Hi.

Online I can find references to the logical xor operation forming a group
on boolean values or bitstrings (binary vectors), but how about the xnor operation, is that also a group on the set of bits or the set of binary vectors?

https://accu.org/journals/overload/20/109/lewin_1915/

"We have already seen that XOR is associative, that the vector (F, … F) is the identity element and that every element has itself as an inverse. It’s easy to see that it is also closed over the set. Hence (S, XOR) is a group."

https://math.stackexchange.com/questions/2599027/is-there-a-logic-gate-nand-or-etc-which-forms-a-group-under-the-set-0-1

On 12/22/2021 7:42 PM, sobriquet wrote:
> Hi.
>
> Online I can find references to the logical xor operation forming a group

00 0
01 1
10 1
11 0

what kind of group ??

> on boolean values or bitstrings (binary vectors), but how about the xnor operation,

00 1
01 0
10 0
11 1

is that also a group on the set of bits or the set of binary vectors?

what kind of group ??

>
> https://accu.org/journals/overload/20/109/lewin_1915/
>
> "We have already seen that XOR is associative, that the vector (F, … F) is the identity element and that every element has itself as an inverse. It’s easy to see that it is also closed over the set. Hence (S, XOR) is a group."

your stuff is out of context.

>
> https://math.stackexchange.com/questions/2599027/is-there-a-logic-gate-nand-or-etc-which-forms-a-group-under-the-set-0-1

On Thursday, December 23, 2021 at 2:55:04 AM UTC+1, Serg io wrote:
> On 12/22/2021 7:42 PM, sobriquet wrote:
> > Hi.
> >
> > Online I can find references to the logical xor operation forming a group
> 00 0
> 01 1
> 10 1
> 11 0
>
> what kind of group ??

A mathematical group.

https://en.wikipedia.org/wiki/Group_(mathematics)

> > on boolean values or bitstrings (binary vectors), but how about the xnor operation,
> 00 1
> 01 0
> 10 0
> 11 1
> is that also a group on the set of bits or the set of binary vectors?
> what kind of group ??
> >
> > https://accu.org/journals/overload/20/109/lewin_1915/
> >
> > "We have already seen that XOR is associative, that the vector (F, … F) is the identity element and that every element has itself as an inverse. It’s easy to see that it is also closed over the set. Hence (S, XOR) is a group."
> your stuff is out of context.

Out of context? What do you mean?
The webpage argues for the xor operation to be a group on binary vectors. Are you saying
that webpage is wrong about that claim?

My question is simply if their line of reasoning would also hold for the xnor operation.

>
>
> >
> > https://math.stackexchange.com/questions/2599027/is-there-a-logic-gate-nand-or-etc-which-forms-a-group-under-the-set-0-1

On 23/12/2021 01:42, sobriquet wrote:
> Hi.
>
> Online I can find references to the logical xor operation forming a group
> on boolean values or bitstrings (binary vectors), but how about the xnor operation, is that also a group on the set of bits or the set of binary vectors?

Yes. You can check out that the operation is closed, associative, and (1,1,1,...1) is the identity,
and every element has itself as its own inverse. Also, it's commutative, so we have an abelian
group. (Checking is much like checking for XOR.)

Mike.

>
> https://accu.org/journals/overload/20/109/lewin_1915/
>
> "We have already seen that XOR is associative, that the vector (F, … F) is the identity element and that every element has itself as an inverse. It’s easy to see that it is also closed over the set. Hence (S, XOR) is a group."
>
> https://math.stackexchange.com/questions/2599027/is-there-a-logic-gate-nand-or-etc-which-forms-a-group-under-the-set-0-1
>

On Thursday, December 23, 2021 at 4:39:55 AM UTC+1, Mike Terry wrote:
> On 23/12/2021 01:42, sobriquet wrote:
> > Hi.
> >
> > Online I can find references to the logical xor operation forming a group
> > on boolean values or bitstrings (binary vectors), but how about the xnor operation, is that also a group on the set of bits or the set of binary vectors?
> Yes. You can check out that the operation is closed, associative, and (1,1,1,...1) is the identity,
> and every element has itself as its own inverse. Also, it's commutative, so we have an abelian
> group. (Checking is much like checking for XOR.)
>
> Mike.

Ok, I think I would agree on that. But then I'm confused by the claim that there supposedly
is only a single group (up to isomorphism) of order two.
So, if we restrict our attention to bits {0,1}, does that mean that the group
of (xor, {0,1}) is isomorphic to the group (xnor, {0,1})?

https://oeis.org/wiki/Number_of_groups_of_order_n

> >
> > https://accu.org/journals/overload/20/109/lewin_1915/
> >
> > "We have already seen that XOR is associative, that the vector (F, … F) is the identity element and that every element has itself as an inverse. It’s easy to see that it is also closed over the set. Hence (S, XOR) is a group."
> >
> > https://math.stackexchange.com/questions/2599027/is-there-a-logic-gate-nand-or-etc-which-forms-a-group-under-the-set-0-1
> >

On 23/12/2021 03:46, sobriquet wrote:
> On Thursday, December 23, 2021 at 4:39:55 AM UTC+1, Mike Terry wrote:
>> On 23/12/2021 01:42, sobriquet wrote:
>>> Hi.
>>>
>>> Online I can find references to the logical xor operation forming a group
>>> on boolean values or bitstrings (binary vectors), but how about the xnor operation, is that also a group on the set of bits or the set of binary vectors?
>> Yes. You can check out that the operation is closed, associative, and (1,1,1,...1) is the identity,
>> and every element has itself as its own inverse. Also, it's commutative, so we have an abelian
>> group. (Checking is much like checking for XOR.)
>>
>> Mike.
>
> Ok, I think I would agree on that. But then I'm confused by the claim that there supposedly
> is only a single group (up to isomorphism) of order two.
> So, if we restrict our attention to bits {0,1}, does that mean that the group
> of (xor, {0,1}) is isomorphic to the group (xnor, {0,1})?

Exactly - the groups are essentially the same, but with the roles of 0 and 1 reversed.

The isomorphism takes (b0, b1, b2, ...bn) to (b0', b1', b2', ...bn'), where 0' = 1, 1' = 0.
Mike.

