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Always use brackets in subscript

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Frederik Hanghøj Iversen 2018-05-07 10:53:22 +02:00
parent 4070e702d7
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2
doc/cubical.tex
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@ -72,7 +72,7 @@ endpoints. I.e.:
\begin{align*} \begin{align*}
p\ 0 & = a_0 \\ p\ 0 & = a_0 \\
p\ 1 & = a_1 p\ 1 & = a_1
\end{align} \end{align*}
% %
The notion of ``homogeneous equalities'' can be recovered by not letting the The notion of ``homogeneous equalities'' can be recovered by not letting the
path-space $P$ depend on it's argument: path-space $P$ depend on it's argument:

92
doc/implementation.tex
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@ -208,14 +208,14 @@ corresponding projections at either endpoint. Thus the element we construct at
$i$ becomes the triple: $i$ becomes the triple:
% %
\begin{equation} \begin{equation}
\begin{alignat}{4} \begin{aligned}
& \var{propIsAssociative} && a.\var{isAssociative}\ & \var{propIsAssociative} && a.\var{isAssociative}\
&& b.\var{isAssociative} && i \\ && b.\var{isAssociative} && i \\
& \var{propIsIdentity} && a.\var{isIdentity}\ & \var{propIsIdentity} && a.\var{isIdentity}\
&& b.\var{isIdentity} && i \\ && b.\var{isIdentity} && i \\
& \var{propArrowsAreSets} && a.\var{arrowsAreSets}\ & \var{propArrowsAreSets} && a.\var{arrowsAreSets}\
&& b.\var{arrowsAreSets} && i && b.\var{arrowsAreSets} && i
\end{alignat} \end{aligned}
\end{equation} \end{equation}
% %
I've found this to be a general pattern when proving things in homotopy type I've found this to be a general pattern when proving things in homotopy type
@ -444,10 +444,10 @@ what we're trying to prove but talks about paths rather than isomorphisms:
\begin{equation} \begin{equation}
\label{eq:coeDomIso} \label{eq:coeDomIso}
\prod_{f \tp \var{Arrow}\ A\ B} \prod_{p \tp A \equiv B} \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)}} \var{coe}\ p_{\var{dom}}\ f \equiv f \lll \inv{(\var{idToIso}\ p)}
\end{equation} \end{equation}
% %
Again $p_\var{dom}$ denotes the path $\var{Arrow}\ A\ X \equiv 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 $p$ \var{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 be given and then invoke based-path-induction. The induction will be based at $A
\tp \var{Object}$, so let $\widetilde{B} \tp \Object$ and $\widetilde{p} \tp \tp \var{Object}$, so let $\widetilde{B} \tp \Object$ and $\widetilde{p} \tp
@ -652,12 +652,15 @@ First, lets proove \ref{eq:equivPropSig}: Let $propP \tp \prod_{a \tp A} \isProp
of $\var{fromIsomorphism}$ it suffices to give an isomorphism between of $\var{fromIsomorphism}$ it suffices to give an isomorphism between
$x \equiv y$ and $\fst\ x \equiv \fst\ y$: $x \equiv y$ and $\fst\ x \equiv \fst\ y$:
% %
\begin{alignat*}{5} %% FIXME: Too much alignement?
