Disjoint Unions

Posted on 2017-12-21 by nbloomf

Tags: literate-haskell, arithmetic-made-difficult

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This page is part of a series on Arithmetic Made Difficult.

This post is literate Haskell; you can load the source into GHCi and play along.

{-# LANGUAGE NoImplicitPrelude, ScopedTypeVariables #-}
module DisjointUnions
  ( Union(..), lft, rgt, either, uswap, upair , uassocL, uassocR, isLft, isRgt
  , _test_disjoint_union, main_disjoint_union
  ) where

import Testing
import Functions
import Flip
import Clone
import Composition
import Booleans

Dual to sets of tuples are disjoint sums.

Let $A$ and $B$ be sets. There is a set $A + B$ together with two functions $\lft : A \rightarrow A + B$ and $\rgt : B \rightarrow A + B$ having the property that if $X$ is a set and $\sigma : A \rightarrow X$ and $\tau : B \rightarrow X$ functions then there is a unique map $\Theta : A + B \rightarrow X$ such that $\Theta \circ \lft = \sigma$ and $\Theta \circ \rgt = \tau$ . That is, there is a unique $\Theta$ such that the following diagram commutes.

$\require{AMScd} \begin{CD} A @>{\lft}>> A + B @<{\rgt}<< B \\ @V{\sigma}VV @VV{\Theta}V @VV{\tau}V \\ X @= X @= X \end{CD}$

We will denote this unique $\Theta : A + B \rightarrow X$ by $\either(\sigma,\tau)$ .

More concretely, if $f : A + B \rightarrow X$ such that $f(\lft(a)) = \sigma(a)$ and $f(\rgt(b)) = \tau(b)$ for all $a \in A$ and $b \in B$ , then $f = \either(\sigma,\tau)$ .

Now every element of $A+B$ is uniquely either of the form $\lft(a)$ or $\rgt(b)$ for some $a \in A$ or $b \in B$ ; this allows us to use case analysis on $A + B$ . This set is sometimes called a tagged union, because it behaves like an ordinary set-theoretic union where elements are “tagged” with the set they came from (so, for instance, if $x \in A$ and $x \in B$ , the $x$ from $A$ and the $x$ from $B$ are distinct in $A+B$ ).

Let $A$ and $B$ be sets.

If $x \in A + B$ , then either $x = \lft(a)$ for some $a \in A$ or $x = \rgt(b)$ for some $b \in B$ .
There does not exist $x \in A + B$ such that $x = \lft(a)$ and $x = \rgt(b)$ for some $a \in A$ and $b \in B$ .
$\lft : A \rightarrow A+B$ and $\rgt : B \rightarrow A+B$ are injective.