>
> https://oeis.org/wiki/Number_of_groups_of_order_n
>
>
>>>
>>> https://accu.org/journals/overload/20/109/lewin_1915/
>>>
>>> "We have already seen that XOR is associative, that the vector (F, … F) is the identity element and that every element has itself as an inverse. It’s easy to see that it is also closed over the set. Hence (S, XOR) is a group."
>>>
>>> https://math.stackexchange.com/questions/2599027/is-there-a-logic-gate-nand-or-etc-which-forms-a-group-under-the-set-0-1
>>>

On 12/22/2021 9:58 PM, Mike Terry wrote:
> On 23/12/2021 03:46, sobriquet wrote:
>> On Thursday, December 23, 2021 at 4:39:55 AM UTC+1, Mike Terry wrote:
>>> On 23/12/2021 01:42, sobriquet wrote:
>>>> Hi.
>>>>
>>>> Online I can find references to the logical xor operation forming a group
>>>> on boolean values or bitstrings (binary vectors), but how about the xnor operation, is that also a group on the set of bits or the set of binary
>>>> vectors?
>>> Yes. You can check out that the operation is closed, associative, and (1,1,1,...1) is the identity,
>>> and every element has itself as its own inverse. Also, it's commutative, so we have an abelian
>>> group. (Checking is much like checking for XOR.)
>>>
>>> Mike.
>>
>> Ok, I think I would agree on that. But then I'm confused by the claim that there supposedly
>> is only a single group (up to isomorphism) of order two.
>> So, if we restrict our attention to bits {0,1}, does that mean that the group
>> of (xor, {0,1}) is isomorphic to the group (xnor, {0,1})?
>
> Exactly - the groups are essentially the same, but with the roles of 0 and 1 reversed.
>
> The isomorphism takes (b0, b1, b2, ...bn) to (b0', b1', b2', ...bn'), where 0' = 1, 1' = 0.
> Mike.

what about AND, or OR ?

>
>>
>> https://oeis.org/wiki/Number_of_groups_of_order_n
>>
>>
>>>>
>>>> https://accu.org/journals/overload/20/109/lewin_1915/
>>>>
>>>> "We have already seen that XOR is associative, that the vector (F, … F) is the identity element and that every element has itself as an inverse.
>>>> It’s easy to see that it is also closed over the set. Hence (S, XOR) is a group."
>>>>
>>>> https://math.stackexchange.com/questions/2599027/is-there-a-logic-gate-nand-or-etc-which-forms-a-group-under-the-set-0-1
>>>>

On Thursday, December 23, 2021 at 5:00:01 AM UTC+1, Mike Terry wrote:
> On 23/12/2021 03:46, sobriquet wrote:
> > On Thursday, December 23, 2021 at 4:39:55 AM UTC+1, Mike Terry wrote:
> >> On 23/12/2021 01:42, sobriquet wrote:
> >>> Hi.
> >>>
> >>> Online I can find references to the logical xor operation forming a group
> >>> on boolean values or bitstrings (binary vectors), but how about the xnor operation, is that also a group on the set of bits or the set of binary vectors?
> >> Yes. You can check out that the operation is closed, associative, and (1,1,1,...1) is the identity,
> >> and every element has itself as its own inverse. Also, it's commutative, so we have an abelian
> >> group. (Checking is much like checking for XOR.)
> >>
> >> Mike.
> >
> > Ok, I think I would agree on that. But then I'm confused by the claim that there supposedly
> > is only a single group (up to isomorphism) of order two.
> > So, if we restrict our attention to bits {0,1}, does that mean that the group
> > of (xor, {0,1}) is isomorphic to the group (xnor, {0,1})?
> Exactly - the groups are essentially the same, but with the roles of 0 and 1 reversed.
>
> The isomorphism takes (b0, b1, b2, ...bn) to (b0', b1', b2', ...bn'), where 0' = 1, 1' = 0.
> Mike.
> >
> > https://oeis.org/wiki/Number_of_groups_of_order_n
> >
> >
> >>>
> >>> https://accu.org/journals/overload/20/109/lewin_1915/
> >>>
> >>> "We have already seen that XOR is associative, that the vector (F, … F) is the identity element and that every element has itself as an inverse. It’s easy to see that it is also closed over the set. Hence (S, XOR) is a group."
> >>>
> >>> https://math.stackexchange.com/questions/2599027/is-there-a-logic-gate-nand-or-etc-which-forms-a-group-under-the-set-0-1
> >>>

I see.. so it would be completely analogous to the abelian group of addition modulo 2 on the set {0,1} being isomorphic to the group of multiplication on the set {-1, 1}? And in fact all four groups being isomorphic.

0 + 0 = 0
0 + 1 = 1
1 + 0 = 1
1 + 1 = 0

-1 * -1 = 1
-1 * 1 = -1
1 * -1 = -1
1 * 1 = 1

On Thursday, December 23, 2021 at 5:26:34 AM UTC+1, Serg io wrote:
> On 12/22/2021 9:58 PM, Mike Terry wrote:
> > On 23/12/2021 03:46, sobriquet wrote:
> >> On Thursday, December 23, 2021 at 4:39:55 AM UTC+1, Mike Terry wrote:
> >>> On 23/12/2021 01:42, sobriquet wrote:
> >>>> Hi.
> >>>>
> >>>> Online I can find references to the logical xor operation forming a group
> >>>> on boolean values or bitstrings (binary vectors), but how about the xnor operation, is that also a group on the set of bits or the set of binary
> >>>> vectors?
> >>> Yes. You can check out that the operation is closed, associative, and (1,1,1,...1) is the identity,
> >>> and every element has itself as its own inverse. Also, it's commutative, so we have an abelian
> >>> group. (Checking is much like checking for XOR.)
> >>>
> >>> Mike.
> >>
> >> Ok, I think I would agree on that. But then I'm confused by the claim that there supposedly
> >> is only a single group (up to isomorphism) of order two.
> >> So, if we restrict our attention to bits {0,1}, does that mean that the group
> >> of (xor, {0,1}) is isomorphic to the group (xnor, {0,1})?
> >
> > Exactly - the groups are essentially the same, but with the roles of 0 and 1 reversed.
> >
> > The isomorphism takes (b0, b1, b2, ...bn) to (b0', b1', b2', ...bn'), where 0' = 1, 1' = 0.
> > Mike.
> what about AND, or OR ?
> >
> >>
> >> https://oeis.org/wiki/Number_of_groups_of_order_n
> >>
> >>
> >>>>
> >>>> https://accu.org/journals/overload/20/109/lewin_1915/
> >>>>
> >>>> "We have already seen that XOR is associative, that the vector (F, … F) is the identity element and that every element has itself as an inverse.
> >>>> It’s easy to see that it is also closed over the set. Hence (S, XOR) is a group."
> >>>>
> >>>> https://math.stackexchange.com/questions/2599027/is-there-a-logic-gate-nand-or-etc-which-forms-a-group-under-the-set-0-1
> >>>>

