Given ~(A->B) how would I reach the conclusion of A&~B

Question

I'm pretty stuck on this question and not sure how to be able to derive the conclusion from the premise. Any help would be appreciated!

My work so far:

Answer 1

To reach $A\&\lnot B$, prove $A$ and $\lnot B$ separately, then use conjunction introduction. To prove them use a reduction to absurdity and an indirect proof, respectively. Both require deriving a contradiction from an assumption, and the only other thing to contradict is the premise, which is a negation of a conditional), so derive that conditional...

The exact rules and format you use will depend on your proof system, but the skeleton will basically be:

$$\def\fitch#1#2{~~~~\begin{array}{|l}#1\\\hline#2\end{array}}\fitch{\lnot (A\to B)}{\fitch{\lnot A}{\fitch{A}{~\vdots\\B}\\A\to B\\\bot}\\\lnot\lnot A\\A\\\fitch{B}{\fitch{A}{~\vdots\\B}\\A\to B\\\bot}\\\lnot B\\A\,\&\,\lnot B}$$

Answer 2

In your attempt, you used ($\lor$ Elim) to show $a\to b$, in order to conclude $\lnot b$, which is correct. However, it's unnecessary, we can just assume $a$, then reit $b$, so we have $a\to b$ hold.

Then you assumed $\lnot a$, but stucked on showing that $a\to b$, a hint for this is that once we assume $a$ under $\lnot a$ it's a contradiction, use this and ($\lnot$ Elim) to derive $b$.

The proof looks like the following $\def\fitch#1#2{\quad\begin{array}{|l}#1\\\hline#2\end{array}}$ $\def\pra#1{\left(#1\right)}$ $$\fitch{1.~\lnot\pra{a\to b}} {\fitch{2.~\lnot a} {\fitch{3.~a} {\fitch{4.~\lnot b}{5.~\lnot a\hspace{5ex}\text{2, R}\\ 6.~a\hspace{6.6ex}\text{3, R}} \\7.~b\hspace{10.3ex}\text{4-6, $\lnot$ E}} \\ 8.~a\to b\hspace{8.9ex}\text{3-7, $\to$ I}\\ 9.~\lnot\pra{a\to b}\hspace{4.6ex}\text{1, R}}\\ 10.~a\hspace{15.9ex}\text{2-9 $\lnot$ E}\\ \fitch{11.~b} { \fitch{12.~a} {13.~b\hspace{9ex}\text{11, R}}\\ 14.~a\to b\hspace{7.6ex}\text{12-13, $\to$ I}\\ 15.~\lnot\pra{a\to b}\hspace{3.5ex}\text{1, R}}\\ 16.~\lnot b\hspace{14.5ex}\text{11-15, $\lnot$ I}\\ 17.~a\land\lnot b\hspace{10.8ex}\text{10,16 $\land$ I}} $$