Suppose to the contrary that there exists some $z \in A+B$ which is not of the form $\lft(a)$ or $\rgt(b)$ for any $a$ or $b$ . Now consider $f = \const(\btrue)$ and $g(x) = \bif{\beq(x,z)}{\bfalse}{\btrue}$ as functions $A+B \rightarrow \bool$ . Note in particular that if $a \in A$ we have $\begin{eqnarray*} & & \compose(f)(\lft)(a) \\ & \href{/posts/arithmetic-made-difficult/Functions.html#def-compose} = & f(\lft(a)) \\ & \let{f = \const(\btrue)} = & \const(\btrue)(\lft(a)) \\ & \href{/posts/arithmetic-made-difficult/Functions.html#def-const} = & \btrue \\ & \hyp{g(\lft(a)) = \btrue} = & g(\lft(a)) \\ & \href{/posts/arithmetic-made-difficult/Functions.html#def-compose} = & \compose(g)(\lft)(a) \end{eqnarray*}$ since $z \neq \lft(a)$ , and if $b \in B$ we have $\begin{eqnarray*} & & \compose(f)(\rgt)(b) \\ & \href{/posts/arithmetic-made-difficult/Functions.html#def-compose} = & f(\rgt(b)) \\ & \let{f = \const(\btrue)} = & \const(\btrue)(\rgt(b)) \\ & \href{/posts/arithmetic-made-difficult/Functions.html#def-const} = & \btrue \\ & \hyp{g(\rgt(b)) = \btrue} = & g(\rgt(b)) \\ & \href{/posts/arithmetic-made-difficult/Functions.html#def-compose} = & \compose(g)(\rgt)(b) \end{eqnarray*}$ similarly. By the universal property of $A + B$ , we thus have $f = \either(\const(\btrue),\const(\btrue)) = g.$ But now we have $\btrue = f(z) = g(z) = \bfalse,$ which is absurd.
Suppose to the contrary that we have $z \in A + B$ with $\lft(a) = z = \rgt(b)$ for some $a \in A$ and $b \in B$ , and consider now $f : A+B \rightarrow \bool$ given by $f = \either(\const(\btrue),\const(\bfalse)).$ We then have $\begin{eqnarray*} & & \btrue \\ & \href{/posts/arithmetic-made-difficult/Functions.html#def-const} = & \const(\btrue)(a) \\ & \href{/posts/arithmetic-made-difficult/DisjointUnions.html#def-either-lft} = & \either(\const(\btrue),\const(\bfalse))(\lft(a)) \\ & \let{z = \lft(a)} = & \either(\const(\btrue),\const(\bfalse))(z) \\ & \let{z = \rgt(b)} = & \either(\const(\btrue),\const(\bfalse))(\rgt(b)) \\ & \href{/posts/arithmetic-made-difficult/DisjointUnions.html#def-either-rgt} = & \const(\bfalse)(b) \\ & \href{/posts/arithmetic-made-difficult/Functions.html#def-const} = & \bfalse \end{eqnarray*}$ which is absurd.
Let $u,v \in A$ , and suppose $\lft(u) = \lft(v)$ . Now define $f : A \rightarrow \bool$ by $f(x) = \bif{\beq(x,u)}{\bfalse}{\btrue},$ and consider $g = \either(f,\const(\btrue))$ . We have $\begin{eqnarray*} & & \bfalse \\ & \href{/posts/arithmetic-made-difficult/Booleans.html#cor-if-true} = & \bif{\btrue}{\bfalse}{\btrue} \\ & \href{/posts/arithmetic-made-difficult/Testing.html#thm-eq-reflexive} = & \bif{\beq(u,u)}{\bfalse}{\btrue} \\ & \let{f(x) = \bif{\beq(x,u)}{\bfalse}{\btrue}} = & f(u) \\ & \href{/posts/arithmetic-made-difficult/DisjointUnions.html#def-either-lft} = & \either(f,\const(\btrue))(\lft(u)) \\ & \hyp{\lft(u) = \lft(v)} = & \either(f,\const(\btrue))(\lft(v)) \\ & \href{/posts/arithmetic-made-difficult/DisjointUnions.html#def-either-lft} = & f(v) \\ & \let{f(x) = \bif{\beq(x,u)}{\bfalse}{\btrue}} = & \bif{\beq(v,u)}{\bfalse}{\btrue} \end{eqnarray*}$ thus $v = u$ as needed. A similar argument holds for $\rgt$ .

The previous results suggest that we can model $A + B$ with the Haskell type Either a b. For simplicity’s sake we’ll redefine this type, and the maps in the universal property, like so.

data Union a b
  = Left a | Right b
  deriving Show

lft :: a -> Union a b
lft a = Left a

rgt :: b -> Union a b
rgt b = Right b

either :: (a -> x) -> (b -> x) -> Union a b -> x
either f _ (Left  a) = f a
either _ g (Right b) = g b

instance (Equal a, Equal b) => Equal (Union a b) where
  eq (Left a1)  (Left a2)  = eq a1 a2
  eq (Left _)   (Right _)  = false
  eq (Right _)  (Left _)   = false
  eq (Right b1) (Right b2) = eq b1 b2

instance (Arbitrary a, Arbitrary b) => Arbitrary (Union a b) where
  arbitrary = do
    s <- arbitrary
    if s
      then arbitrary >>= (return . Left)
      else arbitrary >>= (return . Right)

instance (CoArbitrary a, CoArbitrary b) => CoArbitrary (Union a b) where
  coarbitrary (Left  a) = variant (0 :: Int) . coarbitrary a
  coarbitrary (Right b) = variant (1 :: Int) . coarbitrary b

instance (TypeName a, TypeName b) => TypeName (Union a b) where
  typeName _ =
    "Union " ++ typeName (undefined :: a) ++ " " ++ typeName (undefined :: b)

For example, $\id_{A + B}$ is an $\either$ .

Provided the types match up, we have $\either(\lft,\rgt) = \id_{A + B}.$

If $a \in A$ , we have $\begin{eqnarray*} & & \compose(\id)(\lft)(a) \\ & \href{/posts/arithmetic-made-difficult/Functions.html#def-compose} = & \id(\lft(a)) \\ & \href{/posts/arithmetic-made-difficult/Functions.html#def-id} = & \lft(a) \end{eqnarray*}$ and likewise if $b \in B$ we have $\begin{eqnarray*} & & \compose(\id)(\rgt)(b) \\ & \href{/posts/arithmetic-made-difficult/Functions.html#def-compose} = & \id(\rgt(b)) \\ & \href{/posts/arithmetic-made-difficult/Functions.html#def-id} = & \rgt(b) \end{eqnarray*}$ Since $\either(\lft,\rgt)$ is unique with this property, we have $\either(\lft,\rgt) = \id_{A+B}$ as claimed.

_test_either_lft_rgt :: (Equal a, Equal b)
  => a -> b -> Test (Union a b -> Bool)
_test_either_lft_rgt _ _ =
  testName "either(lft,rgt) == id" $
  \x -> eq (either lft rgt x) x

We define $\uSwap$ on disjoint unions like so.

