If $(A \vee B) \wedge (¬B \vee C)$ is true, then $(A \vee C)$ must be true ... can I argue that? 
If $(A \vee B) \wedge (¬B \vee C)$ is true, then $(A \vee C)$ must be true ... can I argue that?

I don't see how I can argue that $(A \vee C)$ must be true because can't we have 
$(T \vee T) \wedge (F \vee F) \Rightarrow T \wedge T \Rightarrow T$ and it still is true even though both sides of the "or" are true? So how can I argue that $(A \vee C)$ must be true?
 A: $$(A\lor B) \land (\lnot B \lor C)$$
If the premise above is true, then by conjunction elimination, 


*

*$A\lor B$ is true $(1)$

*and $\lnot B \lor C \equiv B\rightarrow C$ is true.$(2)
$(1) \;A:\;$Suppose A is true. Then $A\lor C$ is true ($\lor$-Introduction).
$(1)\;B:\;$ Suppose B is true. Then by modus ponens with $(2): B\rightarrow C$, we have that $C$ is true. If C is true, then so is $A\lor C$, by $\lor$-Introduction.
In either case, $A\lor B$, which being true means $A$ is true or $B$ is true (or both) we have shown that with $(2)$, it follows that $A\lor C$ must be true.
A: You always have  $B\lor \lnot B$.
If $B$ then $\lnot B\lor C$ implies $C$ which on its turn implies $A\lor C$.
If $\lnot B$ then $A\lor B$ implies $A$ which on its turn implies $A\lor C$.
A: It's not an equivalence, it's an implication.
We need to show: $(A\wedge B)\vee(\neg B\wedge C)\to (A\vee C)$
$\begin{align}
& (A\wedge B)\vee(\neg B\wedge C) & \text{Premise}
\\
\iff & ((A\wedge B)\vee\neg B)\wedge ((A\wedge B)\vee C) & \text{Distribution}
\\
\iff & ((A\vee\neg B)\wedge (B\vee\neg B))\wedge ((A\wedge B)\vee C) & \text{Distribution}
\\
\iff & ((A\vee\neg B)\wedge \top)\wedge ((A\wedge B)\vee C) & \text{Complementation}
\\
\iff & (A\vee\neg B)\wedge((A\wedge B)\vee C) & \text{Identity}
\\
\iff & (A\vee\neg B)\wedge((A\vee C)\wedge (B\vee C)) & \text{Distribution}
\\
\iff & (A\vee C)\wedge((A\vee\neg B)\wedge(\wedge (B\vee C)) & \text{Commutivity}
\\[2ex]
\therefore & (A\wedge B)\vee(\neg B\wedge C)\implies (A\vee C)
\end{align}$
