Natural deduction proof $(p→¬p ) ⊢ (p→r)$

Question

I'm just learning natural deduction and I'm struggling how to prove $(p→¬p ) ⊢ (p→r)$ properly. Especially I'm wondering what to get $r$ to the implication.

Premise: $(p→¬p)$

Assume: $p$

Eliminate $→$ from 1 to 2: $¬p$

this is what I have got done so far but where do I get the r so that I get the implication $(p→r)$ at the end?

Here are the rules I'm using for $→$ and $¬$:

Answer 1

You have derived a contradiction: $p, \lnot p$.

Apply $(\lnot \text E)$: $\dfrac { \varphi \ \ \ \lnot \varphi }{\bot}$, to get $\bot$, followed by $(\bot \text E)$: $\dfrac {\bot}{\varphi}$, to get

$r$.

If the symbol $\bot$ is missing, you may use the Ex falso rule:

$\dfrac { \varphi \ \ \ \lnot \varphi }{\psi}.$

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With the rules you have listed, the proof is quite convoluted:

1. $(p \to \lnot p)$ --- premise

2. $p$ --- assumed [a]

3. $\lnot r$ --- assumed [b]

4. $\lnot p$ --- from 1) and 2) by $\to$-elim

5. $p \land \lnot p$ --- from 2) and 4) by $\land$-intro

6. $\lnot \lnot r$ --- from 3) and 5) by $\lnot$-intro, discharging [b]

7. $r$ --- from 6) by $\lnot \lnot$-elim

8. $(p \to r)$ --- from 2) and 7), discharging [a].

Answer 2

Given $\neg p$:

Assume: $p$

$\lor$ Intro: $p \lor r$

$\lor $ Elim (from $\neg p$): $r$

$\to$ Intro: $p\to r$.

This is basically the proof for Principle of Explosion.

Answer 3

$\def\fitch#1#2{\begin{array}{|l}#1 \\ \hline #2\end{array}}$

$\fitch{ 1. P \rightarrow \neg P \qquad \quad Assumption}{ \fitch{ 2. P \qquad \quad \qquad Assumption}{ \fitch{ 3. \neg R \qquad \quad \quad Assumption}{ 4. \neg P \qquad \to \ Elim \ 1,2\\ 5. P \land \neg P \qquad \land \ Intro \ 2,4 } \\ 6. \neg \neg R \qquad \quad \neg \ Intro \ 3-5\\ 7. R \qquad \qquad \neg \ Elim 6}\\ 8. P \to R \qquad \to \ Intro \ 2-7 }$