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Use macros extensively

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Frederik Hanghøj Iversen 2018-05-09 18:24:07 +02:00
parent 179570edf0
commit 9842c66eea
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6
doc/cubical.tex
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@ -118,7 +118,7 @@ B\ a$:
%
\begin{equation}
\label{eq:funExt}
\var{funExt} \tp \prod_{a \tp A} f\ a \equiv g\ a \to f \equiv g
\funExt \tp \prod_{a \tp A} f\ a \equiv g\ a \to f \equiv g
\end{equation}
%
%% p = λ i a → p a i
@ -133,7 +133,7 @@ by the term:
%
\begin{equation}
\label{eq:funExt}
\var{funExt}\ p \defeq λ i\ a → p\ a\ i
\funExt\ p \defeq λ i\ a → p\ a\ i
\end{equation}
%
With this we can now prove the desired equality $f \equiv g$ from section
@ -141,7 +141,7 @@ With this we can now prove the desired equality $f \equiv g$ from section
%
\begin{align*}
p & \tp f \equiv g \\
p & \defeq \var{funExt}\ \lambda n \to \refl
p & \defeq \funExt\ \lambda n \to \refl
\end{align*}
%
Paths have some other important properties, but they are not the focus of

106
doc/implementation.tex
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@ -88,7 +88,7 @@ definitions: If we let $p$ be a witness to the identity law, which formally is:
\id \comp f \equiv f \x f \comp \id \equiv f
\end{equation}
%
Then we can construct the identity isomorphism $\var{idIso} \tp \identity,
Then we can construct the identity isomorphism $\idIso \tp \identity,
\identity, p \tp A \approxeq A$ for any object $A$. Here $\approxeq$ denotes
isomorphism on objects (whereas $\cong$ denotes isomorphism of types). This will
be elaborated further on in sections \S\ref{sec:equiv} and
@ -96,7 +96,7 @@ be elaborated further on in sections \S\ref{sec:equiv} and
an isomorphism from \emph{any} path:
%
\begin{equation}
\var{idToIso} \tp A ≡ B → A ≊ B
\idToIso \tp A ≡ B → A ≊ B
\end{equation}
%
The univalence criterion for categories states that this map must be an
@ -168,9 +168,9 @@ is by way of the `combinators' $\propPi$ and $\propSig$ presented in sections
\S\ref{sec:propPi} and \S\ref{sec:propSig}:
%
\begin{align*}
\var{propPi} & \tp \left(\prod_{a \tp A} \isProp\ (P\ a)\right) \to \isProp\ \left(\prod_{a \tp A} P\ a\right)
\propPi & \tp \left(\prod_{a \tp A} \isProp\ (P\ a)\right) \to \isProp\ \left(\prod_{a \tp A} P\ a\right)
\\
\var{propSig} & \tp \isProp\ A \to \left(\prod_{a \tp A} \isProp\ (P\ a)\right) \to \isProp\ \left(\sum_{a \tp A} P\ a\right)
\propSig & \tp \isProp\ A \to \left(\prod_{a \tp A} \isProp\ (P\ a)\right) \to \isProp\ \left(\sum_{a \tp A} P\ a\right)
\end{align*}
%
So the proof goes like this: We `eliminate' the 3 function abstractions by
@ -202,7 +202,7 @@ Where The definition of $\IsPreCategory$ is the triple:
%
\begin{align*}
\var{isAssociative} & \tp \var{IsAssociative}\\
\var{isIdentity} & \tp \var{IsIdentity}\\
\isIdentity & \tp \var{IsIdentity}\\
\var{arrowsAreSets} & \tp \var{ArrowsAreSets}
\end{align*}
%
@ -245,8 +245,8 @@ $i$ becomes the triple:
\begin{aligned}
& \var{propIsAssociative} && a.\var{isAssociative}\
&& b.\var{isAssociative} && i \\
& \var{propIsIdentity} && a.\var{isIdentity}\
&& b.\var{isIdentity} && i \\
& \propIsIdentity && a.\isIdentity\
&& b.\isIdentity && i \\
& \var{propArrowsAreSets} && a.\var{arrowsAreSets}\
&& b.\var{arrowsAreSets} && i
\end{aligned}
@ -259,7 +259,7 @@ type would already not be possible, since we at least need extensionality (the
projections are all $\prod$-types). Assuming we had functional extensionality
available to us as an axiom, we would use functional extensionality (in
reverse?) to retrieve the equalities in $a$ and $b$, pattern-match on them to
see that they are both $\var{refl}$ and then close the proof with $\var{refl}$.
see that they are both $\refl$ and then close the proof with $\refl$.
Of course this theorem is not so interesting in the setting of ITT since we know
a priori that equality proofs are unique.
