Stuck in a case of proving that $L_1$ is complete First of all, in order to prove that $L_1$ is complete, I have to prove a couple of lemmas, where $L_1$ is like below:
Description of $L_1 :$ 
Primary connectives: $\lnot$ , $\lor$
$A\rightarrow B$ is $\lnot A \lor B$ 
Axioms:
(A1) $B \lor B \rightarrow B$ 
(A2) $B \rightarrow B \lor C$ 
(A3) $B \lor C \rightarrow C \lor B$ 
(A4) $(C \rightarrow D ) \rightarrow (B \lor C \rightarrow B \lor D)$
The only rule of inference is $MP$
But I am stuck in a case of proving one of the lemmas. I am stuck in proving that:
$\lnot B ,\lnot C \vdash_{L_1} \lnot(B \lor C)$
The problem is, none of the formulas that I have proved so far has anything like $\lnot (A \lor B)$ in it. So I am clueless. Do I have to prove some more lemmas or is it something trivial?

I found a proof using prover9. Here is an outline of the proof it finds:

*

*$\lnot B \vdash B \to X$ for any $X$ from A2 using definition of $\to$.

*$\lnot B \vdash (X \lor B) \to (X \lor X)$ for any $X$ from 1 and A4.

*$\vdash Y \lor (X \lor X) \to (Y \lor X)$ for any $X$ and $Y$ from A1 and A4.

*$\lnot B \vdash (X \lor B) \to X$ for any $X$ from 2 and 3 with $Y =\lnot (X \lor B)$ using definition of $\to$.

*$\lnot B \vdash (\lnot(B \lor C) \lor B) \to \lnot (B \lor C)$ from 4 with $X =\lnot (B \lor C)$.

*$\lnot C \vdash (Y \lor C) \to Y$ for any $Y$ proved just like 4 with $C$ in place of $B$.

*$\lnot C \vdash \lnot(B \lor C) \lor B$ from 6 with $Y = B$ and definition of $\to$.

*$\lnot B, \lnot C \vdash \lnot(B \lor C)$ from 5 and 7.

Of course, a proof on more structured lines as suggested in the comments would be more informative.