\begin{equation*}
\begin{aligned}
f & \defeq \congruence\ \fst f & \defeq \congruence\ \fst
&& \tp x && \equiv y && \to \fst\ x && \equiv \fst\ y \\ && \tp x \equiv y && \to \fst\ x \equiv \fst\ y \\
g & \defeq \var{lemSig}\ \var{propP}\ x\ y g & \defeq \var{lemSig}\ \var{propP}\ x\ y
&& \tp \fst\ x && \equiv \fst\ y && \to x && \equiv y && \tp \fst\ x \equiv \fst\ y && \to x \equiv y
\end{alignat*} \end{aligned}
\end{equation*}
% %
\TODO{Is it confusing that I use point-free style here?} Here $\var{lemSig}$ is \TODO{Is it confusing that I use point-free style here?} Here $\var{lemSig}$ is
a lemma that says that if the second component of a pair is a proposition, it a lemma that says that if the second component of a pair is a proposition, it
@ -856,15 +859,15 @@ Which is provable by \TODO{What?} and that the witness to \ref{eq:pairArrowLaw}
for the left-hand-side and the right-hand-side are the same. The type of this for the left-hand-side and the right-hand-side are the same. The type of this
goal is quite involved, and I will not write it out in full, but at present it goal is quite involved, and I will not write it out in full, but at present it
suffices to show the type of the path-space. Note that the arrows in suffices to show the type of the path-space. Note that the arrows in
\ref{eq:productAssoc} are arrows from $\mathcal{A} = (A , a_\pairA , a_\pairB)$ \ref{eq:productAssoc} are arrows from $\mathcal{A} = (A , a_{\pairA} , a_{\pairB})$
to $\mathcal{D} = (D , d_\pairA , d_\pairB)$ where $a_\pairA$, $a_\pairB$, to $\mathcal{D} = (D , d_{\pairA} , d_{\pairB})$ where $a_{\pairA}$, $a_{\pairB}$,
$d_\pairA$ and $d_\pairB$ are arrows in the underlying category. Given that $p$ $d_{\pairA}$ and $d_{\pairB}$ are arrows in the underlying category. Given that $p$
is the chosen proof of \ref{eq:productAssocUnderlying} we then have that the is the chosen proof of \ref{eq:productAssocUnderlying} we then have that the
witness to \ref{eq:pairArrowLaw} vary over the type: witness to \ref{eq:pairArrowLaw} vary over the type:
% %
\begin{align} \begin{align}
\label{eq:productPath} \label{eq:productPath}
λ\ i → d_\pairA \lll p\ i ≡ a_\pairA × d_\pairB \lll p\ i ≡ a_\pairB λ\ i → d_{\pairA} \lll p\ i ≡ 2 a_{\pairA} × d_{\pairB} \lll p\ i ≡ a_{\pairB}
\end{align} \end{align}
% %
And these paths are in the type of the hom-set of the underlying category, so And these paths are in the type of the hom-set of the underlying category, so
@ -874,14 +877,14 @@ stead we generalize \ref{eq:productPath} to:
% %
\begin{align} \begin{align}
\label{eq:productEqPrinc} \label{eq:productEqPrinc}
\prod_{f \tp \Arrow\ X\ Y} \isProp\ \left( y_\pairA \lll f ≡ x_\pairA × y_\pairB \lll f ≡ x_\pairB \right) \prod_{f \tp \Arrow\ X\ Y} \isProp\ \left( y_{\pairA} \lll f ≡ x_{\pairA} × y_{\pairB} \lll f ≡ x_{\pairB} \right)
\end{align} \end{align}
% %
For all objects $X , x_\pairA , x_\pairB$ and $Y , y_\pairA , y_\pairB$, but For all objects $X , x_{\pairA} , x_{\pairB}$ and $Y , y_{\pairA} , y_{\pairB}$,
this follows from pairs preserving homotopical structure and arrows in the but this follows from pairs preserving homotopical structure and arrows in the
underlying category being sets. This gives us an equality principle for arrows underlying category being sets. This gives us an equality principle for arrows
in this category that says that to prove two arrows $f, f_0, f_1$ and $g, g_0, in this category that says that to prove two arrows $f, f_0, f_1$ and $g, g_0,
$g_1$ equal it suffices to give a proof that $f$ and $g$ are equal. g_1$ equal it suffices to give a proof that $f$ and $g$ are equal.
%% % %% %
%% $$ %% $$
%% \prod_{(f, f_0, f_1)\; (g,g_0,g_1) \tp \Arrow\ X\ Y} f \equiv g \to (f, f_0, f_1) \equiv (g,g_0,g_1) %% \prod_{(f, f_0, f_1)\; (g,g_0,g_1) \tp \Arrow\ X\ Y} f \equiv g \to (f, f_0, f_1) \equiv (g,g_0,g_1)
@ -899,13 +902,16 @@ $$
Since pairs preserve homotopical structure this reduces to: Since pairs preserve homotopical structure this reduces to:
% %
$$ $$
\isSet\ (\Arrow_\bC\ X\ Y) \isSet\ (\Arrow_{\bC}\ X\ Y)
$$ $$
% %
Which holds. And Which holds. And
% %
$$ $$
\prod_{f \tp \Arrow\ X\ Y} \isSet\ \left( y_\pairA \lll f ≡ x_\pairA × y_\pairB \lll f ≡ x_\pairB \right) \prod_{f \tp \Arrow\ X\ Y}
\isSet\ \left( y_{\pairA} \lll f ≡ x_{\pairA}
× y_{\pairB} \lll f ≡ x_{\pairB}
\right)
$$ $$
% %
This we get from \ref{eq:productEqPrinc} and the fact that homotopical structure This we get from \ref{eq:productEqPrinc} and the fact that homotopical structure
@ -915,8 +921,8 @@ This finishes the proof that this is a valid pre-category.