@ -287,14 +287,14 @@ latter. This is because $\isProp$ talks about non-dependent paths. So we need to
'promote' the result that univalence is a proposition to a heterogeneous path.
To this end we can use $\lemPropF$, which was introduced in \S\ref{sec:lemPropF}.
In this case $A = \var{IsIdentity}\ \identity$ and $B = \var{Univalent}$. We've
In this case $A = \var{IsIdentity}\ \identity$ and $B = \Univalent$. We've
shown that being a category is a proposition, a result that holds for any choice
of identity proof. Finally we must provide a proof that the identity proofs at
$a$ and $b$ are indeed the same, this we can extract from $p$ by applying
congruence of paths:
%
$$
\congruence\ \var{isIdentity}\ p
\congruence\ \isIdentity\ p
$$
%
And this finishes the proof that being-a-category is a mere proposition
@ -447,8 +447,8 @@ an isomorphism $A \approxeq B$ in some category $\bC$ be given. Name the
isomorphism $\iota \tp A \to B$ and its inverse $\inv{\iota} \tp B \to A$.
Since $\bC$ is a category (and therefore univalent) the isomorphism induces a
path $p \tp A \equiv B$. From this equality we can get two further paths:
$p_{\var{dom}} \tp \var{Arrow}\ A\ X \equiv \var{Arrow}\ B\ X$ and
$p_{\var{cod}} \tp \var{Arrow}\ X\ A \equiv \var{Arrow}\ X\ B$. We
$p_{\var{dom}} \tp \Arrow\ A\ X \equiv \Arrow\ B\ X$ and
$p_{\var{cod}} \tp \Arrow\ X\ A \equiv \Arrow\ X\ B$. We
then have the following two theorems:
%
\begin{align}
@ -465,26 +465,26 @@ I will give the proof of the first theorem here, the second one is analogous.
%
\begin{align*}
\var{coe}\ p_{\var{dom}}\ f
& \equiv f \lll \inv{(\var{idToIso}\ p)} && \text{lemma} \\
& \equiv f \lll \inv{(\idToIso\ p)} && \text{lemma} \\
& \equiv f \lll \inv{\iota}
&& \text{$\var{idToIso}$ and $\var{isoToId}$ are inverses}\\
&& \text{$\idToIso$ and $\isoToId$ are inverses}\\
\end{align*}
%
In the second step we use the fact that $p$ is constructed from the isomorphism
$\iota$ -- $\inv{(\var{idToIso}\ p)}$ denotes the map $B \to A$ induced by the
isomorphism $\var{idToIso}\ p \tp A \cong B$. The helper-lemma is similar to
$\iota$ -- $\inv{(\idToIso\ p)}$ denotes the map $B \to A$ induced by the
isomorphism $\idToIso\ p \tp A \cong B$. The helper-lemma is similar to
what we're trying to prove but talks about paths rather than isomorphisms:
%
\begin{equation}
\label{eq:coeDomIso}
\prod_{f \tp \var{Arrow}\ A\ B} \prod_{p \tp A \equiv B}
\var{coe}\ p_{\var{dom}}\ f \equiv f \lll \inv{(\var{idToIso}\ p)}
\prod_{f \tp \Arrow\ A\ B} \prod_{p \tp A \equiv B}
\var{coe}\ p_{\var{dom}}\ f \equiv f \lll \inv{(\idToIso\ p)}
\end{equation}
%
Again $p_{\var{dom}}$ denotes the path $\var{Arrow}\ A\ X \equiv
\var{Arrow}\ B\ X$ induced by $p$. To prove this statement I let $f$ and $p$ be
Again $p_{\var{dom}}$ denotes the path $\Arrow\ A\ X \equiv
\Arrow\ B\ X$ induced by $p$. To prove this statement I let $f$ and $p$ be
given and then invoke based-path-induction. The induction will be based at $A
\tp \var{Object}$ Let $\widetilde{B} \tp \Object$ and $\widetilde{p} \tp A
\tp \Object$ Let $\widetilde{B} \tp \Object$ and $\widetilde{p} \tp A
\equiv \widetilde{B}$ be given. The family that we perform induction over will
be:
%
@ -494,20 +494,20 @@ D\ \widetilde{B}\ \widetilde{p} \defeq
%% \prod_{\widetilde{p} \tp A \equiv \widetilde{B}}
\var{coe}\ {\widetilde{p}}^*\ f
\equiv
f \lll \inv{(\var{idToIso}\ \widetilde{p})}
f \lll \inv{(\idToIso\ \widetilde{p})}
\end{align}
The base-case therefore becomes
$d \tp \var{coe}\ \refl^*\ f \equiv f \lll \inv{(\var{idToIso}\ \refl)}$
$d \tp \var{coe}\ \refl^*\ f \equiv f \lll \inv{(\idToIso\ \refl)}$
and is inhabited by:
\begin{align*}
\var{coe}\ \refl^*\ f
& \equiv f
&& \text{$\refl$ is a neutral element for $\var{coe}$}\\
& \equiv f \lll \var{identity} \\
& \equiv f \lll \var{subst}\ \var{refl}\ \var{identity}
& \equiv f \lll \identity \\
& \equiv f \lll \var{subst}\ \refl\ \identity
&& \text{$\refl$ is a neutral element for $\var{subst}$}\\
& \equiv f \lll \inv{(\var{idToIso}\ \refl)}
&& \text{By definition of $\var{idToIso}$}\\
& \equiv f \lll \inv{(\idToIso\ \refl)}
&& \text{By definition of $\idToIso$}\\
\end{align*}
%
To close the based-path-induction we must supply the value ``at the other''. In
@ -516,7 +516,7 @@ In summary the proof of \ref{eq:coeDomIso} is the term:
%
\begin{equation}
\label{eq:pathJ-example}
\var{pathJ}\ D\ d\ B\ p
\pathJ\ D\ d\ B\ p
\end{equation}
%
And this finishes the proof of \ref{eq:coeDomIso} and thus \ref{eq:coeDom}.