Let $A$ and $B$ be sets. We define $\uSwap : A + B \rightarrow B + A$ by $\uSwap = \either(\rgt,\lft).$

In Haskell:

uswap :: Union a b -> Union b a
uswap = either rgt lft

Now $\uSwap$ effectively toggles the tag of its argument.

Let $A$ and $B$ be sets. Then we have the following for all $a \in A$ and $b \in B$ .

$\uSwap(\lft(a)) = \rgt(a)$ .
$\uSwap(\rgt(b)) = \lft(b)$ .
$\compose(\uSwap)(\uSwap) = \id$ .

Note that $\begin{eqnarray*} & & \uSwap(\lft(a)) \\ & \href{/posts/arithmetic-made-difficult/DisjointUnions.html#def-uSwap} = & \either(\rgt,\lft)(\lft(a)) \\ & \href{/posts/arithmetic-made-difficult/DisjointUnions.html#def-either-lft} = & \rgt(a) \end{eqnarray*}$ as claimed.
Note that $\begin{eqnarray*} & & \uSwap(\rgt(a)) \\ & \href{/posts/arithmetic-made-difficult/DisjointUnions.html#def-uSwap} = & \either(\rgt,\lft)(\rgt(a)) \\ & \href{/posts/arithmetic-made-difficult/DisjointUnions.html#def-either-rgt} = & \lft(a) \end{eqnarray*}$ as claimed.
Let $x \in A+B$ ; we have two possibilities. If $x = \lft(a)$ , we have $\begin{eqnarray*} & & \compose(\uSwap)(\uSwap)(\lft(a)) \\ & \href{/posts/arithmetic-made-difficult/Functions.html#def-compose} = & \uSwap(\uSwap(\lft(a))) \\ & \href{/posts/arithmetic-made-difficult/DisjointUnions.html#thm-uSwap-lft} = & \uSwap(\rgt(a)) \\ & \href{/posts/arithmetic-made-difficult/DisjointUnions.html#thm-uSwap-rgt} = & \lft(a) \end{eqnarray*}$ and if $x = \rgt(b)$ we have $\begin{eqnarray*} & & \compose(\uSwap)(\uSwap)(\rgt(b)) \\ & \href{/posts/arithmetic-made-difficult/Functions.html#def-compose} = & \uSwap(\uSwap(\rgt(b))) \\ & \href{/posts/arithmetic-made-difficult/DisjointUnions.html#thm-uSwap-rgt} = & \uSwap(\lft(b)) \\ & \href{/posts/arithmetic-made-difficult/DisjointUnions.html#thm-uSwap-lft} = & \rgt(b) \end{eqnarray*}$ as needed.

_test_uswap_lft :: (Equal a, Equal b)
  => a -> b -> Test (a -> Bool)
_test_uswap_lft _ y =
  testName "uswap(left(a)) == right(a)" $
  \a -> eq
    (uswap (lft a))
    ((rgt a) `withTypeOf` (lft y))


_test_uswap_rgt :: (Equal a, Equal b)
  => a -> b -> Test (b -> Bool)
_test_uswap_rgt x _ =
  testName "uswap(right(a)) == left(a)" $
  \b -> eq
    (uswap (rgt b))
    ((lft b) `withTypeOf` (rgt x))


_test_uswap_uswap :: (Equal a, Equal b)
  => a -> b -> Test (Union a b -> Bool)
_test_uswap_uswap _ _ =
  testName "swap(swap(x)) == x" $
  \x -> eq (uswap (uswap x)) x

Next, the utility $\uPair$ facilitates defining functions from one disjoint union to another.

Let $A$ , $B$ , $U$ , and $V$ be sets. We define $\uPair : U^A \times V^B \rightarrow (U + V)^{A + B}$ by $\uPair(f,g) = \either(\compose(\lft)(f),\compose(\rgt)(g)).$

In Haskell:

upair :: (a -> u) -> (b -> v) -> Union a b -> Union u v
upair f g = either (lft . f) (rgt . g)

$\uPair$ has some nice properties.

For all $f$ , $g$ , $h$ , $k$ , $a$ , and $b$ we have the following.

$\uPair(f,g)(\lft(a)) = \lft(f(a))$ .
$\uPair(f,g)(\rgt(b)) = \rgt(g(b))$ .
$\compose(\uPair(f,g))(\uPair(h,k)) = \uPair(\compose(f)(h),\compose(g)(k))$ .

Note that $\begin{eqnarray*} & & \uPair(f,g)(\lft(a)) \\ & \href{/posts/arithmetic-made-difficult/DisjointUnions.html#def-uPair} = & \either(\compose(\lft)(f),\compose(\rgt)(g))(\lft(a)) \\ & \href{/posts/arithmetic-made-difficult/DisjointUnions.html#def-either-lft} = & \compose(\lft)(f)(a) \\ & \href{/posts/arithmetic-made-difficult/Functions.html#def-compose} = & \lft(f(a)) \end{eqnarray*}$ as claimed.
Note that $\begin{eqnarray*} & & \uPair(f,g)(\rgt(b)) \\ & \href{/posts/arithmetic-made-difficult/DisjointUnions.html#def-uPair} = & \either(\compose(\lft)(f),\compose(\rgt)(g))(\rgt(b)) \\ & \href{/posts/arithmetic-made-difficult/DisjointUnions.html#def-either-rgt} = & \compose(\rgt)(g)(b) \\ & \href{/posts/arithmetic-made-difficult/Functions.html#def-compose} = & \rgt(g(b)) \end{eqnarray*}$ as claimed.
We consider two possibilities: note that $\begin{eqnarray*} & & \compose(\uPair(f,g))(\uPair(h,k))(\lft(a)) \\ & \href{/posts/arithmetic-made-difficult/Functions.html#def-compose} = & \uPair(f,g)(\uPair(h,k)(\lft(a))) \\ & \href{/posts/arithmetic-made-difficult/DisjointUnions.html#thm-uPair-lft} = & \uPair(f,g)(\lft(h(a))) \\ & \href{/posts/arithmetic-made-difficult/DisjointUnions.html#thm-uPair-lft} = & \lft(f(h(a))) \\ & \href{/posts/arithmetic-made-difficult/Functions.html#def-compose} = & \lft(\compose(f)(h)(a)) \\ & \href{/posts/arithmetic-made-difficult/DisjointUnions.html#thm-uPair-lft} = & \uPair(\compose(f)(h),\compose(g)(k))(\lft(a)) \end{eqnarray*}$ and $\begin{eqnarray*} & & \compose(\uPair(f,g))(\uPair(h,k))(\rgt(b)) \\ & \href{/posts/arithmetic-made-difficult/Functions.html#def-compose} = & \uPair(f,g)(\uPair(h,k)(\rgt(b))) \\ & \href{/posts/arithmetic-made-difficult/DisjointUnions.html#thm-uPair-rgt} = & \uPair(f,g)(\rgt(k(b))) \\ & \href{/posts/arithmetic-made-difficult/DisjointUnions.html#thm-uPair-rgt} = & \rgt(g(k(b))) \\ & \href{/posts/arithmetic-made-difficult/Functions.html#def-compose} = & \rgt(\compose(g)(k)(b)) \\ & \href{/posts/arithmetic-made-difficult/DisjointUnions.html#thm-uPair-rgt} = & \uPair(\compose(f)(h),\compose(g)(k))(\rgt(b)) \end{eqnarray*}$ as needed.

_test_upair_lft :: (Equal a, Equal b)
  => a -> b -> Test ((a -> a) -> (b -> b) -> a -> Bool)
_test_upair_lft _ y =
  testName "upair(f,g)(lft(a)) == lft(f(a))" $
  \f g a -> eq
    (upair f g ((lft a) `withTypeOf` (rgt y)))
    ((lft (f a)) `withTypeOf` (rgt y))


_test_upair_rgt :: (Equal a, Equal b)
  => a -> b -> Test ((a -> a) -> (b -> b) -> b -> Bool)
_test_upair_rgt x _ =
  testName "upair(f,g)(rgt(b)) == rgt(g(b))" $
  \f g b -> eq
    (upair f g ((rgt b) `withTypeOf` (lft x)))
    ((rgt (g b)) `withTypeOf` (lft x))


_test_upair_upair :: (Equal a, Equal b)
  => a -> b
  -> Test ((a -> a) -> (b -> b) -> (a -> a) -> (b -> b)
      -> Union a b -> Bool)
_test_upair_upair _ _ =
  testName "upair(f,g) . upair(h,k) == upair(f . h, g . k)" $
  \f g h k x ->
    eq
      (upair f g (upair h k x))
      (upair (f . h) (g . k) x)

Finally, note that although as sets $A + (B + C)$ and $(A + B) + C$ cannot possibly be equal to each other in general, they are naturally isomorphic via $\uAssocL$ and $\uAssocR$ .

Let $A$ , $B$ , and $C$ be sets. We define $\uAssocL : A + (B + C) \rightarrow (A + B) + C$ by $\uAssocL = \either(\compose(\lft)(\lft),\either(\compose(\lft)(\rgt),\rgt))$ and define $\uAssocR : (A + B) + C \rightarrow A + (B + C)$ by $\uAssocR = \either(\either(\lft,\compose(\rgt)(\lft)),\compose(\rgt)(\rgt)).$

In Haskell:

uassocL :: Union a (Union b c) -> Union (Union a b) c
uassocL = either (lft . lft) (either (lft . rgt) rgt)

uassocR :: Union (Union a b) c -> Union a (Union b c)
uassocR = either (either lft (rgt . lft)) (rgt . rgt)

Now $\uAssocL$ and $\uAssocR$ have some nice properties.

