Prove $ \{(p \lor q) \land (p \implies r) \land (q \implies r) \} \implies r$ is a tautology using logical properties I spent quite a bit of time on this and have little to no ideas on how to proceed.

Using the conditional laws and De Morgan's law, I got to
$$( \sim p \land \sim q) \lor (p \land  \sim r) \lor(q \land \sim r) \lor r$$.
Hint: the path is clearer if you leave the $\neg(p\vee q)$ as is when you use de Morgan's Law on everything else.

$\begin{array}{l|ll}
1: &amp; \big((p\vee q)\wedge (p\to r)\wedge (q\to r) \big) \to r 
\\ &amp; \Updownarrow \text{Implication equivalence, thrice}
\\ 2: &amp; \neg\big((p\vee q)\wedge (\neg p\vee r)\wedge (\neg q\vee r) \big) \vee r 
\\ &amp; \Updownarrow \text{de Morgan's Negations}
\\ 3: &amp; \neg (p\vee q)\vee (p\wedge \neg r)\vee (q\wedge\neg r) \vee r 
\\ &amp; \Updownarrow \vee \text{ Idempotent Introduction, Commutivity}
\\ 4: &amp; \neg (p\vee q)\vee (p\wedge \neg r)\vee r\vee (q\wedge\neg r) \vee r 
\\  &amp; \Updownarrow \text{Distribution, Complementation, }\wedge\text{ Identity: } (a\wedge\neg b)\vee b= a\vee b  
\\ 5: &amp; \neg (p\vee q)\vee (p\vee r)\vee (q\vee r) 
\\  &amp; \Updownarrow \text{Association, Commutation, }\vee\text{ Idempotence }
\\ 6: &amp; \neg (p\vee q)\vee (p\vee q) \vee r 
\\  &amp; \Updownarrow \text{Complementation: } A\vee \neg A=\top
\\ 7: &amp; \top \vee r 
\\  &amp; \Updownarrow \vee \text{ Annihlation: } \top\vee r=\top
\\ 8: &amp; \top 
\end{array}$
At this point in time you can't prove this as a tautology.  Tautologies get defined as well-formed formulas.  It is not clear if you mean the tautology (in Polish or Lukasiewicz notation)

C(K(K(A(p,q),C(p,r)),C(q,r)),r).

or if you mean the tautology

C(K(A(p,q),K(C(p,r),C(q,r))),r).

The proofs of those well-formed formulas are distinct.