Natural deduction proof: $(A \lor B) \land (A \lor C) \vdash A \lor (B \land C)$

Question

My way that didn't work:

\begin{align*} 1. & \quad (A \lor B) \land (A \lor C) & \text{Premise} \\ 2. & \quad A \lor B & \text{E $\land$ 1} \\ 3. & \quad [A & \text{Hypothesis} \\ 4. & \quad \quad A \lor (B \land C)] & \text{VI 3} \\ 5. & \quad [B & \text{Hypothesis} \\ 6. & \quad \quad [C & \text{Hypothesis} \\ 7. & \quad \quad(B \land C)] & \text{$\land$I 5,6} \\ 8. & \quad A \lor (B \land C)] & \text{VI 3} \\ 9. & \quad A \lor (B \land C) & \text{$\lor$E 2,3-4,5-8} \\ \end{align*}

I couldn't think of another way and the way I did is clearly wrong.

Answer 1

You had the correct approach, and were almost there. You just had to add a few more lines to make the nested $\vee$-elimination work.

$\def\fitch#1#2{\quad\begin{array}{|l}#1\\\hline#2\end{array}} \fitch{~~1.~~(A\vee B)\wedge(A\vee C)}{~~2.~~A\vee B\hspace{16ex}{\wedge}\mathsf E~1\\\fitch{~~3.~~A}{~~4.~~A\vee(B\wedge C)\hspace{6ex}{{\vee}\mathsf I~3}}\\\fitch{~~5.~~B}{\color{red}{~~6.~~A\vee C\hspace{12ex}{{\wedge}\mathsf E~1}}\\\fitch{~~7.~~C}{~~8.~~B\wedge C\hspace{8ex}{{\wedge}\mathsf I~5,7}\\{~~9.~~A\vee(B\wedge C)\hspace{2ex}{{\vee}\mathsf I~8}}}\\\color{red}{10.~~A\vee(B\wedge C)\hspace{5.5ex}{{\vee}\mathsf E~6,3{-}4,7{-}9}}}\\11.~~A\vee(B\wedge C)\hspace{9ex}{{\vee}\mathsf E~2,3{-}4,5{-}10}}$

Answer 2

The idea is: in the second case $B$, you need to use ${\lor}E$ on $A \lor C$. In the first case (under the outer-level second case), you'll again have $A$ from which you can conclude what you need to. In the second case (under the outer-level second case), you now have assumptions of both $B$ and $C$ available and you will be able to conclude what you need to.