@ -552,7 +552,7 @@ Since $\rrr$ is reverse function composition this is just the symmetric version
of associativity.
%
$$
\var{identity} \rrr f \equiv f \x f \rrr identity \equiv f
\identity \rrr f \equiv f \x f \rrr identity \equiv f
$$
%
This is just the swapped version of identity.
@ -570,30 +570,30 @@ B)$ is an isomorphism. Let us denote it's inverse with $\isoToId \tp (A
\approxeq B) \to (A \equiv B)$. If we squint we can see what we need is a way to
go between $\wideoverbar{\approxeq}$ and $\approxeq$.
An inhabitant of $A \approxeq B$ is simply an arrow $f \tp \var{Arrow}\ A\ B$
and it's inverse $g \tp \var{Arrow}\ B\ A$. In the opposite category $g$ will
An inhabitant of $A \approxeq B$ is simply an arrow $f \tp \Arrow\ A\ B$
and it's inverse $g \tp \Arrow\ B\ A$. In the opposite category $g$ will
play the role of the isomorphism and $f$ will be the inverse. Similarly we can
go in the opposite direction. I name these maps $\var{shuffle} \tp (A \approxeq
B) \to (A \wideoverbar{\approxeq} B)$ and $\var{shuffle}^{-1} \tp (A
go in the opposite direction. I name these maps $\shuffle \tp (A \approxeq
B) \to (A \wideoverbar{\approxeq} B)$ and $\shuffle^{-1} \tp (A
\wideoverbar{\approxeq} B) \to (A \approxeq B)$ respectively.
As the inverse of $\wideoverbar{\idToIso}$ I will pick $\wideoverbar{\isoToId}
\defeq \isoToId \comp \var{shuffle}$. The proof that they are inverses go as
\defeq \isoToId \comp \shuffle$. The proof that they are inverses go as
follows:
%
\begin{align*}
\wideoverbar{\isoToId} \comp \wideoverbar{\idToIso} & =
\isoToId \comp \var{shuffle} \comp \wideoverbar{\idToIso}
\isoToId \comp \shuffle \comp \wideoverbar{\idToIso}
\\
%% ≡⟨ cong (λ φ → φ x) (cong (λ φ → η ⊙ shuffle ⊙ φ) (funExt lem)) ⟩ \\
%
& \equiv
\isoToId \comp \var{shuffle} \comp \inv{\var{shuffle}} \comp \idToIso
\isoToId \comp \shuffle \comp \inv{\shuffle} \comp \idToIso
&& \text{lemma} \\
%% ≡⟨⟩ \\
& \equiv
\isoToId \comp \idToIso
&& \text{$\var{shuffle}$ is an isomorphism} \\
&& \text{$\shuffle$ is an isomorphism} \\
& \equiv
\identity
&& \text{$\isoToId$ is an isomorphism}
@ -602,7 +602,7 @@ follows:
The other direction is analogous.
The lemma used in step 2 of this proof states that $\wideoverbar{idToIso} \equiv
\inv{\var{shuffle}} \comp \idToIso$. This is a rather straight-forward proof
\inv{\shuffle} \comp \idToIso$. This is a rather straight-forward proof
since being-an-inverse-of is a proposition, so it suffices to show that their
first components are equal, but this holds judgmentally.
