2025/01/08

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2025-01-08 11:01:01 +0100merijn(~merijn@77.242.116.146) merijn
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2025-01-08 10:55:43 +0100forell(~forell@user/forell) forell
2025-01-08 10:55:22 +0100CiaoSen(~Jura@2a05:5800:213:8800:ca4b:d6ff:fec1:99da) CiaoSen
2025-01-08 10:51:14 +0100merijn(~merijn@77.242.116.146) merijn
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2025-01-08 10:43:31 +0100 <ncf> monochrom: what relation do you use for b? i'd think a function Bool → b is enough
2025-01-08 10:41:21 +0100jespada(~jespada@2800:a4:f9:a300:c8c1:20a5:1a94:f1) jespada
2025-01-08 10:37:49 +0100merijn(~merijn@77.242.116.146) (Ping timeout: 244 seconds)
2025-01-08 10:37:16 +0100m_mari-estel
2025-01-08 10:36:48 +0100mari-estelm_
2025-01-08 10:36:44 +0100tzh(~tzh@c-76-115-131-146.hsd1.or.comcast.net) (Quit: zzz)
2025-01-08 10:34:56 +0100ash3en(~Thunderbi@2a03:7846:b6eb:101:93ac:a90a:da67:f207) ash3en
2025-01-08 10:33:28 +0100eL_Bart0(eL_Bart0@dietunichtguten.org)
2025-01-08 10:31:56 +0100lxsameer(~lxsameer@Serene/lxsameer) lxsameer
2025-01-08 10:28:00 +0100 <lambdabot> (forall x. g . h x = k (f x) . g) => g . fold h y = fold k (g y) . $map f
2025-01-08 10:28:00 +0100 <monochrom> @free fold :: (x -> y -> y) -> y -> [x] -> y
2025-01-08 10:22:49 +0100Smiles(uid551636@id-551636.lymington.irccloud.com) Smiles
2025-01-08 10:18:14 +0100l_k(~student@85.172.76.134) (Read error: Connection reset by peer)
2025-01-08 10:18:11 +0100l__k(~student@85.172.111.143)
2025-01-08 10:10:34 +0100alist(~alist@108-224-153-186.lightspeed.cicril.sbcglobal.net)
2025-01-08 10:05:14 +0100 <monochrom> But the paper doesn't contain such advanced examples. All examples are basically natural transformations like "oh rev is a natural transformation". Those don't need arbitrary relations, just functions.
2025-01-08 10:03:59 +0100 <monochrom> But if you replace Maybe by the System F encoding, i.e., prove that e :: forall a. forall b. b -> ((a -> a) -> b) -> b has only two possibilities, that will require you to use a non-function relation for b.
2025-01-08 10:03:02 +0100lortabac(~lortabac@2a01:e0a:541:b8f0:55ab:e185:7f81:54a4)
2025-01-08 10:02:00 +0100 <monochrom> Err, that one still works with a function relation.
2025-01-08 10:01:00 +0100 <monochrom> Sometimes you really want to use a non-function relation, e.g., when proving that e :: forall a. Maybe (a->a) can only be Nothing or Just id.
2025-01-08 10:00:26 +0100alist(~alist@108-224-153-186.lightspeed.cicril.sbcglobal.net) (Remote host closed the connection)
2025-01-08 09:56:30 +0100machinedgod(~machinedg@d108-173-18-100.abhsia.telus.net) machinedgod
2025-01-08 09:54:57 +0100 <alist> ncf: thanks for the help!!
2025-01-08 09:54:39 +0100Guest5708(~guest5708@117.222.137.69) (Leaving)
2025-01-08 09:53:47 +0100 <ncf> a white horse is a horse
2025-01-08 09:53:26 +0100 <alist> so either way i understand what hes writing better
2025-01-08 09:53:10 +0100 <mari-estel> well then yes, functions are relations, but relations are not functions
2025-01-08 09:53:05 +0100 <alist> well really in the authors definition of relation as just a subset of the cross product they are
2025-01-08 09:52:32 +0100 <mauke> unidirectional relations are valid. we don't discriminate
2025-01-08 09:52:02 +0100 <mari-estel> functions are unidirectional
2025-01-08 09:51:49 +0100 <alist> well they correspond directly
2025-01-08 09:51:37 +0100 <mari-estel> hm sortof
2025-01-08 09:51:16 +0100 <alist> because functions are relations hehe
2025-01-08 09:51:09 +0100 <alist> OH i see
2025-01-08 09:50:29 +0100 <ncf> i.e. functional relations
2025-01-08 09:50:27 +0100 <alist> oh hm
2025-01-08 09:50:14 +0100 <ncf> instead of relations 𝒜 : A ⇔ A', ℬ : B ⇔ B' we use functions a : A → A', b : B → B'
2025-01-08 09:49:52 +0100 <alist> rigfold_AB is the specialization of fold to the types