Study Theorem 3.2.2 #14
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Delete branch "%!s()"
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This is a pretty long proof and involves many moving parts. I don't understand how all of it works
I have most of the proof written down, but the biggest thing that confuses me is:
This means the result is still contingent on already having a value
u : \neg \neg \mathbf{2}
, but the type of the overall theorem is:where the
\neg \neg A
part is still negative.Actually, creating a value of
u : \neg \neg \mathbf{2}
is easy:Still have these parts to prove though: