hott | Global hotkeys for Windows , with node
kandi X-RAY | hott Summary
kandi X-RAY | hott Summary
Global hotkeys for Windows, with node
Support
Quality
Security
License
Reuse
Top functions reviewed by kandi - BETA
Currently covering the most popular Java, JavaScript and Python libraries. See a Sample of hott
hott Key Features
hott Examples and Code Snippets
Community Discussions
Trending Discussions on hott
QUESTION
I am using gcc 10.1.1 on Fedora 32. The following program is to read from a file given by the user and print it into stdout. The code is an example in Modern C by Jens Gusdets.
...ANSWER
Answered 2020-Jul-21 at 12:01argv[0]
is always the program name. You're printing the contents of your executable in addition to any extra files identified by the filenames passed in the arguments by starting your for
loop at i = 0
.
If you start at i = 1
, it will exclude your executable.
QUESTION
Given the definition:
...ANSWER
Answered 2020-May-11 at 08:49I am not completely sure, but it seems to me that what you are trying to prove is no different from forall a (p:a=a), p = eq_refl
. If so, you cannot prove it in Coq, unless you know something about a
, e.g., decidable equality. In that case, you can use the results on UIP (unicity of identity proofs) from the standard library.
QUESTION
I have some data which looks like:
...ANSWER
Answered 2020-Apr-29 at 11:36nchar()
is the function you want:
QUESTION
I am learning HoTT with Agda form beginning. I followed instructions in https://www.cs.bham.ac.uk/~mhe/HoTT-UF-in-Agda-Lecture-Notes/
When I started to type in declaration of dependent sum type induction just like the notes instructed,
...ANSWER
Answered 2020-Mar-06 at 16:15The ability to pattern-match on records in telescopes will be available in the (soon) upcoming version Agda 2.6.1.
Cf. the documentation of the development branch: https://agda.readthedocs.io/en/latest/language/telescopes.html#irrefutable-patterns-in-binding-positions
QUESTION
In HOTT and also in COQ one cannot prove UIP, i.e.
\Prod_{p:a=a} p = refl a
But one can prove:
\Prod_{p:a=a} (a,p) = (a, refl a)
Why is this defined as it is? Is it, because one wants to have a nice homotopy interpretation? Or is there some natural, deeper reason for this definition?
...ANSWER
Answered 2019-May-24 at 19:21Today we know of a good reason for rejecting UIP: it is incompatible with the principle of univalence from homotopy type theory, which roughly says that isomorphic types can be identified. However, as far as I am aware, the reason that Coq's equality does not validate UIP is mostly a historical accident inherited from one of its ancestors: Martin-Löf's intensional type theory, which predates HoTT by many years.
The behavior of equality in ITT was originally motivated by the desire to keep type checking decidable. This is possible in ITT because it requires us to explicitly mark every rewriting step in a proof. (Formally, these rewriting steps correspond to the use of the equality eliminator eq_rect
in Coq.) By contrast, Martin-Löf designed another system called extensional type theory where rewriting is implicit: whenever two terms a
and b
are equal, in the sense that we can prove that a = b
, they can be used interchangeably. This relies on an equality reflection rule which says that propositionally equal elements are also definitionally equal. Unfortunately, there is a price to pay for this convenience: type checking becomes undecidable. Roughly speaking, the type-checking algorithm relies crucially on the explicit rewriting steps of ITT to guide its computation, whereas these hints are absent in ETT.
We can prove UIP easily in ETT because of the equality reflection rule; however, it was unknown for a long time whether UIP was provable in ITT. We had to wait until the 90's for the work of Hofmann and Streicher, which showed that UIP cannot be proved in ITT by constructing a model where UIP is not valid. (Check also these slides by Hofmann, which explain the issue from a historic perspective.)
EditThis doesn' t mean that UIP is incompatible with decidable type checking: it was shown later that it can be derived in other decidable variants of Martin-Löf type theory (such as Agda), and it can be safely added as an axiom in a system like Coq.
QUESTION
I asked the following question on the CS SE:
For example, in the proof of lemma 6.4.1 in the HoTT book, a function inductively defined over a function is simply applied on paths
loop
andrefl
, and then a path betweenloop
andrefl
is used (presumably by congruence viaf
) to construct a path betweenf loop
andf refl
:Suppose that
...loop = refl base
. [...] withx : A
andp : x = x
, there is a functionf : S1 → A
defined byf(base) :≡ x
andf(loop) := p
, we have
ANSWER
Answered 2018-Nov-26 at 09:45See Saizan's answer for a solution along the original lines. Alternatively, there is a simple solution:
QUESTION
According to Homotopy Type Theory (page 49), this is the full induction principle for equality :
...ANSWER
Answered 2018-Sep-02 at 14:55I think you are referring to the fact that the result type of path_induction
mentions the path that is being destructed, whereas the one of eq_rect
does not. This omission is the default for inductive propositions (as opposed to what happens with Type
), because the extra argument is not usually used in proof-irrelevant developments. Nevertheless, you can instruct Coq to generate more complete induction principles with the Scheme
command: https://coq.inria.fr/distrib/current/refman/user-extensions/proof-schemes.html?highlight=minimality. (The Minimality
variant is the one used for propositions by default.)
QUESTION
Is the univalence axiom invertible (modulo paths)? Is it possible to prove, using Agda's Cubical library, to prove the following:
...ANSWER
Answered 2018-Nov-14 at 15:16Sure, ua
is an equivalence, so it's injective. In the HoTT book, the inverse of ua
is idtoeqv
, so by congruence idtoeqv (ua f) ≡ idtoeqv (ua g)
and then by inverses f ≡ g
. I'm not familiar with the contents of cubical Agda prelude but this should be provable since it follows directly from the statement of univalence.
QUESTION
First, I've already looked up into several related materials, including the HoTT book & this question.
But I'm still confused, and I'm wishing for a explanation free from Agda, but directly from its mathematical formula. After removing the dot notation, J axiom then says like this, whose type signature is completely the same in the answer of the question I've mentioned above:
...ANSWER
Answered 2018-Oct-17 at 02:48When you pattern match on p
at refl
in the definition of J
, its type is refined from x ≡ y
to x ≡ x
(since the type of the constructor refl
is ∀ {x} -> x ≡ x
, i.e. it sets both indices to x
), which means we can refine both the left- and the right-hand sides by x ~ y
, which is both why the y
in the pattern becomes x
(or, alternatively, .x
in Agda to make it explicit that it is an inaccessable pattern), and also why c x : C x x refl
passes on the right-hand side for the result type C x y p
.
QUESTION
Say I have two inductively defined datatypes:
...ANSWER
Answered 2018-Sep-19 at 07:08In HoTT with univalence, it is indeed provable that list1 A
is equal to list2 A
for all A
. Given a proof p : list1 A = list2 A
, transport (or subst
) gives you P (list1 A) -> P (list2 A)
for any P
. In cubical type theories, such transporting may also compute as expected. To my knowledge, cubical type theory (CCHM or cartesian) is the only setting where this currently works. cubicaltt
is the most usable (but still not really practical) implementation.
Community Discussions, Code Snippets contain sources that include Stack Exchange Network
Vulnerabilities
No vulnerabilities reported
Install hott
Support
Reuse Trending Solutions
Find, review, and download reusable Libraries, Code Snippets, Cloud APIs from over 650 million Knowledge Items
Find more librariesStay Updated
Subscribe to our newsletter for trending solutions and developer bootcamps
Share this Page