarrow x ) \lor ( z \rightarrow x ) $
I would appreciate any help as to where to get started on these proofs, or what theorems I should be looking at to prove them, as well as any advice on how to prove if one of them is NOT​ distributive (i.e. how do I know that one of the above theorems is false without attempting, and failing, to prove it many, many different ways).
Thank you for any help in advance.
 A: 
How do I know that one of the above statements is false without attempting, and failing, to prove it many, many different ways?

You may simply make a truth table for each of these statements, although that's not the most economic way, it's still one of the most straightforward techniques. Let's take for instance the first one, namely, 
$$x \land ( y \to z ) \equiv ( x \land y ) \to ( x \land z ),$$
which we'll call $\small\tt P_1$. We have:
$$
\begin{array}{|c|c|c|c|c|c|c|c|}\hline
{}
\small  \color{green}{x} &\small  \color{green}{y} &\small  \color{green}{z} &\small  y\to z &\small  x\land(y\to z) &\small  x\land y &\small  x\land z &\small  (x\land y)\to(x\land z) &\small   \tt P_\color{black}{1} \\ \hline
{}
\small  \color{royalblue}{1} &\small  \color{royalblue}{1} &\small  \color{royalblue}{1} &\small  \color{royalblue}{1} &\small  \color{royalblue}{1} &\small  \color{royalblue}{1} &\small  \color{royalblue}{1} &\small  \color{royalblue}{1} &\small  \color{royalblue}{1}  \\ \hline
\small  \color{royalblue}{1} &\small  \color{royalblue}{1} &\small  \color{#C00}{0} &\small  \color{#C00}{0} &\small  \color{#C00}{0} &\small  \color{royalblue}{1} &\small  \color{#C00}{0} &\small  \color{#C00}{0} &\small  \color{royalblue}{1} \\ \hline
\small  \color{royalblue}{1} &\small  \color{#C00}{0} &\small  \color{royalblue}{1} &\small  \color{royalblue}{1} &\small  \color{royalblue}{1} &\small  \color{#C00}{0} &\small  \color{royalblue}{1} &\small  \color{royalblue}{1} &\small  \color{royalblue}{1} \\ \hline
\small  \color{royalblue}{1} &\small  \color{#C00}{0} &\small  \color{#C00}{0} &\small  \color{royalblue}{1} &\small  \color{royalblue}{1} &\small  \color{#C00}{0} &\small  \color{#C00}{0} &\small  \color{royalblue}{1} &\small  \color{royalblue}{1} \\ \hline
\small  \color{#C00}{0} &\small  \color{royalblue}{1} &\small  \color{royalblue}{1} &\small  \color{royalblue}{1} &\small  \color{#C00}{0} &\small  \color{#C00}{0} &\small  \color{#C00}{0} &\small  \color{royalblue}{1} &\small  \color{#C00}{0} \\ \hline
\small  \color{#C00}{0} &\small  \color{royalblue}{1} &\small  \color{#C00}{0} &\small  \color{#C00}{0} &\small  \color{#C00}{0} &\small  \color{#C00}{0} &\small  \color{#C00}{0} &\small  \color{royalblue}{1} &\small  \color{#C00}{0} \\ \hline
\small  \color{#C00}{0} &\small  \color{#C00}{0} &\small  \color{royalblue}{1} &\small  \color{royalblue}{1} &\small  \color{#C00}{0} &\small  \color{#C00}{0} &\small  \color{#C00}{0} &\small  \color{royalblue}{1} &\small  \color{#C00}{0} \\ \hline
\small  \color{#C00}{0} &\small  \color{#C00}{0} &\small  \color{#C00}{0} &\small  \color{royalblue}{1} &\small  \color{#C00}{0} &\small  \color{#C00}{0} &\small  \color{#C00}{0} &\small  \color{royalblue}{1} &\small  \color{#C00}{0} \\ \hline
\end{array} 
$$
Hence $\small\tt P_1$ is certainly not a tautology. The use of a truth table is unnecessary if we observe from the beginning that we do not have an equivalence when $x$ is false. You can do the same for the remaining ones,
$$\begin{align}
x \lor ( y \to z ) &\equiv ( x \lor y ) \to ( x \lor z ),
\\
( y \lor z ) \to x &\equiv ( y \to x ) \lor ( z \to x ),
\end{align}$$
which we will respectively denote as  $\small\tt P_2$ and $\small\tt P_3$. You'll see that $\small\tt P_2$ is the only tautology.
Now, to show that this is indeed the case $-$ without relying on a truth table $-$ we'll start with the RHS,
$$\begin{align}
(y\to x)\lor(z\to x) & \equiv (\lnot y\lor x)\lor(\lnot z\lor x) \\
                     & \equiv (\lnot y\lor\lnot z)\lor x         \\
                     & \equiv \lnot(y\land z)\lor x              \\
                     & \equiv (y\land z)\to x. \tag*{$\small\square$}
\end{align}
$$
A: Prove of the above equation
p -> (q ^ r)  <-> (p -> q) ^ (p -> r)
https://i.stack.imgur.com/D7im1.jpg
