Proof using Laws of Logic Using the laws of logic, prove:
p → (q ∧ r) ≡ (p → q) ∧ (p → r)

My attempt to prove this:
p → (q ∧ r)

Implication Law: ¬p ∨ (q ∧ r)

Distribution Law: (¬p ∨ q) ∧ (¬q ∨ r)

I am unsure how to correctly apply the laws of logic (without the use of Truth Tables) so that:
p → (q ∧ r) ≡ (p → q) ∧ (p → r)

 A: You didn't apply the Distribution Law correctly.
You should go from:
$$\neg p \lor (q \land r)$$
to:
$$(\neg p \lor q) \land (\color{red}{\neg p} \lor r)$$
and from there you use Implication twice to get:
$$(p \rightarrow q) \land (p \rightarrow r)$$
A: Note that 
$$p → (q ∧ r) ≡$$ 
$$¬p ∨ (q ∧ r) ≡ $$
$$(¬p ∨ q) ∧ (¬p ∨ r) ≡$$
$$(p → q) ∧ (p → r)$$
Thus $$ p → (q ∧ r)≡(p → q) ∧ (p → r)$$
A: $\def\fitch#1#2{\begin{array}{|l}#1\\\hline#2\end{array}}$By natural deduction (classical and intuitive logic)
Take $p\to(q\wedge r)$ as a premise.   Assuming $p$ infers $q\wedge r$ which entails $q$, and so we conclude $p\to q$. Likewise we conclude $p\to r$.   Thus the premise entails $(p\to q)\wedge (p\to r)$.
Conversely take $(p\to q)\wedge (p\to r)$ as a premise.   This entails both $p\to q$ and $p\to r$.   Assuming $p$ thereby infers $q$ and $r$.   Thus the premise entails $p\to (q\wedge r)$.
Therefore proving the equivalence. $p\to (q\wedge r) \dashv\vdash (p\to q)\wedge(p\to r)$ 
${\fitch{p\to(q\wedge r)}{\fitch{p}{q\wedge r\\q}\\p\to q\\\fitch{p}{q\wedge r\\r}\\p\to r\\ (p\to q)\wedge (p\to r)}\\ \\[2ex] \fitch{(p\to q)\wedge (p\to r)}{p\to q\\p\to r\\\fitch{p}{q\\r\\q\wedge r}\\p\to (q\wedge r)}}$
Identifying which of the Rules of Inference were used is left to the student.