The following hold whenever everything has the appropriate type.

$\uAssocL(\lft(a)) = \lft(\lft(a))$ .
$\uAssocL(\rgt(\lft(b))) = \lft(\rgt(b))$ .
$\uAssocL(\rgt(\rgt(c))) = \rgt(c)$ .
$\uAssocR(\lft(\lft(a))) = \lft(a)$ .
$\uAssocR(\lft(\rgt(b))) = \rgt(\lft(b))$ .
$\uAssocR(\rgt(c)) = \rgt(\rgt(c))$ .
$\compose(\uAssocR)(\uAssocL) = \id_{A + (B + C)}$ .
$\compose(\uAssocL)(\uAssocR) = \id_{(A + B) + C}$ .

We have $\begin{eqnarray*} & & \uAssocL(\lft(a)) \\ & \href{/posts/arithmetic-made-difficult/DisjointUnions.html#def-uAssocL} = & \either(\compose(\lft)(\lft),\either(\compose(\lft)(\rgt),\rgt))(\lft(a)) \\ & \href{/posts/arithmetic-made-difficult/DisjointUnions.html#def-either-lft} = & \compose(\lft)(\lft)(a) \\ & \href{/posts/arithmetic-made-difficult/Functions.html#def-compose} = & \lft(\lft(a)) \end{eqnarray*}$ as claimed.
We have $\begin{eqnarray*} & & \uAssocL(\rgt(\lft(b))) \\ & \href{/posts/arithmetic-made-difficult/DisjointUnions.html#def-uAssocL} = & \either(\compose(\lft)(\lft),\either(\compose(\lft)(\rgt),\rgt))(\rgt(\lft(b))) \\ & \href{/posts/arithmetic-made-difficult/DisjointUnions.html#def-either-rgt} = & \either(\compose(\lft)(\rgt),\rgt)(\lft(b)) \\ & \href{/posts/arithmetic-made-difficult/DisjointUnions.html#def-either-lft} = & \compose(\lft)(\rgt)(b) \\ & \href{/posts/arithmetic-made-difficult/Functions.html#def-compose} = & \lft(\rgt(b)) \end{eqnarray*}$ as claimed.
We have $\begin{eqnarray*} & & \uAssocL(\rgt(\rgt(c))) \\ & \href{/posts/arithmetic-made-difficult/DisjointUnions.html#def-uAssocL} = & \either(\compose(\lft)(\lft),\either(\compose(\lft)(\rgt),\rgt))(\rgt(\rgt(c))) \\ & \href{/posts/arithmetic-made-difficult/DisjointUnions.html#def-either-rgt} = & \either(\compose(\lft)(\rgt),\rgt)(\rgt(c)) \\ & \href{/posts/arithmetic-made-difficult/DisjointUnions.html#def-either-rgt} = & \rgt(c) \end{eqnarray*}$ as claimed.
We have $\begin{eqnarray*} & & \uAssocR(\lft(\lft(a))) \\ & \href{/posts/arithmetic-made-difficult/DisjointUnions.html#def-uAssocR} = & \either(\either(\lft,\compose(\rgt)(\lft)),\compose(\rgt)(\rgt))(\lft(\lft(a))) \\ & \href{/posts/arithmetic-made-difficult/DisjointUnions.html#def-either-lft} = & \either(\lft,\compose(\rgt)(\lft))(\lft(a)) \\ & \href{/posts/arithmetic-made-difficult/DisjointUnions.html#def-either-lft} = & \lft(a) \end{eqnarray*}$ as claimed.
We have $\begin{eqnarray*} & & \uAssocR(\lft(\rgt(b))) \\ & \href{/posts/arithmetic-made-difficult/DisjointUnions.html#def-uAssocR} = & \either(\either(\lft,\compose(\rgt)(\lft)),\compose(\rgt)(\rgt))(\lft(\rgt(b))) \\ & \href{/posts/arithmetic-made-difficult/DisjointUnions.html#def-either-lft} = & \either(\lft,\compose(\rgt)(\lft))(\rgt(b)) \\ & \href{/posts/arithmetic-made-difficult/DisjointUnions.html#def-either-rgt} = & \compose(\rgt)(\lft)(b) \\ & \href{/posts/arithmetic-made-difficult/Functions.html#def-compose} = & \rgt(\lft(b)) \end{eqnarray*}$ as claimed.
We have $\begin{eqnarray*} & & \uAssocR(\rgt(c)) \\ & \href{/posts/arithmetic-made-difficult/DisjointUnions.html#def-uAssocR} = & \either(\either(\lft,\compose(\rgt)(\lft)),\compose(\rgt)(\rgt))(\rgt(c)) \\ & \href{/posts/arithmetic-made-difficult/DisjointUnions.html#def-either-rgt} = & \compose(\rgt)(\rgt)(c) \\ & \href{/posts/arithmetic-made-difficult/Functions.html#def-compose} = & \rgt(\rgt(c)) \end{eqnarray*}$ as claimed.
