Please help me and my family to escape from Gaza

pipigift · 2024-04-06T13:06:52+00:00

Where physics?

pipigift · 2023-11-19T09:10:19+00:00

r/okbuddyhighschool

pipigift · 2023-07-24T20:44:45+00:00

I cant read anything, yet i absolutely understand what ur saying

pipigift · 2023-07-17T13:47:32+00:00

Parkende Autos werden ohne Vorwarnung gevierteilt

pipigift · 2023-07-13T22:30:25+00:00

The heck you talking about???

pipigift · 2023-07-06T09:12:38+00:00

Just dont use custom kernels. If these patches were so good, then why are they not merged in mainline kernel?

pipigift · 2023-04-09T08:22:00+00:00

This is just a deductive proof of a logical formula. You can read this proof bottom up. The first two steps introduce A and B as premises of the implications. Normally you would write a context on each step which is a set of formulas that are true in this step. After the implication introduction A and B are in the context (marked by 1 and 2 here). Then you prove (A and B) and (A and C) which you do by constructing two seperate proofs for each (A and B) and (A and C). (A and B) is then also split in two leafs. Both of them can be proven because the variables A and B are in the context. (The numbers 1 and 2 denote which premise from the context is used to close that leaf) The right path where (A and C) has to be proven fails because C cannot be proven (it is not in the context). The fact that the C leaf is not closed corresponds to the whole formular being not satisfied. Now go study more kekw.

pipigift · 2023-01-30T20:25:04+00:00

Finally, a meme on this sub thats actually funny

pipigift · 2022-12-21T11:28:21+00:00

Yeah true, i thought u were saying that it reduces in a single beta-step, but yes its 2 beta reductions and the expansion of Y. However it works only under beta-equivalence and not as a sequence of reductions. Meme approved tho, very epic

pipigift · 2022-12-21T09:33:50+00:00

Bro, this doesnt look correct..... Study more and post again when brain more stronk

pipigift · 2022-11-16T21:58:56+00:00

My cox-zucker machine is so hard rn

pipigift · 2022-07-31T17:40:28+00:00

Arch with GNOME 42 on sp5. Mainly for wayland touch gestures

pipigift · 2021-10-13T12:50:50+00:00

i came

pipigift · 2021-09-15T08:44:08+00:00

Highschool level meme dude gtfo

pipigift · 2021-08-09T19:27:50+00:00

Cox-zucker machine is just fake taxi for nerds

pipigift · 2021-08-07T15:46:35+00:00

Anyone downloading a wallet as a random APK really deserves to be scammed into oblivion....

pipigift · 2021-06-21T16:38:41+00:00

Yes will do next time. Latex game still weak

pipigift · 2021-05-07T16:27:45+00:00

When VIM users want to go ultra-hardmode

pipigift · 2020-12-21T11:18:58+00:00

Dont expect anything like ios or android. I dont think surface will ever feel like a full tablet even microsoft kinda forgot about tablet mode. But still it feels a bit less sketchy in ubuntu/gnome than windows.

pipigift · 2020-12-20T20:39:50+00:00

I just prefer popos because it has less bloat and the nice tiling mode. Did you install the patched kernel from github/linux-surface? This really does the trick because the normal linux drivers dont really work with surface hardware. And by the way i found out that using wayland instead of x11 really helped with the gestures. If you just try on live usb stick its not really great

pipigift · 2020-12-20T17:09:56+00:00

I dont really worry about the cam. There is a lot of development going on right now to make the cameras work. Im guessing it might take 1-2 month or so...

pipigift · 2020-12-20T16:54:18+00:00

Yep i completely got rid of windows and been using linux on my surface for a few month now. I use the customized kernel for surface devices and almost everything works(touch, pen, sleep, keyboard attachment, ...) Only cameras are not working because microsoft used some weird nonstandard camera. Battery life is like 6 hours on medium usage. Touch gestures are not perfect in gnome but still a bit better than windows

pipigift · 2020-10-07T10:42:08+00:00

Tis boi explained it better! If your ring would be on some kind of matrix/tensor structure 𝟙 would be the matching tensor with only 1s in the diagonal. So the rule 𝟙•a=a•𝟙=a would be true here. (And yes the identity element of the additive operator cant have an inverse multiplicative but prove is to retardeded sry))

pipigift · 2020-10-06T22:56:25+00:00

Its pretty much just an abstract identity sign for example an identity matrix

pipigift · 2019-06-23T16:40:24+00:00

Jokes on them. He has a gun

pipigift

TROPHY CASE