\subsubsection{Univalence} \subsubsection{Univalence}
To prove that this is a proper category it must be shown that it is univalent. To prove that this is a proper category it must be shown that it is univalent.
That is, for any two objects $\mathcal{X} = (X, x_\mathcal{A} , x_\mathca{B})$ That is, for any two objects $\mathcal{X} = (X, x_{\mathcal{A}} , x_{\mathcal{B}})$
and $\mathcal{Y} = Y, y_\mathcal{A}, y_\mathcal{B}$ I will show: and $\mathcal{Y} = Y, y_{\mathcal{A}}, y_{\mathcal{B}}$ I will show:
% %
\begin{align} \begin{align}
(\mathcal{X} \equiv \mathcal{Y}) \cong (\mathcal{X} \approxeq \mathcal{Y}) (\mathcal{X} \equiv \mathcal{Y}) \cong (\mathcal{X} \approxeq \mathcal{Y})
@ -928,7 +934,7 @@ The first type is:
% %
\begin{align} \begin{align}
\label{eq:univ-0} \label{eq:univ-0}
(X , x_\mathcal{A} , x_\mathcal{B}) ≡ (Y , y_\mathcal{A} , y_\mathcal{B}) (X , x_{\mathcal{A}} , x_{\mathcal{B}}) ≡ (Y , y_{\mathcal{A}} , y_{\mathcal{B}})
\end{align} \end{align}
% %
The next types will be the triple: The next types will be the triple:
@ -937,8 +943,8 @@ The next types will be the triple:
\label{eq:univ-1} \label{eq:univ-1}
\begin{split} \begin{split}
p \tp & X \equiv Y \\ p \tp & X \equiv Y \\
& \Path\ (λ i → \Arrow\ (p\ i)\ \mathcal{A})\ x_\mathcal{A}\ y_\mathcal{A} \\ & \Path\ (λ i → \Arrow\ (p\ i)\ \mathcal{A})\ x_{\mathcal{A}}\ y_{\mathcal{A}} \\
& \Path\ (λ i → \Arrow\ (p\ i)\ \mathcal{B})\ x_\mathcal{B}\ y_\mathcal{B} & \Path\ (λ i → \Arrow\ (p\ i)\ \mathcal{B})\ x_{\mathcal{B}}\ y_{\mathcal{B}}
\end{split} \end{split}
%% \end{split} %% \end{split}
\end{align} \end{align}
@ -950,8 +956,8 @@ isomorphism, and create a path from this:
\label{eq:univ-2} \label{eq:univ-2}
\begin{split} \begin{split}
\var{iso} \tp & X \cong Y \\ \var{iso} \tp & X \cong Y \\
& \Path\ (λ i → \Arrow\ (\widetilde{p}\ i)\ \mathcal{A})\ x_\mathcal{A}\ y_\mathcal{A} \\ & \Path\ (λ i → \Arrow\ (\widetilde{p}\ i)\ \mathcal{A})\ x_{\mathcal{A}}\ y_{\mathcal{A}} \\
& \Path\ (λ i → \Arrow\ (\widetilde{p}\ i)\ \mathcal{B})\ x_\mathcal{B}\ y_\mathcal{B} & \Path\ (λ i → \Arrow\ (\widetilde{p}\ i)\ \mathcal{B})\ x_{\mathcal{B}}\ y_{\mathcal{B}}
\end{split} \end{split}
\end{align} \end{align}
% %
@ -961,7 +967,7 @@ Finally we have the type:
% %
\begin{align} \begin{align}
\label{eq:univ-3} \label{eq:univ-3}
(X , x_\mathcal{A} , x_\mathcal{B}) ≊ (Y , y_\mathcal{A} , y_\mathcal{B}) (X , x_{\mathcal{A}} , x_{\mathcal{B}}) ≊ (Y , y_{\mathcal{A}} , y_{\mathcal{B}})
\end{align} \end{align}
\emph{Proposition} \ref{eq:univ-0} is isomorphic to \ref{eq:univ-1}: This is \emph{Proposition} \ref{eq:univ-0} is isomorphic to \ref{eq:univ-1}: This is
@ -976,7 +982,7 @@ the implementation for the details).