If $x \in A + (B + C)$ , we have three possibilities. If $x = \lft(a)$ , note that $\begin{eqnarray*} & & \compose(\uAssocR)(\uAssocL)(x) \\ & \let{x = \lft(a)} = & \compose(\uAssocR)(\uAssocL)(\lft(a)) \\ & \href{/posts/arithmetic-made-difficult/Functions.html#def-compose} = & \uAssocR(\uAssocL(\lft(a))) \\ & \href{/posts/arithmetic-made-difficult/DisjointUnions.html#thm-uAssocL-lft} = & \uAssocR(\lft(\lft(a))) \\ & \href{/posts/arithmetic-made-difficult/DisjointUnions.html#thm-uAssocR-lft-lft} = & \lft(a) \\ & \let{x = \lft(a)} = & x \end{eqnarray*}$ if $x = \rgt(\lft(b))$ , note that $\begin{eqnarray*} & & \compose(\uAssocR)(\uAssocL)(x) \\ & \let{x = \rgt(\lft(b))} = & \compose(\uAssocR)(\uAssocL)(\rgt(\lft(b))) \\ & \href{/posts/arithmetic-made-difficult/Functions.html#def-compose} = & \uAssocR(\uAssocL(\rgt(\lft(b)))) \\ & \href{/posts/arithmetic-made-difficult/DisjointUnions.html#thm-uAssocL-rgt-lft} = & \uAssocR(\lft(\rgt(b))) \\ & \href{/posts/arithmetic-made-difficult/DisjointUnions.html#thm-uAssocR-lft-rgt} = & \rgt(\lft(b)) \\ & \let{x = \rgt(\lft(b))} = & x \end{eqnarray*}$ and if $x = \rgt(\rgt(c))$ , note that $\begin{eqnarray*} & & \compose(\uAssocR)(\uAssocL)(x) \\ & \let{x = \rgt(\rgt(c))} = & \compose(\uAssocR)(\uAssocL)(\rgt(\rgt(c))) \\ & \href{/posts/arithmetic-made-difficult/Functions.html#def-compose} = & \uAssocR(\uAssocL(\rgt(\rgt(c)))) \\ & \href{/posts/arithmetic-made-difficult/DisjointUnions.html#thm-uAssocL-rgt-rgt} = & \uAssocR(\rgt(c)) \\ & \href{/posts/arithmetic-made-difficult/DisjointUnions.html#thm-uAssocR-rgt} = & \rgt(\rgt(c)) \\ & \let{x = \rgt(\rgt(c))} = & x \end{eqnarray*}$ as needed.
If $x \in (A + B) + C$ , we have three possibilities. If $x = \lft(\lft(a))$ , note that $\begin{eqnarray*} & & \compose(\uAssocL)(\uAssocR)(x) \\ & \let{x = \lft(\lft(a))} = & \compose(\uAssocL)(\uAssocR)(\lft(\lft(a))) \\ & \href{/posts/arithmetic-made-difficult/Functions.html#def-compose} = & \uAssocL(\uAssocR(\lft(\lft(a)))) \\ & \href{/posts/arithmetic-made-difficult/DisjointUnions.html#thm-uAssocR-lft-lft} = & \uAssocL(\lft(a)) \\ & \href{/posts/arithmetic-made-difficult/DisjointUnions.html#thm-uAssocL-lft} = & \lft(\lft(a)) \\ & \let{x = \lft(\lft(a))} = & x \end{eqnarray*}$ if $x = \lft(\rgt(b))$ , note that $\begin{eqnarray*} & & \compose(\uAssocL)(\uAssocR)(x) \\ & \let{x = \lft(\rgt(b))} = & \compose(\uAssocL)(\uAssocR)(\lft(\rgt(b))) \\ & \href{/posts/arithmetic-made-difficult/Functions.html#def-compose} = & \uAssocL(\uAssocR(\lft(\rgt(b)))) \\ & \href{/posts/arithmetic-made-difficult/DisjointUnions.html#thm-uAssocR-lft-rgt} = & \uAssocL(\rgt(\lft(b))) \\ & \href{/posts/arithmetic-made-difficult/DisjointUnions.html#thm-uAssocL-rgt-lft} = & \lft(\rgt(b)) \\ & \let{x = \lft(\rgt(b))} = & x \end{eqnarray*}$ and if $x = \rgt(c)$ , note that $\begin{eqnarray*} & & \compose(\uAssocL)(\uAssocR)(x) \\ & \let{x = \rgt(c)} = & \compose(\uAssocL)(\uAssocR)(\rgt(c)) \\ & \href{/posts/arithmetic-made-difficult/Functions.html#def-compose} = & \uAssocL(\uAssocR(\rgt(c))) \\ & \href{/posts/arithmetic-made-difficult/DisjointUnions.html#thm-uAssocR-rgt} = & \uAssocL(\rgt(\rgt(c))) \\ & \href{/posts/arithmetic-made-difficult/DisjointUnions.html#thm-uAssocL-rgt-rgt} = & \rgt(c) \\ & \let{x = \rgt(c)} = & x \end{eqnarray*}$ as needed.

