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
$$( \sim p \land \sim q) \lor (p \land  \sim r) \lor(q \land \sim r) \lor r$$.
 A: 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.
$\begin{array}{l|ll}
1: & \big((p\vee q)\wedge (p\to r)\wedge (q\to r) \big) \to r 
\\ & \Updownarrow \text{Implication equivalence, thrice}
\\ 2: & \neg\big((p\vee q)\wedge (\neg p\vee r)\wedge (\neg q\vee r) \big) \vee r 
\\ & \Updownarrow \text{de Morgan's Negations}
\\ 3: & \neg (p\vee q)\vee (p\wedge \neg r)\vee (q\wedge\neg r) \vee r 
\\ & \Updownarrow \vee \text{ Idempotent Introduction, Commutivity}
\\ 4: & \neg (p\vee q)\vee (p\wedge \neg r)\vee r\vee (q\wedge\neg r) \vee r 
\\  & \Updownarrow \text{Distribution, Complementation, }\wedge\text{ Identity: } (a\wedge\neg b)\vee b= a\vee b  
\\ 5: & \neg (p\vee q)\vee (p\vee r)\vee (q\vee r) 
\\  & \Updownarrow \text{Association, Commutation, }\vee\text{ Idempotence }
\\ 6: & \neg (p\vee q)\vee (p\vee q) \vee r 
\\  & \Updownarrow \text{Complementation: } A\vee \neg A=\top
\\ 7: & \top \vee r 
\\  & \Updownarrow \vee \text{ Annihlation: } \top\vee r=\top
\\ 8: & \top 
\end{array}$
A: Laws used,
$$a\implies b=\overline a\lor b$$
$$a\lor (b\land c)=(a\lor b)\land (a\lor c)$$
$$a\lor \overline a=1$$
$$1\lor a=1$$
$$
\overline{(p\lor q)\land (\overline p\lor r)\land (\overline q\lor r)}\lor r
$$
$$
=\overline{(p\lor q)}\lor \overline {(\overline p\lor r)}\lor \overline {(\overline q\lor r)}\lor r
$$
$$
=(\overline p\land \overline q)\lor (p \land \overline q)\lor (q\land \overline r)\lor r
$$
$$
=(\overline q\land (\overline p\lor p)) \lor (q\land \overline r)\lor r
$$
$$
=\overline q\lor (q\land \overline r)\lor r
$$
$$
=((\overline q\lor q)\land (\overline q\lor \overline r))\lor r
$$
$$
=\overline q\lor \overline r\lor r
$$
$$
=\overline q\lor 1
$$
$$
=1
$$
A: Starting from where you correctly arrived, namely
$$( \sim p \land \sim q) \lor (p \land  \sim r) \lor(q \land \sim r) \lor r,\tag{1}$$
one can expand it using the distributive law into a conjunction of eight terms of the form $a \lor b \lor c \lor d,$ where in each case $d=r,$ while $a,b,c$ are each obtained by choosing one of the two terms in the first, second, or third parentheses of $(1).$ For example one of these terms is $\sim p \lor \sim r \lor q \lor r.$ If one can use the commutative and associative laws, and the two laws $x \lor \sim x=1$ and $y \lor 1 =1,$ then each of the eight terms in the conjunction can be shown to be $1$ since each contains at least one case of the type $x \lor \sim x$ (once they are moved next to each other). So after this application of the distributive law, your formula becomes the conjunction of eight terms which are all provably $1$, and the result is the conjunction of $1$'s so winds up as $1$ meaning it's a tautology.
A: Based on the Wiki definition --
A formula of propositional logic is a tautology if the formula itself is always true regardless of which valuation is used for the propositional variables. --
Here is another challenging solution: The truth table:

A: 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.
A: \begin{align}
(p\land\lnot r)\lor(q\land\lnot r)\lor r\equiv ((p\lor q)\land\lnot r)\lor r\equiv p\lor q \\
(\lnot p\land \lnot q)\lor(p\land\lnot r)\lor(q\land\lnot r)\lor r\equiv (\lnot p\land\lnot q)\lor (p\lor q) \equiv (\lnot p\land \lnot q) \lor p \lor q\equiv \top
\end{align}