If you look at the multiplication tables for finite groups:

https://nathancarter.github.io/group-explorer/GroupExplorer.html

You can see that the only binary logical operations that could possibly qualify as a group operation are xor and xnor (given that every row as well as column in the multiplication table has to feature all elements of the set for the binary operation to operate on).

http://web.archive.org/web/20131231075948/http://ibbu.nl/~nsprakel/

On 12/22/2021 10:39 PM, sobriquet wrote:
> On Thursday, December 23, 2021 at 5:26:34 AM UTC+1, Serg io wrote:
>> On 12/22/2021 9:58 PM, Mike Terry wrote:
>>> On 23/12/2021 03:46, sobriquet wrote:
>>>> On Thursday, December 23, 2021 at 4:39:55 AM UTC+1, Mike Terry wrote:
>>>>> On 23/12/2021 01:42, sobriquet wrote:
>>>>>> Hi.
>>>>>>
>>>>>> Online I can find references to the logical xor operation forming a group
>>>>>> on boolean values or bitstrings (binary vectors), but how about the xnor operation, is that also a group on the set of bits or the set of binary
>>>>>> vectors?
>>>>> Yes. You can check out that the operation is closed, associative, and (1,1,1,...1) is the identity,
>>>>> and every element has itself as its own inverse. Also, it's commutative, so we have an abelian
>>>>> group. (Checking is much like checking for XOR.)
>>>>>
>>>>> Mike.
>>>>
>>>> Ok, I think I would agree on that. But then I'm confused by the claim that there supposedly
>>>> is only a single group (up to isomorphism) of order two.
>>>> So, if we restrict our attention to bits {0,1}, does that mean that the group
>>>> of (xor, {0,1}) is isomorphic to the group (xnor, {0,1})?
>>>
>>> Exactly - the groups are essentially the same, but with the roles of 0 and 1 reversed.
>>>
>>> The isomorphism takes (b0, b1, b2, ...bn) to (b0', b1', b2', ...bn'), where 0' = 1, 1' = 0.
>>> Mike.
>> what about AND, or OR ?
>>>
>>>>
>>>> https://oeis.org/wiki/Number_of_groups_of_order_n
>>>>
>>>>
>>>>>>
>>>>>> https://accu.org/journals/overload/20/109/lewin_1915/
>>>>>>
>>>>>> "We have already seen that XOR is associative, that the vector (F, … F) is the identity element and that every element has itself as an inverse.
>>>>>> It’s easy to see that it is also closed over the set. Hence (S, XOR) is a group."
>>>>>>
>>>>>> https://math.stackexchange.com/questions/2599027/is-there-a-logic-gate-nand-or-etc-which-forms-a-group-under-the-set-0-1
>>>>>>
>
> If you look at the multiplication tables for finite groups:
>
> https://nathancarter.github.io/group-explorer/GroupExplorer.html

that is a cool link! thanks! (just have to figure out what it all means...)

>
> You can see that the only binary logical operations that could possibly qualify as a group operation are xor and xnor (given that every row as well as column in the multiplication table has to feature all elements of the set for the binary operation to operate on).
>
> http://web.archive.org/web/20131231075948/http://ibbu.nl/~nsprakel/

excellent links! there are related fields of math having to do with encoding and decoding binary bit streams

On Wednesday, December 22, 2021 at 8:42:43 PM UTC-5, sobriquet wrote:
> Hi.
>
> Online I can find references to the logical xor operation forming a group
> on boolean values or bitstrings (binary vectors), but how about the xnor operation, is that also a group on the set of bits or the set of binary vectors?
>
> https://accu.org/journals/overload/20/109/lewin_1915/
>
> "We have already seen that XOR is associative, that the vector (F, … F) is the identity element and that every element has itself as an inverse. It’s easy to see that it is also closed over the set. Hence (S, XOR) is a group."
>
> https://math.stackexchange.com/questions/2599027/is-there-a-logic-gate-nand-or-etc-which-forms-a-group-under-the-set-0-1

Is it possible that simply running the n-ary form will inform us of the integrity?
Then your notion of 'binary vectors' in plural form does away with the Cartesian operator?
That three bits can XNOR as well as two bits can XNOR? Do the two exist simultaneously to the three? Then the four, and so fourth? And then fifth of course whereupon the junction to the sixth occurs.

On Thursday, December 23, 2021 at 12:20:52 PM UTC-5, Timothy Golden wrote:
> On Wednesday, December 22, 2021 at 8:42:43 PM UTC-5, sobriquet wrote:
> > Hi.
> >
> > Online I can find references to the logical xor operation forming a group
> > on boolean values or bitstrings (binary vectors), but how about the xnor operation, is that also a group on the set of bits or the set of binary vectors?
> >
> > https://accu.org/journals/overload/20/109/lewin_1915/
> >
> > "We have already seen that XOR is associative, that the vector (F, … F) is the identity element and that every element has itself as an inverse. It’s easy to see that it is also closed over the set. Hence (S, XOR) is a group."
> >
> > https://math.stackexchange.com/questions/2599027/is-there-a-logic-gate-nand-or-etc-which-forms-a-group-under-the-set-0-1
> Is it possible that simply running the n-ary form will inform us of the integrity?
> Then your notion of 'binary vectors' in plural form does away with the Cartesian operator?
> That three bits can XNOR as well as two bits can XNOR? Do the two exist simultaneously to the three? Then the four, and so fourth? And then fifth of course whereupon the junction to the sixth occurs.

Ahhh.... So we engage a crux of:
XNOR( a, b, c )
versus
a XNOR b XNOR c
and should we ask which did you mean or was it stated or if unstated can it be that a split exists that lacks the courage to speak?

