tag:blogger.com,1999:blog-176770731979379806.post938631135653771283..comments2019-02-15T16:41:42.253+01:00Comments on Palmström: The Lambda Calculus for Absolute Dummies (like myself)Joscha Bachhttps://plus.google.com/109564100423483405944noreply@blogger.comBlogger7125tag:blogger.com,1999:blog-176770731979379806.post-2112245918752057862013-02-12T05:12:11.659+01:002013-02-12T05:12:11.659+01:00The b is part of the inner expression (λ sz.z) bc,...The b is part of the inner expression (λ sz.z) bc, which is shorthand for (λ s.λ z.z) bc. We resolve the first lambda by replacing all occurrences of s in λ z.z with b. Since there are none, nothing is replaced with b, and we get (λ z.z) c, which resolves to c.Joscha Bachhttps://www.blogger.com/profile/08974461728804933391noreply@blogger.comtag:blogger.com,1999:blog-176770731979379806.post-10291792230158525242013-02-11T07:33:09.416+01:002013-02-11T07:33:09.416+01:00Could you clarify why it was okay to throw b away ...Could you clarify why it was okay to throw b away in the second step of calculating the successor of zero? Perhaps it's the use of shorthand, but I can't map that example back to the cut'n'paste rules you described.Sheldon Hearnhttps://www.blogger.com/profile/07377059190562437963noreply@blogger.comtag:blogger.com,1999:blog-176770731979379806.post-89080743388394051022013-01-12T11:34:36.827+01:002013-01-12T11:34:36.827+01:00Try to perform the resolution of the successor fun...Try to perform the resolution of the successor function applied to 0!Joscha Bachhttps://www.blogger.com/profile/08974461728804933391noreply@blogger.comtag:blogger.com,1999:blog-176770731979379806.post-72813535209151329172013-01-12T04:51:43.565+01:002013-01-12T04:51:43.565+01:00I'm still confused as why if 0 = (lambda sz.z)...I'm still confused as why if 0 = (lambda sz.z) then 1 would be (lambda sz.s(z)) ?<br /><br />tim phamhttps://www.blogger.com/profile/01431885064625482019noreply@blogger.comtag:blogger.com,1999:blog-176770731979379806.post-44812234383937066472012-11-12T15:30:42.884+01:002012-11-12T15:30:42.884+01:00Thank you! (Corrected now.)Thank you! (Corrected now.)Joscha Bachhttps://www.blogger.com/profile/08974461728804933391noreply@blogger.comtag:blogger.com,1999:blog-176770731979379806.post-78105550981023250352012-11-12T15:17:33.506+01:002012-11-12T15:17:33.506+01:00you state that boolean-OR is :
λ ab.a (λ xy.x) a
...you state that boolean-OR is :<br /><br />λ ab.a (λ xy.x) a<br /><br />it should actually be:<br /><br />λ ab.a (λ xy.x) bAnonymousnoreply@blogger.comtag:blogger.com,1999:blog-176770731979379806.post-81690165075369564842012-05-09T23:16:35.266+02:002012-05-09T23:16:35.266+02:00If this post was too elaborate for your (for insta...If this post was too elaborate for your (for instance, because you are a computer scientist), or you want to progress, here's a very condensed summary, and an extension into Lisp: http://forums.xkcd.com/viewtopic.php?f=40&t=46449#p1827434Joscha Bachhttps://www.blogger.com/profile/08974461728804933391noreply@blogger.com