@ -610,7 +610,7 @@ This finished the proof that the opposite category is in fact a category. Now,
to prove that that opposite-of is an involution we must show:
%
$$
\prod_{\bC \tp \var{Category}} \left(\bC^{\var{Op}}\right)^{\var{Op}} \equiv \bC
\prod_{\bC \tp \Category} \left(\bC^{\var{Op}}\right)^{\var{Op}} \equiv \bC
$$
%
As we've seen the laws in $\left(\bC^{\var{Op}}\right)^{\var{Op}}$ get quite
@ -685,11 +685,11 @@ lemma:
Which says that if two type-families are equivalent at all points, then pairs
with identical first components and these families as second components will
also be equivalent. For our purposes $P \defeq \isEquiv\ A\ B$ and $Q \defeq
\var{Isomorphism}$. So we must finally prove:
\Isomorphism$. So we must finally prove:
%
\begin{align}
\label{eq:equivIso}
\prod_{f \tp A \to B} \left( \isEquiv\ A\ B\ f \simeq \var{Isomorphism}\ f \right)
\prod_{f \tp A \to B} \left( \isEquiv\ A\ B\ f \simeq \Isomorphism\ f \right)
\end{align}
First, lets prove \ref{eq:equivPropSig}: Let $propP \tp \prod_{a \tp A} \isProp (P\ a)$ and $x\;y \tp \sum_{a \tp A} P\ a$ be given. Because
@ -732,9 +732,9 @@ choose:
%
\begin{align*}
\var{toIso}
& \tp \isEquiv\ f \to \var{Isomorphism}\ f \\
& \tp \isEquiv\ f \to \Isomorphism\ f \\
\var{fromIso}
& \tp \var{Isomorphism}\ f \to \isEquiv\ f
& \tp \Isomorphism\ f \to \isEquiv\ f
\end{align*}
%
As mentioned in section \S\ref{sec:equiv}. These maps are not in general inverses
@ -748,11 +748,11 @@ For this we can use the fact that being-an-equivalence is a mere proposition.
For the other direction:
%
\begin{align*}
\var{toIso} \comp \var{fromIso} \equiv \identity_{\var{Isomorphism}\ f}
\var{toIso} \comp \var{fromIso} \equiv \identity_{\Isomorphism\ f}
\end{align*}
%
We will show that $\var{Isomorphism}\ f$ is also a mere proposition. To this
end, let $X\;Y \tp \var{Isomorphism}\ f$ be given. Name the maps $x\;y \tp B
We will show that $\Isomorphism\ f$ is also a mere proposition. To this
end, let $X\;Y \tp \Isomorphism\ f$ be given. Name the maps $x\;y \tp B
\to A$ respectively. Now, the proof that $X$ and $Y$ are the same is a pair of
paths: $p \tp x \equiv y$ and $\Path\ (\lambda\; i \to
\var{AreInverses}\ f\ (p\ i))\ \mathcal{X}\ \mathcal{Y}$ where $\mathcal{X}$
@ -1008,7 +1008,7 @@ isomorphism, and create a path from this:
\end{split}
\end{align}
%
Where $\widetilde{p} \defeq \var{isoToId}\ \var{iso} \tp X \equiv Y$.
Where $\widetilde{p} \defeq \isoToId\ \var{iso} \tp X \equiv Y$.
Finally we have the type:
%
@ -1248,8 +1248,8 @@ The monoidal formulation of monads consists of the following data:
\label{eq:monad-monoidal-data}
\begin{split}
\EndoR & \tp \Endo ℂ \\
\var{pure} & \tp \NT{\EndoR^0}{\EndoR} \\
\var{join} & \tp \NT{\EndoR^2}{\EndoR}
\pure & \tp \NT{\EndoR^0}{\EndoR} \\
\join & \tp \NT{\EndoR^2}{\EndoR}
\end{split}
\end{align}
%
@ -1264,10 +1264,10 @@ following laws:
\begin{align}
\label{eq:monad-monoidal-laws}
\begin{split}
\var{join} \lll \fmap\ \var{join}
& ≡ \var{join} \lll \var{join}\ \fmap \\
\var{join} \lll \var{pure}\ \fmap & ≡ \identity \\
\var{join} \lll \fmap\ \var{pure} & ≡ \identity
\join \lll \fmap\ \join
& ≡ \join \lll \join\ \fmap \\
\join \lll \pure\ \fmap & ≡ \identity \\
\join \lll \fmap\ \pure & ≡ \identity
\end{split}
\end{align}
%

1
doc/macros.tex
View file

@ -45,6 +45,7 @@
\newcommand{\id}{\var{id}}
\newcommand{\isEquiv}{\var{isEquiv}}
\newcommand{\idToIso}{\var{idToIso}}
\newcommand{\idIso}{\var{idIso}}
\newcommand{\isSet}{\var{isSet}}
\newcommand{\isContr}{\var{isContr}}
\newcommand{\isGroupoid}{\var{isGroupoid}}