On Thursday, December 23, 2021 at 1:32:56 PM UTC-5, Timothy Golden wrote:
> On Thursday, December 23, 2021 at 12:20:52 PM UTC-5, Timothy Golden wrote:
> > On Wednesday, December 22, 2021 at 8:42:43 PM UTC-5, sobriquet wrote:
> > > Hi.
> > >
> > > Online I can find references to the logical xor operation forming a group
> > > on boolean values or bitstrings (binary vectors), but how about the xnor operation, is that also a group on the set of bits or the set of binary vectors?
> > >
> > > https://accu.org/journals/overload/20/109/lewin_1915/
> > >
> > > "We have already seen that XOR is associative, that the vector (F, … F) is the identity element and that every element has itself as an inverse. It’s easy to see that it is also closed over the set. Hence (S, XOR) is a group."
> > >
> > > https://math.stackexchange.com/questions/2599027/is-there-a-logic-gate-nand-or-etc-which-forms-a-group-under-the-set-0-1
> > Is it possible that simply running the n-ary form will inform us of the integrity?
> > Then your notion of 'binary vectors' in plural form does away with the Cartesian operator?
> > That three bits can XNOR as well as two bits can XNOR? Do the two exist simultaneously to the three? Then the four, and so fourth? And then fifth of course whereupon the junction to the sixth occurs.
> Ahhh.... So we engage a crux of:
> XNOR( a, b, c )
> versus
> a XNOR b XNOR c
> and should we ask which did you mean or was it stated or if unstated can it be that a split exists that lacks the courage to speak?

And as we engage them in n should they deserve the n-ary awareness or does it speak for itself? In the spirit of the Cartesian product a dimensional result is purveyed yet the mathematician will cast doubt and insist upon a return to the one dimensional interpretation of the result.

For the sum certainly this behavior is appropriate, but for the initial assumption of the Cartesian product. Here I like the latter and not the former and there (the product) clearly the latter and the former as coherent but nat, not the latter as a singular dimension anymore. This would be known to man as the dimensional product I suppose in the ordinary sense, yet even there the sense is in two; not in one; and so humanity as off by one and stuck in the rut commences.

I'd like to declare that we are engaged in a progression. And yet stuck in a rut is their state. Good-bye to you and hello to your new form. Us and them and I oh I, owe noone anything. Nothing gained; Nothing lost; Giving it away here for free. For all to see. Anything else would be criminality. Pure form takes none other than free or freely or easily or effortlessly; so much so that to stoop down and brush ones boot against a branch did put such a spring in ones sole that he jumped from the pole and passed yards beyond comprehensible. Some love a tale and I'm not generally one to do it. But here I do feel capable or at least out of a good place do try to make some interesting content while I promote a substantial branch of information theory. Well, it is pure mathematics. The overlaps are tremendous. The leadins are stupendous. Breaking away from the two and yonder on to the three one feels empowered. In the four some ordinary things but a kind of break in reality. At five I was quite bright and well behaved but for the influence of an older brother goading me on. No it's not true. I was a fighter. I was not a bully though, but maybe once or twice, well so some is probably forgotten, too. Here I bark and I bight.

On 12/23/2021 12:32 PM, Timothy Golden wrote:
> On Thursday, December 23, 2021 at 12:20:52 PM UTC-5, Timothy Golden wrote:
>> On Wednesday, December 22, 2021 at 8:42:43 PM UTC-5, sobriquet wrote:
>>> Hi.
>>>
>>> Online I can find references to the logical xor operation forming a group
>>> on boolean values or bitstrings (binary vectors), but how about the xnor operation, is that also a group on the set of bits or the set of binary vectors?
>>>
>>> https://accu.org/journals/overload/20/109/lewin_1915/
>>>
>>> "We have already seen that XOR is associative, that the vector (F, … F) is the identity element and that every element has itself as an inverse. It’s easy to see that it is also closed over the set. Hence (S, XOR) is a group."
>>>
>>> https://math.stackexchange.com/questions/2599027/is-there-a-logic-gate-nand-or-etc-which-forms-a-group-under-the-set-0-1
>> Is it possible that simply running the n-ary form will inform us of the integrity?
>> Then your notion of 'binary vectors' in plural form does away with the Cartesian operator?
>> That three bits can XNOR as well as two bits can XNOR? Do the two exist simultaneously to the three? Then the four, and so fourth? And then fifth of course whereupon the junction to the sixth occurs.
>
> Ahhh.... So we engage a crux of:
> XNOR( a, b, c )
> versus
> a XNOR b XNOR c

use the proper notion, you dweebe.

On Thursday, December 23, 2021 at 6:20:52 PM UTC+1, timba...@gmail.com wrote:
> On Wednesday, December 22, 2021 at 8:42:43 PM UTC-5, sobriquet wrote:
> > Hi.
> >
> > Online I can find references to the logical xor operation forming a group
> > on boolean values or bitstrings (binary vectors), but how about the xnor operation, is that also a group on the set of bits or the set of binary vectors?
> >
> > https://accu.org/journals/overload/20/109/lewin_1915/
> >
> > "We have already seen that XOR is associative, that the vector (F, … F) is the identity element and that every element has itself as an inverse. It’s easy to see that it is also closed over the set. Hence (S, XOR) is a group."
> >
> > https://math.stackexchange.com/questions/2599027/is-there-a-logic-gate-nand-or-etc-which-forms-a-group-under-the-set-0-1
> Is it possible that simply running the n-ary form will inform us of the integrity?
> Then your notion of 'binary vectors' in plural form does away with the Cartesian operator?
> That three bits can XNOR as well as two bits can XNOR? Do the two exist simultaneously to the three? Then the four, and so fourth? And then fifth of course whereupon the junction to the sixth occurs.

The concept of a group simply involves both a set and a binary operation that maps
combinations of two elements from that set to single elements of that set.

https://www.youtube.com/playlist?list=PLwV-9DG53NDxU337smpTwm6sef4x-SCLv

But the use of the word binary is confusing in this context. Since when
we're speaking of a binary operation, it means the operation operates on a combination
of two things.
So multiplication as a binary operation on a set like the complex numbers is a binary
operation in the sense that it takes a combination of two complex numbers as
an input and yields a single complex number as an output, even though complex
numbers themselves are not binary.
But when speaking about binary vectors, we're simply talking about bitstrings,
so linear sequences of 2-valued elements.

