2025/10/24

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2025-10-24 16:50:17 +0200Lord_of_Life(~Lord@user/lord-of-life/x-2819915) Lord_of_Life
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2025-10-24 16:23:48 +0200srazkvt(~sarah@user/srazkvt) (Ping timeout: 260 seconds)
2025-10-24 16:22:30 +0200 <__monty__> Hmm, you're right.
2025-10-24 16:18:57 +0200 <tomsmeding> there's probably more recent material
2025-10-24 16:18:49 +0200 <tomsmeding> ("Theorems for free!" by Wadler)
2025-10-24 16:18:27 +0200 <tomsmeding> in any case, they should look up free theorems, but they left already
2025-10-24 16:17:52 +0200 <tomsmeding> __monty__: would it not? If m is bottom, wouldn't all three expressions evaluate to bottom?
2025-10-24 16:17:35 +0200Square2(~Square@user/square) Square
2025-10-24 16:08:36 +0200paul424(~lektor@2a01:111f:1301:c700:146c:2766:6a2d:c6cd) (Quit: Leaving)
2025-10-24 16:05:22 +0200 <__monty__> paul424: Without further constraints on the problem that's not provable. `m` could be bottom in Haskell, fulfilling the signature but not commuting with map.
2025-10-24 16:01:40 +0200AntiRembaneRembane
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2025-10-24 15:55:45 +0200 <paul424> I have function of type : m :: forall a b. (a -> b) -> [a] -> [b] . How to prove from that : m f xs = map f (m id xs) = m id (map f xs)
2025-10-24 15:54:29 +0200paul424(~lektor@2a01:111f:1301:c700:146c:2766:6a2d:c6cd)
2025-10-24 15:50:45 +0200Googulator81(~Googulato@2a01-036d-0106-03fa-d161-d36f-e0e5-1b0a.pool6.digikabel.hu)
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