will swho two corrolaries of \ref{eq:coeCod}: For an isomorphism $(\iota, will swho two corrolaries of \ref{eq:coeCod}: For an isomorphism $(\iota,
\inv{\iota}, \var{inv}) \tp A \cong B$, arrows $f \tp \Arrow\ A\ X$, $g \tp \inv{\iota}, \var{inv}) \tp A \cong B$, arrows $f \tp \Arrow\ A\ X$, $g \tp
\Arrow\ B\ X$ and a heterogeneous path between them, $q \tp \Path\ (\lambda i \Arrow\ B\ X$ and a heterogeneous path between them, $q \tp \Path\ (\lambda i
\to p_\var{dom}\ i)\ f\ g$, where $p_\var{dom} \tp \Arrow\ A\ X \equiv \to p_{\var{dom}}\ i)\ f\ g$, where $p_{\var{dom}} \tp \Arrow\ A\ X \equiv
\Arrow\ B\ X$ is a path induced by $\var{iso}$, we have the following two \Arrow\ B\ X$ is a path induced by $\var{iso}$, we have the following two
results results
% %
@ -993,14 +999,14 @@ Now we can prove the equiavalence in the following way: Given $(f, \inv{f},
\var{inv}_f) \tp X \cong Y$ and two heterogeneous paths \var{inv}_f) \tp X \cong Y$ and two heterogeneous paths
% %
\begin{align*} \begin{align*}
p_\mathcal{A} & \tp \Path\ (\lambda i \to p_\var{dom}\ i)\ x_\mathcal{A}\ y_\mathcal{A}\\ p_{\mathcal{A}} & \tp \Path\ (\lambda i \to p_{\var{dom}}\ i)\ x_{\mathcal{A}}\ y_{\mathcal{A}}\\
% %
q_\mathcal{B} & \tp \Path\ (\lambda i \to p_\var{dom}\ i)\ x_\mathcal{B}\ y_\mathcal{B} q_{\mathcal{B}} & \tp \Path\ (\lambda i \to p_{\var{dom}}\ i)\ x_{\mathcal{B}}\ y_{\mathcal{B}}
\end{align*} \end{align*}
% %
all as in \ref{eq:univ-2}. I use $p_\var{dom}$ here again to mean the path all as in \ref{eq:univ-2}. I use $p_{\var{dom}}$ here again to mean the path
induced by the isomorphism $f, \inv{f}$. I must now construct an isomorphism induced by the isomorphism $f, \inv{f}$. I must now construct an isomorphism
$(X, x_\mathcal{A}, x_\mathcal{B}) \approxeq (Y, y_\mathcal{A}, y_\mathcal{B}$ $(X, x_{\mathcal{A}}, x_{\mathcal{B}}) \approxeq (Y, y_{\mathcal{A}}, y_{\mathcal{B}})$
as in \ref{eq:univ-3}. That is, an isomorphism in the present category. I remind as in \ref{eq:univ-3}. That is, an isomorphism in the present category. I remind
the reader that such a gadget is a triple. The first component shall be: the reader that such a gadget is a triple. The first component shall be:
% %
@ -1012,10 +1018,10 @@ To show that this choice fits the bill I must now verify that it satisfies
\ref{eq:pairArrowLaw}, which in this case becomes: \ref{eq:pairArrowLaw}, which in this case becomes:
% %
\begin{align} \begin{align}
y_\mathcal{A} \lll f ≡ x_\mathcal{A} × y_\mathcal{B} \lll f ≡ x_\mathcal{B} y_{\mathcal{A}} \lll f ≡ x_{\mathcal{A}} × y_{\mathcal{B}} \lll f ≡ x_{\mathcal{B}}
\end{align} \end{align}
% %
Which, since $f$ is an isomorphism and $p_\mathcal{A}$ (resp. $p_\mathcal{B}$) Which, since $f$ is an isomorphism and $p_{\mathcal{A}}$ (resp. $p_{\mathcal{B}}$)
is a path varying according to a path constructed from this isomorphism, this is is a path varying according to a path constructed from this isomorphism, this is
exactly what \ref{eq:domain-twist-0} gives us. exactly what \ref{eq:domain-twist-0} gives us.