On Thursday, December 23, 2021 at 4:45:50 PM UTC-5, sobriquet wrote:
> On Thursday, December 23, 2021 at 6:20:52 PM UTC+1, timba...@gmail.com wrote:
> > On Wednesday, December 22, 2021 at 8:42:43 PM UTC-5, sobriquet wrote:
> > > Hi.
> > >
> > > Online I can find references to the logical xor operation forming a group
> > > on boolean values or bitstrings (binary vectors), but how about the xnor operation, is that also a group on the set of bits or the set of binary vectors?
> > >
> > > https://accu.org/journals/overload/20/109/lewin_1915/
> > >
> > > "We have already seen that XOR is associative, that the vector (F, … F) is the identity element and that every element has itself as an inverse. It’s easy to see that it is also closed over the set. Hence (S, XOR) is a group."
> > >
> > > https://math.stackexchange.com/questions/2599027/is-there-a-logic-gate-nand-or-etc-which-forms-a-group-under-the-set-0-1
> > Is it possible that simply running the n-ary form will inform us of the integrity?
> > Then your notion of 'binary vectors' in plural form does away with the Cartesian operator?
> > That three bits can XNOR as well as two bits can XNOR? Do the two exist simultaneously to the three? Then the four, and so fourth? And then fifth of course whereupon the junction to the sixth occurs.
> The concept of a group simply involves both a set and a binary operation that maps
> combinations of two elements from that set to single elements of that set..
>
> https://www.youtube.com/playlist?list=PLwV-9DG53NDxU337smpTwm6sef4x-SCLv
>
> But the use of the word binary is confusing in this context. Since when
> we're speaking of a binary operation, it means the operation operates on a combination
> of two things.
> So multiplication as a binary operation on a set like the complex numbers is a binary
> operation in the sense that it takes a combination of two complex numbers as
> an input and yields a single complex number as an output, even though complex
> numbers themselves are not binary.
> But when speaking about binary vectors, we're simply talking about bitstrings,
> so linear sequences of 2-valued elements.

elementally speaking this seems like a wealthier element than the reals. Sort of like the digits keep flying at you and no final value is of interest.

On Thursday, December 23, 2021 at 10:52:57 PM UTC+1, timba...@gmail.com wrote:
> On Thursday, December 23, 2021 at 4:45:50 PM UTC-5, sobriquet wrote:
> > On Thursday, December 23, 2021 at 6:20:52 PM UTC+1, timba...@gmail.com wrote:
> > > On Wednesday, December 22, 2021 at 8:42:43 PM UTC-5, sobriquet wrote:
> > > > Hi.
> > > >
> > > > Online I can find references to the logical xor operation forming a group
> > > > on boolean values or bitstrings (binary vectors), but how about the xnor operation, is that also a group on the set of bits or the set of binary vectors?
> > > >
> > > > https://accu.org/journals/overload/20/109/lewin_1915/
> > > >
> > > > "We have already seen that XOR is associative, that the vector (F, … F) is the identity element and that every element has itself as an inverse. It’s easy to see that it is also closed over the set. Hence (S, XOR) is a group."
> > > >
> > > > https://math.stackexchange.com/questions/2599027/is-there-a-logic-gate-nand-or-etc-which-forms-a-group-under-the-set-0-1
> > > Is it possible that simply running the n-ary form will inform us of the integrity?
> > > Then your notion of 'binary vectors' in plural form does away with the Cartesian operator?
> > > That three bits can XNOR as well as two bits can XNOR? Do the two exist simultaneously to the three? Then the four, and so fourth? And then fifth of course whereupon the junction to the sixth occurs.
> > The concept of a group simply involves both a set and a binary operation that maps
> > combinations of two elements from that set to single elements of that set.
> >
> > https://www.youtube.com/playlist?list=PLwV-9DG53NDxU337smpTwm6sef4x-SCLv
> >
> > But the use of the word binary is confusing in this context. Since when
> > we're speaking of a binary operation, it means the operation operates on a combination
> > of two things.
> > So multiplication as a binary operation on a set like the complex numbers is a binary
> > operation in the sense that it takes a combination of two complex numbers as
> > an input and yields a single complex number as an output, even though complex
> > numbers themselves are not binary.
> > But when speaking about binary vectors, we're simply talking about bitstrings,
> > so linear sequences of 2-valued elements.
> elementally speaking this seems like a wealthier element than the reals. Sort of like the digits keep flying at you and no final value is of interest.

In practice there is little difference between a rational number that is so big that even expressing a tiny fraction of it would require way more than the total amount of matter/energy in the observable universe and a real number.
So the distinction between the infinite and the finite is moot for finite quantities that are inconceivably large.

On Thursday, December 23, 2021 at 4:52:57 PM UTC-5, Timothy Golden wrote:
> On Thursday, December 23, 2021 at 4:45:50 PM UTC-5, sobriquet wrote:
> > On Thursday, December 23, 2021 at 6:20:52 PM UTC+1, timba...@gmail.com wrote:
> > > On Wednesday, December 22, 2021 at 8:42:43 PM UTC-5, sobriquet wrote:
> > > > Hi.
> > > >
> > > > Online I can find references to the logical xor operation forming a group
> > > > on boolean values or bitstrings (binary vectors), but how about the xnor operation, is that also a group on the set of bits or the set of binary vectors?
> > > >
> > > > https://accu.org/journals/overload/20/109/lewin_1915/
> > > >
> > > > "We have already seen that XOR is associative, that the vector (F, … F) is the identity element and that every element has itself as an inverse. It’s easy to see that it is also closed over the set. Hence (S, XOR) is a group."
> > > >
> > > > https://math.stackexchange.com/questions/2599027/is-there-a-logic-gate-nand-or-etc-which-forms-a-group-under-the-set-0-1
> > > Is it possible that simply running the n-ary form will inform us of the integrity?
> > > Then your notion of 'binary vectors' in plural form does away with the Cartesian operator?
> > > That three bits can XNOR as well as two bits can XNOR? Do the two exist simultaneously to the three? Then the four, and so fourth? And then fifth of course whereupon the junction to the sixth occurs.
> > The concept of a group simply involves both a set and a binary operation that maps
> > combinations of two elements from that set to single elements of that set.
> >
> > https://www.youtube.com/playlist?list=PLwV-9DG53NDxU337smpTwm6sef4x-SCLv
> >
> > But the use of the word binary is confusing in this context. Since when
> > we're speaking of a binary operation, it means the operation operates on a combination
> > of two things.
> > So multiplication as a binary operation on a set like the complex numbers is a binary
> > operation in the sense that it takes a combination of two complex numbers as
> > an input and yields a single complex number as an output, even though complex
> > numbers themselves are not binary.
> > But when speaking about binary vectors, we're simply talking about bitstrings,
> > so linear sequences of 2-valued elements.
> elementally speaking this seems like a wealthier element than the reals. Sort of like the digits keep flying at you and no final value is of interest.

If XNOR is general then it would operate on a triple, right? Is this then a triple stream or three uniquely defined streams? No guarantee all are same length. There is ambiguity here in tying your concept to xnor isn't there?

