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!
Why are there infinitely many proofs for any tautology in Natural Deduction?
106 Views Asked by Bumbble Comm https://math.techqa.club/user/bumbble-comm/detail AtThere are 2 best solutions below
On
Comment
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 $P \lor \lnot P$.
Because nothing prevents you from additionally talking irrelevant nonsense during your proof.