% %
@ -1033,7 +1039,7 @@ For the other direction we're given just given the isomorphism
$$ $$
(f, \inv{f}, \var{inv}_f) (f, \inv{f}, \var{inv}_f)
\tp \tp
(X, x_\mathcal{A}, x_\mathcal{B}) \approxeq (Y, y_\mathcal{A}, y_\mathcal{B}) (X, x_{\mathcal{A}}, x_{\mathcal{B}}) \approxeq (Y, y_{\mathcal{A}}, y_{\mathcal{B}})
$$ $$
% %
Projecting out the first component gives us the isomorphism Projecting out the first component gives us the isomorphism
@ -1048,8 +1054,8 @@ This gives rise to the following paths:
\begin{align} \begin{align}
\begin{split} \begin{split}
\widetilde{p} & \tp X \equiv Y \\ \widetilde{p} & \tp X \equiv Y \\
\widetilde{p}_\mathcal{A} & \tp \Arrow\ X\ \mathcal{A} \equiv \Arrow\ Y\ \mathcal{A} \\ \widetilde{p}_{\mathcal{A}} & \tp \Arrow\ X\ \mathcal{A} \equiv \Arrow\ Y\ \mathcal{A} \\
\widetilde{p}_\mathcal{B} & \tp \Arrow\ X\ \mathcal{B} \equiv \Arrow\ Y\ \mathcal{B} \widetilde{p}_{\mathcal{B}} & \tp \Arrow\ X\ \mathcal{B} \equiv \Arrow\ Y\ \mathcal{B}
\end{split} \end{split}
\end{align} \end{align}
% %
@ -1058,8 +1064,8 @@ It then remains to construct the two paths:
\begin{align} \begin{align}
\begin{split} \begin{split}
\label{eq:product-paths} \label{eq:product-paths}
& \Path\ (λ i → \widetilde{p}_\mathcal{A}\ i)\ x_\mathcal{A}\ y_\mathcal{A}\\ & \Path\ (λ i → \widetilde{p}_{\mathcal{A}}\ i)\ x_{\mathcal{A}}\ y_{\mathcal{A}}\\
& \Path\ (λ i → \widetilde{p}_\mathcal{B}\ i)\ x_\mathcal{B}\ y_\mathcal{B} & \Path\ (λ i → \widetilde{p}_{\mathcal{B}}\ i)\ x_{\mathcal{B}}\ y_{\mathcal{B}}
\end{split} \end{split}
\end{align} \end{align}
% %
@ -1077,17 +1083,17 @@ Which is used without proof. See the implementation for the details.
\begin{align} \begin{align}
\begin{split} \begin{split}
\label{eq:product-paths} \label{eq:product-paths}
\var{coe}\ \widetilde{p}_\mathcal{A}\ x_\mathcal{A} ≡ y_\mathcal{A}\\ \var{coe}\ \widetilde{p}_{\mathcal{A}}\ x_{\mathcal{A}} ≡ y_{\mathcal{A}}\\
\var{coe}\ \widetilde{p}_\mathcal{B}\ x_\mathcal{B} ≡ y_\mathcal{B} \var{coe}\ \widetilde{p}_{\mathcal{B}}\ x_{\mathcal{B}} ≡ y_{\mathcal{B}}
\end{split} \end{split}
\end{align} \end{align}
% %
The proof of the first one is: The proof of the first one is:
% %
\begin{align*} \begin{align*}
\var{coe}\ \widetilde{p}_\mathcal{A}\ x_\mathcal{A} \var{coe}\ \widetilde{p}_{\mathcal{A}}\ x_{\mathcal{A}}
& ≡ x_\mathcal{A} \lll \fst\ \inv{f} && \text{$\var{coeDom}$ and the isomorphism $f, \inv{f}$} \\ & ≡ x_{\mathcal{A}} \lll \fst\ \inv{f} && \text{$\var{coeDom}$ and the isomorphism $f, \inv{f}$} \\
& ≡ y_\mathcal{A} && \text{\ref{eq:pairArrowLaw} for $\inv{f}$} & ≡ y_{\mathcal{A}} && \text{\ref{eq:pairArrowLaw} for $\inv{f}$}
\end{align*} \end{align*}
% %
We have now constructed the maps between \ref{eq:univ-0} and \ref{eq:univ-1}. It We have now constructed the maps between \ref{eq:univ-0} and \ref{eq:univ-1}. It

20
doc/introduction.tex
View file

@ -1,7 +1,7 @@
\chapter{Introduction} \chapter{Introduction}
Functional extensionality and univalence is not expressible in Functional extensionality and univalence is not expressible in
\nomen{Intensional Martin Löf Type Theory} (ITT). This poses a severe limitation \nomen{Intensional Martin Löf Type Theory} (ITT). This poses a severe limitation
on both 1) what is \emph{provable} and 2) the \emph{reusability} of proofs. on both i. what is \emph{provable} and ii. the \emph{reusability} of proofs.