As for me I am caught wondering if the mod-3 xnor is well defined?
A: 0 1 0 1
B: 0 0 1 1
op: 1 0 0 1 = A xnor B ( read as a table vertically )

A: 01010101
B: 00110011
C: 00001111
o1:11101001 = ( A xnor B ) xnor C
o2:01101001 = A xnor( B xnor C )
o3:10000001 = xnor( A, B, C )

I don't think there are any others. Here as I complain of the Cartesian product I see the possibility space, but I do not see the results as in that possibility space, at least not for the sum. For the product I do, and this puts them at odds with each other. An inverse XNOR operation does not have predictable results does it? Call this IXNOR:
o3: 1 0 0 1 = A IXNOR B
A: 0 1 0 1
Bd: 0 0 1 1 = (notA AND notB) 0R (B AND A)
Perfectly fine.

Now for A IXNOR B IXNOR C and it so naturally wants to flow as:
( notA AND notB AND notC ) OR ( A AND B AND C )
and all is well in the land of operator theory but for the binary confusion within operator theory.

A IXNOR B IXNOR C = C IXNOR B IXNOR A = B IXNOR C IXNOR A = ...

As to who was fundamental and which was the inverse of the other: to some this is a matter of confusion. Indeed the lack of teaching of IXNOR exposes a gap. Your group lays with IXNOR.

On Thursday, December 23, 2021 at 2:49:59 PM UTC-8, sobriquet wrote:

> In practice there is little difference between a rational number that is so big that even expressing a tiny fraction of it would require way more than the total amount of matter/energy in the observable universe and a real number.
> So the distinction between the infinite and the finite is moot for finite quantities that are inconceivably large.

[T]he distinction between finite and infinite is not as relevant as the distinction between realistic and unrealistic.
-- Donald Knuth

https://www.science.org/doi/10.1126/science.194.4271.1235

On 23/12/2021 04:32, sobriquet wrote:
> On Thursday, December 23, 2021 at 5:00:01 AM UTC+1, Mike Terry wrote:
>> On 23/12/2021 03:46, sobriquet wrote:
>>> On Thursday, December 23, 2021 at 4:39:55 AM UTC+1, Mike Terry wrote:
>>>> On 23/12/2021 01:42, sobriquet wrote:
>>>>> Hi.
>>>>>
>>>>> Online I can find references to the logical xor operation forming a group
>>>>> on boolean values or bitstrings (binary vectors), but how about the xnor operation, is that also a group on the set of bits or the set of binary vectors?
>>>> Yes. You can check out that the operation is closed, associative, and (1,1,1,...1) is the identity,
>>>> and every element has itself as its own inverse. Also, it's commutative, so we have an abelian
>>>> group. (Checking is much like checking for XOR.)
>>>>
>>>> Mike.
>>>
>>> Ok, I think I would agree on that. But then I'm confused by the claim that there supposedly
>>> is only a single group (up to isomorphism) of order two.
>>> So, if we restrict our attention to bits {0,1}, does that mean that the group
>>> of (xor, {0,1}) is isomorphic to the group (xnor, {0,1})?
>> Exactly - the groups are essentially the same, but with the roles of 0 and 1 reversed.
>>
>> The isomorphism takes (b0, b1, b2, ...bn) to (b0', b1', b2', ...bn'), where 0' = 1, 1' = 0.
>> Mike.
>>>
>>> https://oeis.org/wiki/Number_of_groups_of_order_n
>>>
>>>
>>>>>
>>>>> https://accu.org/journals/overload/20/109/lewin_1915/
>>>>>
>>>>> "We have already seen that XOR is associative, that the vector (F, … F) is the identity element and that every element has itself as an inverse. It’s easy to see that it is also closed over the set. Hence (S, XOR) is a group."
>>>>>
>>>>> https://math.stackexchange.com/questions/2599027/is-there-a-logic-gate-nand-or-etc-which-forms-a-group-under-the-set-0-1
>>>>>
>
> I see.. so it would be completely analogous to the abelian group of addition modulo 2 on the set {0,1} being isomorphic to the group of multiplication on the set {-1, 1}? And in fact all four groups being isomorphic.
>
> 0 + 0 = 0
> 0 + 1 = 1
> 1 + 0 = 1
> 1 + 1 = 0
>
> -1 * -1 = 1
> -1 * 1 = -1
> 1 * -1 = -1
> 1 * 1 = 1
>

Yes. Sometimes our knowledge of particular examples (operations like addition/multiplication/other
combined with the meanings we already know for particular elements) can get in the way of seeing the
underlying structure. In the case of XOR, XNOR perhaps the way to be convinced they are
"structurally" the same is to (1) write out the operation in a table, and (2) the further rewrite
those tables using a neutral a,b for the elements, with a representing the identity (and b the other
element of course):

XOR:
0 1
--+------
0 | 0 1
1 | 1 0

using a=identity=0 , b=1:

a b
--+------
a | a b
b | b a

XNOR:
0 1
--+------
0 | 1 0
1 | 0 1

using a=identity=1 , b=0:

b a
--+------
b | a b
a | b a

(same as for XOR, but just in a different order)

Mike.

sobriquet expressed precisely :
> On Thursday, December 23, 2021 at 6:20:52 PM UTC+1, timba...@gmail.com wrote:
>> On Wednesday, December 22, 2021 at 8:42:43 PM UTC-5, sobriquet wrote:
>>> Hi.
>>>
>>> Online I can find references to the logical xor operation forming a group
>>> on boolean values or bitstrings (binary vectors), but how about the xnor
>>> operation, is that also a group on the set of bits or the set of binary
>>> vectors?
>>>
>>> https://accu.org/journals/overload/20/109/lewin_1915/
>>>
>>> "We have already seen that XOR is associative, that the vector (F, … F) is
>>> the identity element and that every element has itself as an inverse. It’s
>>> easy to see that it is also closed over the set. Hence (S, XOR) is a
>>> group."
>>>
>>> https://math.stackexchange.com/questions/2599027/is-there-a-logic-gate-nand-or-etc-which-forms-a-group-under-the-set-0-1
>> Is it possible that simply running the n-ary form will inform us of the
>> integrity? Then your notion of 'binary vectors' in plural form does away
>> with the Cartesian operator? That three bits can XNOR as well as two bits
>> can XNOR? Do the two exist simultaneously to the three? Then the four, and
>> so fourth? And then fifth of course whereupon the junction to the sixth
>> occurs.
>
> The concept of a group simply involves both a set and a binary operation that
> maps combinations of two elements from that set to single elements of that
> set.

It takes more than that to form a group.

