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Can anyone help me with this basic derivation with natural rules of inference: $$ Q \rightarrow R \vdash (P \rightarrow Q) \rightarrow (P \rightarrow R) $$ I can use the follwing: $\wedge I, \wedge E, \vee I, \vee E, \rightarrow I, \rightarrow E, \leftrightarrow I, \leftrightarrow E, RAA $

I am struggling to understand where to go from assumption of line 2:

$$ 1. Q \rightarrow R \qquad\qquad\qquad given $$

$$ \qquad 2. P \rightarrow Q \qquad\qquad\quad assume $$ $$ \qquad 3. P \rightarrow R \qquad\qquad \rightarrow E $$

$$ Get: (P \rightarrow Q) \rightarrow (P \rightarrow R) $$

Am I going the right direction? I wanted to introduce another arrow elimination to get Q, R and P but I'm not too sure if that's correct.

My apologies for such a silly question - I'm only the beginner.

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  • $\begingroup$ Only someone with your textbook knows its rules of natural inference. So this is probably not the place to get a solution. $\endgroup$ – GEdgar Apr 1 '16 at 13:08
  • $\begingroup$ Thank you GEdgar for your comment, I added allowed rules of inference. Although I understand the rules, I'm finding difficult to actually apply them to given scenarios. If it's impossible to get help for this particular example, do you maybe have any suggestions how can I improve on this? $\endgroup$ – qwerty Apr 1 '16 at 13:18
  • $\begingroup$ What GEdgar said - adding those codes doesn't really help, since we don't know what rules they denote. But assuming it's a fairly typical natural deduction system: After assuming $Q\to R$ and $P\to Q$ you want to prove $P\to R$. In order to do that you're allowed to assume $P$, and now you just have to prove $R$, right? $\endgroup$ – David C. Ullrich Apr 1 '16 at 13:22
  • $\begingroup$ Sorry David C. Ullrich. And thank you - your suggestion really helped. $\endgroup$ – qwerty Apr 1 '16 at 13:36
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The derivation must be :

1) $Q→R$ --- premise

2) $P→Q$ --- assumed [a]

3) $P$ --- assumed [b]

4) $Q$ --- from 3) and 2) by $\to$-E

5) $R$ --- from 4) and 1) by $\to$-E

6) $P→R$ --- from 3) and 5) by $\to$-I, discharging [b]

7) $(P→Q) \to (P→R)$ --- from 2) and 6) by $\to$-I, discharging [a].

Now, from 1) and 7) we have :

$Q→R ⊢ (P→Q)→(P→R)$.

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