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I'm trying to prove the following:

$\vdash (p\to q)\lor(q\to r)$

using only intuitionistically valid rules. I've tried a few different ways, and I think my problem is that I'm not sure what assumptions to make. Can anyone help? Thanks!

Léreau
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mj_indefinite
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  • What's the relationship between $q$ and $r$? – Luigi D. Mar 13 '15 at 11:50
  • I'm not sure what you mean... I think there isn't one? What's in the sequent is all I have to go on. – mj_indefinite Mar 13 '15 at 11:54
  • You want to assert that $p$ implies either $q$ or $r$ (or both, from what we know). But if we don't know any other pre-existing concept connecting the three, then yours is just a statement and there is nothing to prove. – Luigi D. Mar 13 '15 at 12:03
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    What if $p$ is true but $q,r$ both false? Then the thing to the right of the $\vdash$ would be false. – coffeemath Mar 13 '15 at 12:03
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    You cannot prove it because it is not a tautology. But, in general, you are "mixing" to different concepts : with Natural Deduction a derivation necessarily have to start with some assumptions; if the formula you are trying to prove is a tautology, you can complete the derivation having discharged all the assumptions. – Mauro ALLEGRANZA Mar 13 '15 at 12:10
  • Say it was a tautology... what could I do then? – mj_indefinite Mar 13 '15 at 12:19
  • Here you van find an example of a derivation concluded with no open assumptions left. – Mauro ALLEGRANZA Mar 13 '15 at 12:36
  • sorry, there was a typo, I've corrected what the sequent should've been. Any ideas now that it is right? – mj_indefinite Mar 14 '15 at 04:12
  • I think I have it, just by assuming a contradiction and $q$, thanks for the help! – mj_indefinite Mar 14 '15 at 05:25
  • Since it's just propositional, why not just check it in a model, like a Kripke model or a Heyting algebra? Or am I missing something? – ShyPerson May 18 '21 at 03:54

3 Answers3

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This statement cannot be shown by intuitionistically valid rules, as it is equivalent to the law of excluded middle, which does not hold intuitionistically.

To be more precise we have:

$~\vdash X \lor \neg X$ for every $X$ is equivalent to having $~\vdash (A \rightarrow B) \lor (B \rightarrow C)$ for every $A, B, C.$

Proof:

  • Assume we need to show $~\vdash (A \rightarrow B) \lor (B \rightarrow C)$ then by $\mathsf{LEM}$ we know $\vdash B \lor \neg B$. In the case where $B$ holds, we can certainly show $A \rightarrow B$ and in the other case we can show $\neg B \vdash B \rightarrow C$ since $\neg B, B \vdash \bot$.
  • Assume we have $~\vdash (A \rightarrow B) \lor (B \rightarrow C)$ for every $A,B,C$ and try to show $~\vdash X \lor \neg X$ for very $X$. For this we simply choose $A := (X \rightarrow X)$, $B := X$, $C := \bot$.

For a list of even more statements that are equivalent to $\mathsf{LEM}$ in intuitionistic logic, have a look at this paper: Classifying Material Implications over Minimal Logic.

Léreau
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You can prove it in classical logic.

1) $q$ --- assumed [a]

2) $p \to q$ --- $\to$-intro

3) $(p \to q) \lor (q \to r)$ --- $\lor$-intro

4) $\lnot q$ --- assumed [b]

5) $\bot$ --- from 1) and 4)

6) $r$ --- from 5)

7) $q \to r$ --- $\to$-intro

8) $(p \to q) \lor (q \to r)$ --- $\lor$-intro

9) $\vdash q \lor \lnot q$ --- LEM

10) $(p \to q) \lor (q \to r)$ --- from 1)-3) and 4)-8) with 9) by $\lor$-elimination, discharging assumptions [a] and [b].

I suppose that we cannot prove it intuitionistically ...

Mauro ALLEGRANZA
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@Lereau gives the technical answer why your wff is not intuitionistically provable.

But faced with a question like this, it is always worth stepping back and asking yourself whether a given wff ought to be intuitionistically provable. Does it seem the sort of thing which is constructively provable?

Well obviously not so in this case, for a disjunction is only constructively provable if one of the disjuncts is constructively provable. And evidently in this case $(p \to q)$ is not constructively provable (with no background assumptions to play with). And similarly for the other disjunct. So even in advance of the technical answer you already should know that the wff is not constructively provable. (Though it is classically provable as @Mauro notes, or as immediately follows from @Lereau's answer.)

Peter Smith
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