Add fixity for (==) and (/=)

[Dannyu NDos]

Given that they are instantized for `Bool`s, they are associative, so it
seems approvable to give them a fixity.

(Sidenote: For the monoid over (==) I suggested on last May, I stated that
it determines if there is an even number of `True`s, but that's wrong. It
determines if there is an odd number of `False`s.)

[David Feuer]

Please write out the associativity proofs!

Post by Dannyu NDos
Given that they are instantized for `Bool`s, they are associative, so it
seems approvable to give them a fixity.
(Sidenote: For the monoid over (==) I suggested on last May, I stated that
it determines if there is an even number of `True`s, but that's wrong. It
determines if there is an odd number of `False`s.)
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[Dannyu NDos]

---------- Forwarded message ---------
From: Dannyu NDos <***@gmail.com>
Date: 2018년 9월 17음 (월) 였후 6:18
Subject: Re: Add fixity for (==) and (/=)
To: <***@gmail.com>


Proof by truth table (F is False, T is True):
p q r (p == q) (q == r) ((p == q) == r) (p == (q == r))
F F F T T F F
F F T T F T T
F T F F F T T
F T T F T F F
T F F F T T T
T F T F F F F
T T F T F F F
T T T T T T T
That proves associativity of (==).

Also note that p /= q ≡ not p == q. Proof:
p q (p /= q) (not p) (not p == q)
F F F T F
F T T T T
T F T F T
T T F F F
And by symmetry of (/=), p /= q ≡ p == not q.

Then:
(p /= q) /= r ≡ (not p == q) == not r ≡ not p == (q == not r) ≡ p /= (q /=
r).
Hence (/=) is associative.

Q.E.D.

[David Feuer]

Looks good to me! Do you have an opinion about infixl vs infixr?

Post by Dannyu NDos
---------- Forwarded message ---------
Date: 2018년 9월 17음 (월) 였후 6:18
Subject: Re: Add fixity for (==) and (/=)
p q r (p == q) (q == r) ((p == q) == r) (p == (q == r))
F F F T T F F
F F T T F T T
F T F F F T T
F T T F T F F
T F F F T T T
T F T F F F F
T T F T F F F
T T T T T T T
That proves associativity of (==).
p q (p /= q) (not p) (not p == q)
F F F T F
F T T T T
T F T F T
T T F F F
And by symmetry of (/=), p /= q ≡ p == not q.
(p /= q) /= r ≡ (not p == q) == not r ≡ not p == (q == not r) ≡ p /= (q /=
r).
Hence (/=) is associative.
Q.E.D.
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[Dannyu NDos]

Well, infixr is friendlier to parsers.

Post by David Feuer
Looks good to me! Do you have an opinion about infixl vs infixr?
Libraries mailing list

Post by Dannyu NDos
http://mail.haskell.org/cgi-bin/mailman/listinfo/libraries

[C Maeder]

Hi,

infixr seeems right for an equivalence (==) since implication is usually
also right associative.

Implication "==>" corresponds to "<=" on Bool, which might be confusing.
Reverse implication (>=) should be left associative, then.

Cheers Christian

Post by Dannyu NDos
Well, infixr is friendlier to parsers.
Looks good to me! Do you have an opinion about infixl vs infixr?
Libraries mailing list
http://mail.haskell.org/cgi-bin/mailman/listinfo/libraries
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[Eric Mertens]

Do we have any other good examples where we've got an operator that is
considered associative where the result type isn't identical to the
argument types? It's much more common to allow the types to vary when
there's only one associativity that makes sense for the way an operator is
intended to be used.

a -> a -> a

While it's true that focusing on Bool, (==) satisfies associativity in its
truth table, the types don't work out so cleanly. Outside of Bool it starts
to matter which associativity you pick.

(\x y z -> (x == y) == z) :: Eq a => a -> a -> Bool -> Bool

(\x y z -> x == (y == z)) :: Eq a => Bool -> a -> a -> Bool

Making == associative is just going to lead to harder to understand code
and will require people to memorize which arbitrary choice about the
associativity of the operation was selected by the mailing list in order to
make sense of the types of code using multiple ==.

I see no gain here, and I'd prefer to leave == as is.

