True == 1 weirdness

[Jussi Piitulainen]

Ian Kelly writes:
> On Thu, Sep 17, 2015 at 1:06 AM, Jussi Piitulainen wrote:
>> Ordinary binary operators not only combine things of the same type,
>> they also produce a thing of that same type. So 'in' does not fit
>> among them either.
>>
>> I feel it's _more_ at home among comparison operators. (Hm. That's
>> 'operator' in a different sense.)
>
> Comparison operators *are* binary operators. All that "binary" means
> is that it takes two arguments.

I confused two words. It's operation I meant, not operator. A binary
_operation_ is defined as any map X * X -> X (by Lawvere and Schanuel,
Lawvere and Rosebrugh, at least; let's allow division as close enough).
(The asterisk stands for the Cartesian product. I'm too lazy to look up
the proper Unicode character for it and not willing to see a non-ASCII
symbol come back to me in a mangled form again anyway. Not today.)

Then comparisons are not binary _operations_ except in a very restricted
case. Their types are of the form X * X -> W, where W stands for a set
of truth values, {True, False}. A comparison of truth-values could be
taken as a binary operation, W * W -> W; other comparisons, under this
definition, not.

And I'm saying 'in', being truth-valued, is more like a comparison than
a proper binary operation that has its value in the same set as its two
arguments. Proper binary operations produce results that can be fed back
to them, as in either ((x - y) - z) or (x - (y -z)).

((x < y) < z) or (x < (y < z)) does not usually make sense.

((x in y) in z) more or less fails.

(x in (y in z)) fails.

Just trying to explain what I had in mind when I said that I feel that
'in' is more at home with comparisons (where it is now) than with, hm,
arithmetic operations.