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I'm trying to prove that $p\implies q \vdash \lnot p \lor b$ only using rules of natural deduction, but am having a lot of difficulty. I was able to prove the other direction. Could anyone show me or point me to the right direction?

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    $\begingroup$ Are you sure about $b$ ? $\endgroup$ – Mauro ALLEGRANZA Feb 8 at 12:56
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I assume that you are trying to prove : $p \to q \vdash \lnot p \lor q$.

1) $p \to q$ --- premise

2) $\lnot (\lnot p \lor q)$ --- assumed [a]

3) $\lnot p$ --- assumed [b]

4) $\lnot p \lor q$ --- from 3) by $\lor$-intro

5) $\bot$ --- from 2) and 4)

6) $p$ --- from 3) and 5) by $\lnot$-elim, discharging [b]

7) $q$ --- from 1) and6) by $\to$-elim

8) $\lnot p \lor q$ --- from 7) by $\lor$-intro

9) $\bot$ --- from 2) and 8)

10) $\lnot p \lor q$ --- from 2) and 9) by $\lnot$-elim, discharging [a].

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