_test_uassocL_lft :: (Equal a, Equal b, Equal c)
  => a -> b -> c -> Test (a -> Bool)
_test_uassocL_lft _ y z =
  testName "uassocL(lft(a)) == lft(lft(a))" $
  \a ->
    let w = (lft y) `withTypeOf` (rgt z)
    in eq
      (uassocL ((lft a) `withTypeOf` (rgt w)))
      (lft (lft a))


_test_uassocL_rgt_lft :: (Equal a, Equal b, Equal c)
  => a -> b -> c -> Test (b -> Bool)
_test_uassocL_rgt_lft x _ z =
  testName "uassocL(rgt(lft(b))) == lft(rgt(b))" $
  \b -> eq
    (uassocL 
      ((rgt ((lft b) `withTypeOf` (rgt z)) `withTypeOf` (lft x))))
    (lft (rgt b))


_test_uassocL_rgt_rgt :: (Equal a, Equal b, Equal c)
  => a -> b -> c -> Test (c -> Bool)
_test_uassocL_rgt_rgt x y _ =
  testName "uassocL(rgt(rgt(c))) == rgt(c)" $
  \c -> eq
    (uassocL
      ((rgt ((rgt c) `withTypeOf` (lft y)) `withTypeOf` (lft x))))
    (rgt c)


_test_uassocR_lft_lft :: (Equal a, Equal b, Equal c)
  => a -> b -> c -> Test (a -> Bool)
_test_uassocR_lft_lft _ y z =
  testName "uassocR(lft(lft(a))) == lft(a)" $
  \a -> eq
    (uassocR
      ((lft ((lft a) `withTypeOf` (rgt y)) `withTypeOf` (rgt z))))
    (lft a)


_test_uassocR_lft_rgt :: (Equal a, Equal b, Equal c)
  => a -> b -> c -> Test (b -> Bool)
_test_uassocR_lft_rgt x _ z =
  testName "uassocR(lft(rgt(b))) == rgt(lft(b))" $
  \b -> eq
    (uassocR
      ((lft ((rgt b) `withTypeOf` (lft x)) `withTypeOf` (rgt z))))
    (rgt (lft b))


_test_uassocR_rgt :: (Equal a, Equal b, Equal c)
  => a -> b -> c -> Test (c -> Bool)
_test_uassocR_rgt x y _ =
  testName "uassocR(rgt(c)) == rgt(rgt(c))" $
  \c ->
    let w = (lft x) `withTypeOf` (rgt y)
    in eq
      (uassocR ((rgt c) `withTypeOf` (lft w)))
      (rgt (rgt c))


_test_uassocR_uassocL :: (Equal a, Equal b, Equal c)
  => a -> b -> c -> Test (Union a (Union b c) -> Bool)
_test_uassocR_uassocL _ _ _ =
  testName "uassocR . uassocL == id" $
  \x -> eq (uassocR (uassocL x)) x


_test_uassocL_uassocR :: (Equal a, Equal b, Equal c)
  => a -> b -> c -> Test (Union (Union a b) c -> Bool)
_test_uassocL_uassocR _ _ _ =
  testName "uassocL . uassocR == id" $
  \x -> eq (uassocL (uassocR x)) x

We also define some helper functions which detect whether an element of $A + B$ is a $\lft$ or a $\rgt$ .

Let $A$ and $B$ be sets. We define $\isLft : A + B \rightarrow \bool$ by $\isLft = \either(\const(\btrue),\const(\bfalse))$ and $\isRgt : A + B \rightarrow \bool$ by $\isRgt = \either(\const(\bfalse),\const(\btrue)).$

In Haskell:

isLft :: Union a b -> Bool
isLft = either (const true) (const false)

isRgt :: Union a b -> Bool
isRgt = either (const false) (const true)

Now $\isLft$ and $\isRgt$ have some nice properties.

Let $A$ and $B$ be sets. Then we have the following for all $a \in A$ and $b \in B$ .

$\isLft(\lft(a)) = \btrue$ .
$\isLft(\rgt(b)) = \bfalse$ .
$\isRgt(\lft(a)) = \bfalse$ .
$\isRgt(\rgt(b)) = \btrue$ .