On Thursday, December 23, 2021 at 6:18:39 PM UTC-5, Timothy Golden wrote:
> On Thursday, December 23, 2021 at 4:52:57 PM UTC-5, Timothy Golden wrote:
> > On Thursday, December 23, 2021 at 4:45:50 PM UTC-5, sobriquet wrote:
> > > On Thursday, December 23, 2021 at 6:20:52 PM UTC+1, timba...@gmail.com wrote:
> > > > On Wednesday, December 22, 2021 at 8:42:43 PM UTC-5, sobriquet wrote:
> > > > > Hi.
> > > > >
> > > > > Online I can find references to the logical xor operation forming a group
> > > > > on boolean values or bitstrings (binary vectors), but how about the xnor operation, is that also a group on the set of bits or the set of binary vectors?
> > > > >
> > > > > https://accu.org/journals/overload/20/109/lewin_1915/
> > > > >
> > > > > "We have already seen that XOR is associative, that the vector (F, … F) is the identity element and that every element has itself as an inverse. It’s easy to see that it is also closed over the set. Hence (S, XOR) is a group."
> > > > >
> > > > > https://math.stackexchange.com/questions/2599027/is-there-a-logic-gate-nand-or-etc-which-forms-a-group-under-the-set-0-1
> > > > Is it possible that simply running the n-ary form will inform us of the integrity?
> > > > Then your notion of 'binary vectors' in plural form does away with the Cartesian operator?
> > > > That three bits can XNOR as well as two bits can XNOR? Do the two exist simultaneously to the three? Then the four, and so fourth? And then fifth of course whereupon the junction to the sixth occurs.
> > > The concept of a group simply involves both a set and a binary operation that maps
> > > combinations of two elements from that set to single elements of that set.
> > >
> > > https://www.youtube.com/playlist?list=PLwV-9DG53NDxU337smpTwm6sef4x-SCLv
> > >
> > > But the use of the word binary is confusing in this context. Since when
> > > we're speaking of a binary operation, it means the operation operates on a combination
> > > of two things.
> > > So multiplication as a binary operation on a set like the complex numbers is a binary
> > > operation in the sense that it takes a combination of two complex numbers as
> > > an input and yields a single complex number as an output, even though complex
> > > numbers themselves are not binary.
> > > But when speaking about binary vectors, we're simply talking about bitstrings,
> > > so linear sequences of 2-valued elements.
> > elementally speaking this seems like a wealthier element than the reals.. Sort of like the digits keep flying at you and no final value is of interest.
> If XNOR is general then it would operate on a triple, right? Is this then a triple stream or three uniquely defined streams? No guarantee all are same length. There is ambiguity here in tying your concept to xnor isn't there?
>
> As for me I am caught wondering if the mod-3 xnor is well defined?
> A: 0 1 0 1
> B: 0 0 1 1
> op: 1 0 0 1 = A xnor B ( read as a table vertically )
>
> A: 01010101
> B: 00110011
> C: 00001111
> o1:01101001 = ( A xnor B ) xnor C

uh-oh I got this wrong
> o2:01101001 = A xnor( B xnor C )
> o3:10000001 = xnor( A, B, C )
>
> I don't think there are any others. Here as I complain of the Cartesian product I see the possibility space, but I do not see the results as in that possibility space, at least not for the sum. For the product I do, and this puts them at odds with each other. An inverse XNOR operation does not have predictable results does it? Call this IXNOR:
IXNOR: = (notA AND notB) 0R (B AND A)
> Perfectly fine.
>
> Now for A IXNOR B IXNOR C and it so naturally wants to flow as:
> ( notA AND notB AND notC ) OR ( A AND B AND C )
> and all is well in the land of operator theory but for the binary confusion within operator theory.
>
> A IXNOR B IXNOR C = C IXNOR B IXNOR A = B IXNOR C IXNOR A = ...
>
> As to who was fundamental and which was the inverse of the other: to some this is a matter of confusion. Indeed the lack of teaching of IXNOR exposes a gap. Your group lays with IXNOR.

And we see that :
( A ixnor B ) ixnor C = A ixnor ( B ixnor C ) = ixnor( A, B, C )

I guess it is true that not( ixnor ) = xor

On Friday, December 24, 2021 at 1:18:54 AM UTC+1, FromTheRafters wrote:
> sobriquet expressed precisely :
> > On Thursday, December 23, 2021 at 6:20:52 PM UTC+1, timba...@gmail.com wrote:
> >> On Wednesday, December 22, 2021 at 8:42:43 PM UTC-5, sobriquet wrote:
> >>> Hi.
> >>>
> >>> Online I can find references to the logical xor operation forming a group
> >>> on boolean values or bitstrings (binary vectors), but how about the xnor
> >>> operation, is that also a group on the set of bits or the set of binary
> >>> vectors?
> >>>
> >>> https://accu.org/journals/overload/20/109/lewin_1915/
> >>>
> >>> "We have already seen that XOR is associative, that the vector (F, … F) is
> >>> the identity element and that every element has itself as an inverse. It’s
> >>> easy to see that it is also closed over the set. Hence (S, XOR) is a
> >>> group."
> >>>
> >>> https://math.stackexchange.com/questions/2599027/is-there-a-logic-gate-nand-or-etc-which-forms-a-group-under-the-set-0-1
> >> Is it possible that simply running the n-ary form will inform us of the
> >> integrity? Then your notion of 'binary vectors' in plural form does away
> >> with the Cartesian operator? That three bits can XNOR as well as two bits
> >> can XNOR? Do the two exist simultaneously to the three? Then the four, and
> >> so fourth? And then fifth of course whereupon the junction to the sixth
> >> occurs.
> >
> > The concept of a group simply involves both a set and a binary operation that
> > maps combinations of two elements from that set to single elements of that
> > set.
> It takes more than that to form a group.

Yeah, they have to satisfy some conditions (the group axioms), but those are the
basic structures that constitute the group.