Recent developments have, however, resulted in \nomen{Cubical Type Theory} (CTT) Recent developments have, however, resulted in \nomen{Cubical Type Theory} (CTT)
which permits a constructive proof of these two important notions. which permits a constructive proof of these two important notions.
@ -19,9 +19,13 @@ limitations inherent in ITT and -- by extension -- Agda.
Consider the functions: Consider the functions:
% %
\begin{multicols}{2} \begin{multicols}{2}
$f \defeq (n : \bN) \mapsto (0 + n : \bN)$ \noindent
\begin{equation*}
$g \defeq (n : \bN) \mapsto (n + 0 : \bN)$ f \defeq (n : \bN) \mapsto (0 + n : \bN)
\end{equation*}
\begin{equation*}
g \defeq (n : \bN) \mapsto (n + 0 : \bN)
\end{equation*}
\end{multicols} \end{multicols}
% %
$n + 0$ is \nomen{definitionally} equal to $n$, which we write as $n + 0 = n$. $n + 0$ is \nomen{definitionally} equal to $n$, which we write as $n + 0 = n$.
@ -160,7 +164,7 @@ Another approach is to use the \emph{setoid interpretation} of type theory
equivalence relation $\sim \tp X \to X \to \MCU$ on that type. Under the setoid equivalence relation $\sim \tp X \to X \to \MCU$ on that type. Under the setoid
interpretation the equivalence relation serve as a sort of ``local'' interpretation the equivalence relation serve as a sort of ``local''
propositional equality. This approach has other drawbacks; it does not satisfy propositional equality. This approach has other drawbacks; it does not satisfy
all propositional equalites of type theory (\TODO{Citation needed}, is all propositional equalites of type theory (\TODO{Citation needed}), is
cumbersome to work with in practice (\cite[p. 4]{huber-2016}) and makes cumbersome to work with in practice (\cite[p. 4]{huber-2016}) and makes
equational proofs less reusable since equational proofs $a \sim_{X} b$ are equational proofs less reusable since equational proofs $a \sim_{X} b$ are
inherently `local' to the extensional set $(X , \sim)$. inherently `local' to the extensional set $(X , \sim)$.
@ -192,9 +196,9 @@ Name & Agda & Notation \\
Function, morphism, map & \texttt{A → B} & $A → B$ \\ Function, morphism, map & \texttt{A → B} & $A → B$ \\
Dependent- ditto & \texttt{(a : A) → B} & $∏_{a \tp A} → B$ \\ Dependent- ditto & \texttt{(a : A) → B} & $∏_{a \tp A} → B$ \\
\nomen{Arrow} & \texttt{Arrow A B} & $\Arrow\ A\ B$ \\ \nomen{Arrow} & \texttt{Arrow A B} & $\Arrow\ A\ B$ \\
\nomen{Object} & \texttt{C.Object} & $̱ℂ.Object$ \nomen{Object} & \texttt{C.Object} & $̱ℂ.Object$ \\
Definition & \texttt{=} & $̱\defeq$ Definition & \texttt{=} & $̱\defeq$ \\
Judgmental equality & \null & $̱=$ Judgmental equality & \null & $̱=$ \\
Propositional equality & \null & $̱\equiv$ Propositional equality & \null & $̱\equiv$
\end{tabular} \end{tabular}
\end{center} \end{center}