We have $\begin{eqnarray*} & & \isLft(\lft(a)) \\ & \href{/posts/arithmetic-made-difficult/DisjointUnions.html#dfn-isLft} = & \either(\const(\btrue),\const(\bfalse))(\lft(a)) \\ & \href{/posts/arithmetic-made-difficult/DisjointUnions.html#def-either-lft} = & \const(\btrue)(a) \\ & \href{/posts/arithmetic-made-difficult/Functions.html#def-const} = & \btrue \end{eqnarray*}$ as claimed.
We have $\begin{eqnarray*} & & \isLft(\rgt(b)) \\ & \href{/posts/arithmetic-made-difficult/DisjointUnions.html#dfn-isLft} = & \either(\const(\btrue),\const(\bfalse))(\rgt(b)) \\ & \href{/posts/arithmetic-made-difficult/DisjointUnions.html#def-either-rgt} = & \const(\bfalse)(b) \\ & \href{/posts/arithmetic-made-difficult/Functions.html#def-const} = & \bfalse \end{eqnarray*}$ as claimed.
We have $\begin{eqnarray*} & & \isRgt(\lft(a)) \\ & \href{/posts/arithmetic-made-difficult/DisjointUnions.html#dfn-isRgt} = & \either(\const(\bfalse),\const(\btrue))(\lft(a)) \\ & \href{/posts/arithmetic-made-difficult/DisjointUnions.html#def-either-lft} = & \const(\bfalse)(a) \\ & \href{/posts/arithmetic-made-difficult/Functions.html#def-const} = & \bfalse \end{eqnarray*}$ as claimed.
We have $\begin{eqnarray*} & & \isRgt(\rgt(b)) \\ & \href{/posts/arithmetic-made-difficult/DisjointUnions.html#dfn-isRgt} = & \either(\const(\bfalse),\const(\btrue))(\rgt(b)) \\ & \href{/posts/arithmetic-made-difficult/DisjointUnions.html#def-either-rgt} = & \const(\btrue)(b) \\ & \href{/posts/arithmetic-made-difficult/Functions.html#def-const} = & \btrue \end{eqnarray*}$ as claimed.

_test_isLft_lft :: (Equal a, Equal b)
  => a -> b -> Test (a -> Bool)
_test_isLft_lft _ y =
  testName "isLft(lft(a)) == true" $
  \a -> eq (isLft ((lft a) `withTypeOf` (rgt y))) true


_test_isLft_rgt :: (Equal a, Equal b)
  => a -> b -> Test (b -> Bool)
_test_isLft_rgt x _ =
  testName "isLft(rgt(b)) == false" $
  \b -> eq (isLft ((rgt b) `withTypeOf` (lft x))) false


_test_isRgt_lft :: (Equal a, Equal b)
  => a -> b -> Test (a -> Bool)
_test_isRgt_lft _ y =
  testName "isRgt(lft(a)) == false" $
  \a -> eq (isRgt ((lft a) `withTypeOf` (rgt y))) false


_test_isRgt_rgt :: (Equal a, Equal b)
  => a -> b -> Test (b -> Bool)
_test_isRgt_rgt x _ =
  testName "isRgt(rgt(b)) == true" $
  \b -> eq (isRgt ((rgt b) `withTypeOf` (lft x))) true

Testing

Suite:

_test_disjoint_union
  :: ( Equal a, Show a, Arbitrary a, CoArbitrary a, TypeName a
     , Equal b, Show b, Arbitrary b, CoArbitrary b, TypeName b
     , Equal c, Show c, Arbitrary c, CoArbitrary c, TypeName c
  ) => Int -> Int -> a -> b -> c -> IO ()
_test_disjoint_union size cases a b c = do
  testLabel3 "Disjoint Union" a b c
  let args = testArgs size cases

  runTest args (_test_either_lft_rgt a b)

  runTest args (_test_uswap_lft a b)
  runTest args (_test_uswap_rgt a b)
  runTest args (_test_uswap_uswap a b)

  runTest args (_test_upair_lft a b)
  runTest args (_test_upair_rgt a b)
  runTest args (_test_upair_upair a b)

  runTest args (_test_uassocL_lft a b c)
  runTest args (_test_uassocL_rgt_lft a b c)
  runTest args (_test_uassocL_rgt_rgt a b c)
  runTest args (_test_uassocR_lft_lft a b c)
  runTest args (_test_uassocR_lft_rgt a b c)
  runTest args (_test_uassocR_rgt a b c)
  runTest args (_test_uassocR_uassocL a b c)
  runTest args (_test_uassocL_uassocR a b c)

  runTest args (_test_isLft_lft a b)
  runTest args (_test_isLft_rgt a b)
  runTest args (_test_isRgt_lft a b)
  runTest args (_test_isRgt_rgt a b)

Main:

main_disjoint_union :: IO ()
main_disjoint_union = do
  _test_disjoint_union 20 100 (true :: Bool) (true :: Bool) (true :: Bool)

  let a = lft true :: Union Bool Bool

  _test_functions 20 100 a a a a
  _test_flip      20 100 a a a a a a a
  _test_clone     20 100 a a
  _test_compose   20 100 a a a a a a a