Problem: Prove that $(A \land B)' \land (C' \land A)' \land (C \land B')' \to A'$.
What I have done so far:
$R \equiv(A'\lor B')\land(C\lor A')\land(C'\lor B) $
$\equiv(A'\lor(B'\land C))\land(B\lor C') $
$\equiv((A'\lor(B'\land C))\land B)\lor ((A'\lor(B'\land C))\land C')$
$\equiv((B\land A')\lor (B\land (B'\land C)) \lor((C'\land A')\lor (C'\land (B'\land C))$
Since $(B\land (B'\land C)) \equiv 0$ and $(C'\land (B'\land C)) \equiv 0$, We have that :
$R \equiv (B\land A')\lor (C'\land A') \equiv A' \land(B\lor C')$
So you can say that if $R\equiv 1$, then $A'\equiv 1$.
Consider the following: Let $$ \Omega\equiv\neg(A\land B)\land\neg(\neg C\land A)\land\neg(C\land\neg B)\to\neg A. $$ Your job, then, is to show that $\Omega$ is a tautology: \begin{align} \Omega&\equiv\neg(A\land B)\land\neg(\neg C\land A)\land\neg(C\land\neg B)\to\neg A\tag{by definition}\\[1em] &\equiv (\neg A\lor\neg B)\land(C\lor\neg A)\land(\neg C\lor B)\to\neg A\tag{DeMorgan}\\[1em] &\equiv[A\land(B\lor\neg C)]\lor(C\land\neg B)\lor\neg A\tag{mat. imp., distrib.}\\[1em] &\equiv [(A\lor\neg A)\land(B\lor\neg C\lor\neg A)]\lor(C\land\neg B)\tag{distrib.}\\[1em] &\equiv (B\lor\neg C\lor\neg A\lor C)\land(B\lor\neg C\lor\neg A\lor\neg B)\tag{distrib.}\\[1em] &\equiv \mathbf{T}\land\mathbf{T}\\[1em] &\equiv \mathbf{T}. \end{align} The biggest "jump" is from the second to the third step. If you can see that, then you should be well on your way.
Good so far.
You have shown that $(A \land B)' \land (C' \land A)' \land (C \land B')' \equiv (A' \lor B') \land (C \lor A') \land (C' \lor B)$
You can then write $(A' \lor B') \land (C \lor A') \land (C' \lor B) \equiv (A' \lor B') \land \left [(C \lor A') \land C' \lor (C \lor A') \land B) \right]$ (distributive law)
Then consider smaller parts like $(C \lor A') \land C' \equiv [C \land C'] \lor [A' \land C'] \equiv 0 \lor [A' \land C'] \equiv A' \land C'$
