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I've being trying to understand more about logic. My reference book is "A Concise Introduction to Mathematical Logic" by Wolfgang Rautenberg. Now I'm having troubles with exercise 3 in section 1.4. In the solutions to the exercises is stated that the following easily follows from the $\lnot$ - rules:

$\vdash\top$

$\lnot \alpha \vdash \lnot (\alpha \land \beta)$

Well, not so easy for me. If you can please help. Thanks

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    $\begingroup$ Hey! Could you cite the whole exercise? $\endgroup$ – Vinyl_cape_jawa Nov 20 '17 at 9:42
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$\bot$ is defined as a contradiction, namely: $(p_1 ∧ ¬p_1)$ and $\top$ is defined as $\lnot \bot$ (see page 5).

The proof of the first one:

1) $\lnot \bot \vdash \lnot \bot$ --- by (IS)

2) $\bot \vdash p_1 \land \lnot p_1$ --- by abbreviation

3) $\bot \vdash p_1,\lnot p_1$ --- by ($\land_2$)

4) $\bot \vdash \lnot \bot$ --- by ($\lnot_1$)

5) $\vdash \lnot \bot$ --- from 1) and 4) by ($\lnot_2$), i.e. $\vdash \top$.


For the second one:

1) $\lnot \alpha, \lnot (\alpha \land \beta) \vdash \lnot (\alpha \land \beta)$ --- by (IS) and (MR)

2) $\lnot \alpha, \alpha \land \beta \vdash \lnot (\alpha \land \beta)$ --- by (IS), (MR), ($\land_2$) and ($\lnot_1$)

3) $\lnot \alpha \vdash \lnot (\alpha \land \beta)$ --- by ($\lnot_2$).

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