AuthorWrite something about yourself. No need to be fancy, just an overview. ArchivesCategories |
Back to Blog
Falso quod libet1/26/2024 which should seem quite plausible.Īs Graham Kemp points out in his answer, the convention of using 'efq' to discharge an assumption is a little bit non-standard and usually this is called RAA or framed as double negation elimination. (And this was certainly not an unconditional proof of an absurdity, it was under specific assumptions that were framed up in the larger structure of the proof.) Perhaps a better way to think about it is that what we wanted to prove was $P\to Q$ under the assumption that $P$ is false. So we did assume $\lnot P$ and assume $P$ and get nonsense, but the assumptions had an orderly motivation and nonsense was exactly what we needed to prove $Q,$ which was a goal we had at that moment in the proof. There is a contradiction between our assumption of $\lnot P$ and our assumption of $P.$. ![]() We will prove $Q$ by proving a contradiction, from which we can prove anything (ex falso.).We will prove $P\to Q$ by assuming $P$ and then proving $Q.$.For information on how to build, install or test Quod Libet. We will derive a contradiction by proving $P\to Q,$ which contradicts with our assumption of $\lnot(P\to Q).$ Ex Falso is a bare-bones tag editor with the same editing interface as Quod Libet.We will prove $P$ by assuming $\lnot P$ and deriving a contradiction.We want to prove $\lnot(P\to Q)\to P,$ so we assume $\lnot(P\to Q)$ and our goal is to prove $P$.Here's how we can organize things (informally) to make a bit more sense. Here we see that a contradiction can be derived on raising the third assumption, and so we may validly derive $Q$ within that context (because whatever $Q$ may be, it is at least as true as an absurdity).Įx falso Quodlibet: If we can say "false is true", then we may say anything is true. In this notation we keep track of the order in which assumptions are raised and discharged by indentation. To show $P$, we will use proof by contradiction and we will assume $\neg P$. ![]()
0 Comments
Read More
Leave a Reply. |