I would like to prove
$\{ \neg N,\neg N \to L,D \leftrightarrow \neg N \} \vdash (L \lor A) \land D $.
My work until now is as follows: $$ \begin{array}{l|ll} 1 & \neg N & \text{Premise} \\ 2 & (\neg N \to L) \land (D \leftrightarrow \neg N) & \text{Premise} \\ 3 & \vdash (L \lor A) \wedge D & \text{Premise} \\ 4 & \neg N \to L & \text{$ \land $-Elimination from $ 2 $} \\ 5 & (D \leftrightarrow \neg N) & \text{$ \land $-Elimination from $ 2 $ and $ 4 $} \end{array} $$ What’s next? This is where I’ve been for the past $ 3 $ hours. Help me out if you can, and please suggest me books for Logic and Natural Deduction.
Thanks.
The conclusion is not a premise. You can't use it to prove itself; that's circular reasoning. Circular reasoning is bad because circular reasoning is bad.
$$\begin{array}{r|ll} 1- & ¬N &\text{Premise} \\ 2- & (¬N → L) \wedge (D ↔ ¬N) &\text{Premise} \\ 3- & ¬N → L &\wedge \text{ Elimination from }2 \\ 4- & & \text{Modus Ponens from }1, 3 \\ 5- & & \vee \text{ Introduction from }4 \\ 6- & (D ↔ ¬N) & \wedge \text{ Elimination from }2 \\ 7- & (D\to \neg N)\wedge (\neg N \to D) & \text{Equivalent restatement of }6 \\ 8- & & \wedge \text{ Elimination from }7 \\ 9- & & \text{Modus Ponens from }1, 8 \\ 10 - & (L\vee A)\wedge D & \wedge \text{ Introduction of }5, 9 \\ \hline \therefore & \{\neg N, (\neg N\to L)\wedge (D\leftrightarrow \neg N)\}\vdash (L\vee A)\wedge D & \Box \end{array} $$
Can you fill in the blanks?