Natural deduction $\{A \vee L, A \leftrightarrow N, L \rightarrow N\} \vdash N$ Natural deduction $\{A \vee L, A \leftrightarrow N, L \rightarrow N\} \vdash N$
my work
1- $A \vee L$
2- $A \vee$ Elim FROM 1
3- $A \leftrightarrow N$
4- $A \leftrightarrow$ Elim FROM 3
5- $L \rightarrow N$
6- $L \rightarrow$ Elim FROM 5
is that correct or wrong, need guide please
 A: The disjunction elimination rule allows to eliminate a disjunctive statement by reasoning that if $A$ implies $C$ and $B$ implies $C$, then if $A$ or $B$ is asserted, we can conclude $C$ directly. Symbolically:
$$\frac{\begin{align}
A \rightarrow C, B \rightarrow C, A \lor B
\end{align}}{\begin{align} C \end{align}}$$
That is, whenever both disjunctors, separately, imply a certain sentence, we can just conclude it.
That being said, we can apply it in the following:

  
*
  
*$A ↔ N$, Premise
  
*$A \rightarrow N$, 1, $↔$E
  
*$L \rightarrow N$, Premise
  
*$A \lor L$, Premise
  
*$N$, 2,3,4, $\lor$E
  

This shows that $A \lor L, A ↔ N, L → N ⊢ N$ (In fact, note that we don't even need $A ↔ N$, we only need one side of it, $A \rightarrow N$)
A: That's not how the elimination rules work.


*

*Conjunction elimination ($\wedge$ elim.) is the rule: $$\;P\wedge Q\vdash P$$ However...

*Disjunction elimination ($\vee$ elim.) is the rule: $$P\vee Q, P\to R, Q\to R\vdash R$$

*Biconditional elimination ($\leftrightarrow$ elim.) is the rule: $$P\leftrightarrow Q \vdash P\to Q$$

*Implication elimination ($\to$ elim.) , also known as Modus Ponens (MP), is the rule: $$P, P\to Q \vdash Q$$


*

*Contrapositive elimination, also known as Modus Tollens (MT), the rule: $$\neg Q, P\to Q\vdash \neg P$$




tl;dr Just use Biconditional Elimination and follow with Disjunction Elimination
$$\frac{\begin{align}
A\leftrightarrow N      & \vdash A\to N 
\\
A\vee L, A\to N, L\to N & \vdash N
\end{align}}{\begin{align}A\vee L, A\leftrightarrow N, L\to N & \vdash N\end{align}}$$