Best regards,
Eric Mertens

Post by C Maeder
Hi,
infixr seeems right for an equivalence (==) since implication is usually
also right associative.
Implication "==>" corresponds to "<=" on Bool, which might be confusing.
Reverse implication (>=) should be left associative, then.
Cheers Christian

Post by Dannyu NDos
Well, infixr is friendlier to parsers.
Looks good to me! Do you have an opinion about infixl vs infixr?
Libraries mailing list
http://mail.haskell.org/cgi-bin/mailman/listinfo/libraries
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[Dannyu NDos]

Well, the motivation to make them associative was that (==) is logical
XNOR, and (/=) is logical XOR. Perhaps we want an alias for
Bool-instantization of them.

Post by Eric Mertens
Do we have any other good examples where we've got an operator that is
considered associative where the result type isn't identical to the
argument types? It's much more common to allow the types to vary when
there's only one associativity that makes sense for the way an operator is
intended to be used.
a -> a -> a
While it's true that focusing on Bool, (==) satisfies associativity in its
truth table, the types don't work out so cleanly. Outside of Bool it starts
to matter which associativity you pick.
(\x y z -> (x == y) == z) :: Eq a => a -> a -> Bool -> Bool
(\x y z -> x == (y == z)) :: Eq a => Bool -> a -> a -> Bool
Making == associative is just going to lead to harder to understand code
and will require people to memorize which arbitrary choice about the
associativity of the operation was selected by the mailing list in order to
make sense of the types of code using multiple ==.
I see no gain here, and I'd prefer to leave == as is.
Best regards,
Eric Mertens

[Eric Mertens]

Well, the motivation to make them associative was that (==) is logical XNOR, and (/=) is logical XOR. Perhaps we want an alias for Bool-instantization of them.

We already have such an binary operation in base that is associative: xor.

import Data.Bits
:i xor

class Eq a => Bits a where
...
xor :: a -> a -> a
...
-- Defined in ‘Data.Bits’

True `xor` False `xor` True

False

The thing that would be missing is an xnor operation in Data.Bits.

[Richard Eisenberg]

+1 to Eric's argument here. I would prefer to leave these fixities out. Of course, defining Bool-specific instantiations with fixities is a fine idea.

Richard

Do we have any other good examples where we've got an operator that is considered associative where the result type isn't identical to the argument types? It's much more common to allow the types to vary when there's only one associativity that makes sense for the way an operator is intended to be used.
a -> a -> a
While it's true that focusing on Bool, (==) satisfies associativity in its truth table, the types don't work out so cleanly. Outside of Bool it starts to matter which associativity you pick.
(\x y z -> (x == y) == z) :: Eq a => a -> a -> Bool -> Bool
(\x y z -> x == (y == z)) :: Eq a => Bool -> a -> a -> Bool
Making == associative is just going to lead to harder to understand code and will require people to memorize which arbitrary choice about the associativity of the operation was selected by the mailing list in order to make sense of the types of code using multiple ==.
I see no gain here, and I'd prefer to leave == as is.
Best regards,
Eric Mertens
Hi,
infixr seeems right for an equivalence (==) since implication is usually
also right associative.
Implication "==>" corresponds to "<=" on Bool, which might be confusing.
Reverse implication (>=) should be left associative, then.
Cheers Christian

Post by Dannyu NDos
Well, infixr is friendlier to parsers.
Looks good to me! Do you have an opinion about infixl vs infixr?
Libraries mailing list
http://mail.haskell.org/cgi-bin/mailman/listinfo/libraries <http://mail.haskell.org/cgi-bin/mailman/listinfo/libraries>
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http://mail.haskell.org/cgi-bin/mailman/listinfo/libraries <http://mail.haskell.org/cgi-bin/mailman/listinfo/libraries>

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[Lennart Augustsson]

The (==) and (/=) operators are non-associative on purpose. Writing
x==y==z is much more likely to be a typo than a legitimate use case. We
decided to make them non-associative, even though they are associative.
(Note that they have fixity already.)

Post by Dannyu NDos
Given that they are instantized for `Bool`s, they are associative, so it
seems approvable to give them a fixity.
(Sidenote: For the monoid over (==) I suggested on last May, I stated that
it determines if there is an even number of `True`s, but that's wrong. It
determines if there is an odd number of `False`s.)
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