How do I simplify (SA ∧SB ∧SC)→(B∧¬A∧¬C)?

Question

It is given that

1. $SA = B \land \neg C$
2. $SB =A \to C$
3. $SC = \neg C \land (A \lor B)$

How do I get $(SA \land SB \land SC) \to (B \land \neg A \land \neg C)$?

I did this (below) but now I'm stuck.

$$B \land \neg C \land A \to C \land \neg C \land (A \lor B) \\ = \neg C \land (B \land (A \lor B)) \land \neg A \lor C$$

Answer 1

So you have:

$\begin{align}&(B\wedge\neg C)\wedge(A\to C)\wedge(\neg C\wedge (A\vee B)) \\&(B\wedge\neg C)\wedge(\neg A\vee C)\wedge(\neg C\wedge(A\vee B))&&\text{by conditional equivalence}\\&\neg C\wedge\neg C\wedge B\wedge (A\vee B)\wedge (\neg A\vee C)&&\text{by association and commutivity}\\&\neg C\wedge B\wedge (A\vee B)\wedge (\neg A\vee C)&&\text{by idempotence}\end{align}$

That checks out okay. So, now use absorption: $B\wedge (A\vee B)~=~B$ $$\begin{align}&B\wedge (A\vee B)\\&(B\vee \bot)\wedge(B\vee A)\\&B\vee(\bot\wedge A)\\&B\vee\bot\\&B\end{align}$$

Next use distribution, and the path should be clear from then on.