# Prove TRUE is derivable (or provable) from the Empty Set (Ex 3, Sec 1.4 in “A Concise Introduction to Mathematical Logic” by Rautenberg.)

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:

$\vdash\top$

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

• Hey! Could you cite the whole exercise? – Vinyl_cape_jawa Nov 20 '17 at 9:42

$\bot$ 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 ($\lnot_2$).