From $(\neg A \land \neg B)$ I will deduce $\neg(A\lor B)$
1.If $(\neg A \land \neg B)$:
2. $\neg A$[and elim]
3. $\neg B$[and elim]
4. If A:
$\neg A$
$\bot$
5. If B:
$\neg B$
$\bot$
6. $A \implies \bot$
7. $B \implies \bot$
8. $(A \lor B) \implies \bot$[or elim]
9. $\neg(A \lor B)$
I didnt write all the steps but is the framework of this proof correct?
The perpendicular sign means contradiction; I am pretty sure my proof is a little wrong even though i can’t find my mistake since my teacher said it’s wrong
That is mostly it.
However, to use disjunction elimination (OR elim, ${\lor}\mathsf E$), you must have a disjunction to eliminate. So assume one. After all, when a contradiction is derived from an assumption, you may use negation introduction (NOT intro, $\lnot\mathsf I$) to deduce its negation. As is desired.
$\def\fitch#1#2{~~~~\begin{array}{|l}#1\\\hline#2\end{array}}\fitch{~~1.~\lnot A\land\lnot B\qquad\mathsf P}{~~2.~\lnot A\qquad\qquad{\land}\mathsf E~1\\~~3.~\lnot B\qquad\qquad{\land}\mathsf E~1\\\fitch{~~4.~A\lor B\qquad\mathsf A}{\fitch{~~5.~A\qquad\mathsf A}{~~6.~\bot\qquad\lnot\mathsf E~5,2}\\\fitch{~~7.~B\qquad\mathsf A}{~~8.~\bot\qquad\lnot\mathsf E~7,3}\\~~9.~\bot\qquad\qquad{\lor}\mathsf E~~4,5{-}6,7{-}8}\\10.~\lnot(A\lor B)\qquad\lnot\mathsf I~4{-}9}$