I have this issue in natural deduction:
B -> (A & ~A) ├ ~B
I am having a really hard time understanding the rules in natural deduction - but this is what I got so far - can someone help me to the end?
B / premis
(A&~A) /premis
B-> ^ /1, 2 ^E
B /1
^ /3->E
~B /4,5 ->I
Thank you!
$\endgroup$ 32 Answers
$\begingroup$For a start, neither (1) nor (2) are given to you as premisses. Your premiss is the given conditional, and then your proof should proceed by assuming $B$ for reductio ...
To be honest, your proof attempt suggests you really haven't understood ND at all. For a gentle and I hope very accessible intro to ND (which is quite freely available) try the relevant chapters of my book Introduction to Formal Logic which you can download from
$\endgroup$ $\begingroup$Natural Deduction's rules of inference revolve around 'introducing' or 'eliminating' logical connectives. So that is where we start.$\def\fitch#1#2{~~~~~\begin{array}{|l}#1\\\hline#2\end{array}}$
You are given a premise $B\to(A\land\lnot A)$ and wish to infer $\lnot B$. Well there is a conditional and conjunction to eliminate in the premise, and a negation to introduce in the conclusion. Clearly these rules need to be understood (at least).
$$\dfrac{\phi\to\chi\quad\phi}{\chi}{~\small{\to}\mathsf e}\qquad\dfrac{\gamma\wedge\beta}{\gamma}{~\small{\land}\mathsf e}\qquad\dfrac{\gamma\wedge\beta}{\beta}{~\small{\land}\mathsf e}\qquad\dfrac{[\psi]\\~\vdots\\\bot}{\lnot\psi~} {~\small\lnot\mathsf i}$$
So... to successfully apply negation introduction, an assumption must be made, under which a contradiction must be derived. For this proof that assumption must be $B$, and this and the premise then allows conditional elimination to be applied, thus deriving $A\land\lnot A$. Then from this... well...
$$\fitch{~~1.~~B\to(A\land\lnot A)\hspace{7ex}\textsf{Premise}}{\fitch{~~2.~~B\hspace{16ex}\textsf{Assumption}}{~~3.~~A\land\lnot A\hspace{10ex}{\to}\mathsf e~1,2\\~~4.~~A\hspace{16ex}{\land}\mathsf e~3\\~~5.~~\lnot A\hspace{14.5ex}{\land}\mathsf e~3\\~~~~\vdots\\~~\mathrm m.~~\bot}\\~~\mathrm n.~~\lnot B\hspace{17ex}\lnot\mathsf i~2{-}\mathrm m}$$
So all that is left is to derive that contradiction.
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