I've recently come across the claim that there are infinitely many derivations in Natural Deduction for any given tautology. Intuitively, this seems odd, because there is, say, a perfectly straightforward way of proving 'P or not-P' (assume P, use disjunction introduction to get 'P or not-P', etc). How would I go about making an argument to the effect that there are infinitely many other ways of proving, say, 'P or not-P' in Natural Deduction? I'm a bit stuck here!


Because nothing prevents you from additionally talking irrelevant nonsense during your proof.



The following :

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

2) $P \lor \lnot P$ --- from 1) by $\lor$-intro

is not a derivation of $P \lor \lnot P$, because the assumption [a] has not been discharged; it is a proof of :

$P \vdash P \lor \lnot P$.

Now for the real question.

Here is a proof :

1) $\lnot (P \lor \lnot P)$ --- ssumed [a]

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

3) $P \lor \lnot P$ --- from 2) by $\lor$-intro

4) $\bot$ --- from 1) and 3)

5) $\lnot P$ --- from 2) and 4), discharging [b]

6) $P \lor \lnot P$ --- from 5) by $\lor$-intro

7) $\bot$ --- from 1) and 6)

8) $P \lor \lnot P$ --- from 1) and 7), by Double Negation, discharging [a].

After step 6) we can insert a detour : use $\lor$-elimination to derive $P \lor \lnot P$ from 6) (i.e. from : $P \lor \lnot P$) and then go on with 7).

The new derivation is formally correct and thus is a new ("longer") derivation of $P \lor \lnot P$.

  • $\begingroup$ That's great, thanks! One question though: how does one generalise this result? My original question was why the statement that there are infinitely proofs for any tautology holds true, and P or not-P is only one instance of this. And of course not all proofs in ND need to use the disjunction rule, so what works in the case of 'P or not P' won't work in all cases (say for 'not (P and not-P). $\endgroup$ – user211309 Dec 3 '15 at 13:41

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