On Friday, December 24, 2021 at 1:38:43 AM UTC+1, timba...@gmail.com wrote:
> On Thursday, December 23, 2021 at 6:18:39 PM UTC-5, Timothy Golden wrote:
> > On Thursday, December 23, 2021 at 4:52:57 PM UTC-5, Timothy Golden wrote:
> > > On Thursday, December 23, 2021 at 4:45:50 PM UTC-5, sobriquet wrote:
> > > > On Thursday, December 23, 2021 at 6:20:52 PM UTC+1, timba...@gmail.com wrote:
> > > > > On Wednesday, December 22, 2021 at 8:42:43 PM UTC-5, sobriquet wrote:
> > > > > > Hi.
> > > > > >
> > > > > > Online I can find references to the logical xor operation forming a group
> > > > > > on boolean values or bitstrings (binary vectors), but how about the xnor operation, is that also a group on the set of bits or the set of binary vectors?
> > > > > >
> > > > > > https://accu.org/journals/overload/20/109/lewin_1915/
> > > > > >
> > > > > > "We have already seen that XOR is associative, that the vector (F, … F) is the identity element and that every element has itself as an inverse. It’s easy to see that it is also closed over the set.. Hence (S, XOR) is a group."
> > > > > >
> > > > > > https://math.stackexchange.com/questions/2599027/is-there-a-logic-gate-nand-or-etc-which-forms-a-group-under-the-set-0-1
> > > > > Is it possible that simply running the n-ary form will inform us of the integrity?
> > > > > Then your notion of 'binary vectors' in plural form does away with the Cartesian operator?
> > > > > That three bits can XNOR as well as two bits can XNOR? Do the two exist simultaneously to the three? Then the four, and so fourth? And then fifth of course whereupon the junction to the sixth occurs.
> > > > The concept of a group simply involves both a set and a binary operation that maps
> > > > combinations of two elements from that set to single elements of that set.
> > > >
> > > > https://www.youtube.com/playlist?list=PLwV-9DG53NDxU337smpTwm6sef4x-SCLv
> > > >
> > > > But the use of the word binary is confusing in this context. Since when
> > > > we're speaking of a binary operation, it means the operation operates on a combination
> > > > of two things.
> > > > So multiplication as a binary operation on a set like the complex numbers is a binary
> > > > operation in the sense that it takes a combination of two complex numbers as
> > > > an input and yields a single complex number as an output, even though complex
> > > > numbers themselves are not binary.
> > > > But when speaking about binary vectors, we're simply talking about bitstrings,
> > > > so linear sequences of 2-valued elements.
> > > elementally speaking this seems like a wealthier element than the reals. Sort of like the digits keep flying at you and no final value is of interest.
> > If XNOR is general then it would operate on a triple, right? Is this then a triple stream or three uniquely defined streams? No guarantee all are same length. There is ambiguity here in tying your concept to xnor isn't there?
> >
> > As for me I am caught wondering if the mod-3 xnor is well defined?
> > A: 0 1 0 1
> > B: 0 0 1 1
> > op: 1 0 0 1 = A xnor B ( read as a table vertically )
> >
> > A: 01010101
> > B: 00110011
> > C: 00001111
> > o1:01101001 = ( A xnor B ) xnor C
>
> uh-oh I got this wrong
> > o2:01101001 = A xnor( B xnor C )
> > o3:10000001 = xnor( A, B, C )
> >
> > I don't think there are any others. Here as I complain of the Cartesian product I see the possibility space, but I do not see the results as in that possibility space, at least not for the sum. For the product I do, and this puts them at odds with each other. An inverse XNOR operation does not have predictable results does it? Call this IXNOR:
> IXNOR: = (notA AND notB) 0R (B AND A)
> > Perfectly fine.
> >
> > Now for A IXNOR B IXNOR C and it so naturally wants to flow as:
> > ( notA AND notB AND notC ) OR ( A AND B AND C )
> > and all is well in the land of operator theory but for the binary confusion within operator theory.
> >
> > A IXNOR B IXNOR C = C IXNOR B IXNOR A = B IXNOR C IXNOR A = ...
> >
> > As to who was fundamental and which was the inverse of the other: to some this is a matter of confusion. Indeed the lack of teaching of IXNOR exposes a gap. Your group lays with IXNOR.
> And we see that :
> ( A ixnor B ) ixnor C = A ixnor ( B ixnor C ) = ixnor( A, B, C )
>
> I guess it is true that not( ixnor ) = xor

Logically, xnor is like logical equivalence, while xor is like logical unequivalence.
so I would think of xnor(a,b,c) to be indeed true if a,b,c all have the same truthvalue.
xor(a,b,c) would be problematic, since there are only two truth values, so a, b and
c can't all have a different truthvalue, but I guess we can still retain the concept of
exclusive or in the sense that at least one and at most one of the variables is true.

The bitstrings should all have the same length, but perhaps the definition of how
logical operators act on bitstrings of equal length can be extended so that they
can also work on bitstrings of unequal length.

I think it's interesting to see how these logical spaces can be embedded recursively.

Like we might associate numbers between 0 and 1 with bitstrings, such that
for instance, the variables are always associated with the same number, regardless of
the context involved (the number of variables).

For instance, we could have 0=false and 1=true for the case there are no variables (bitstrings
of length 1).
If we now move to a context where we have a variable x (bitstrings of length 2), we can
associate not(x) with 1/3, x with 2/3 and false=0/3 and true is 3/3.
In a subsequent step, where we have variables x and y (bitstrings of length 4), we
can associate all possible cases with the numbers 0/15, 1/15, ..., 14/15 and 15/15,
where x is still 2/3, but associated with the number 10/15.
So the idea is that logical concepts, like the variable x or the expression (x implies y)
retain their associated numbers regardless of the context in which they are considered.
As opposed to having different bitstrings associated with the same logical expression,
depending on the context of the total number of available variables.

On Friday, December 24, 2021 at 12:45:52 AM UTC+1, Mike Terry wrote:
> On 23/12/2021 04:32, sobriquet wrote:
> > On Thursday, December 23, 2021 at 5:00:01 AM UTC+1, Mike Terry wrote:
> >> On 23/12/2021 03:46, sobriquet wrote:
> >>> On Thursday, December 23, 2021 at 4:39:55 AM UTC+1, Mike Terry wrote:
> >>>> On 23/12/2021 01:42, sobriquet wrote:
> >>>>> Hi.
> >>>>>
> >>>>> Online I can find references to the logical xor operation forming a group
> >>>>> on boolean values or bitstrings (binary vectors), but how about the xnor operation, is that also a group on the set of bits or the set of binary vectors?
> >>>> Yes. You can check out that the operation is closed, associative, and (1,1,1,...1) is the identity,
> >>>> and every element has itself as its own inverse. Also, it's commutative, so we have an abelian
> >>>> group. (Checking is much like checking for XOR.)
> >>>>
> >>>> Mike.
> >>>
> >>> Ok, I think I would agree on that. But then I'm confused by the claim that there supposedly
> >>> is only a single group (up to isomorphism) of order two.
> >>> So, if we restrict our attention to bits {0,1}, does that mean that the group
> >>> of (xor, {0,1}) is isomorphic to the group (xnor, {0,1})?
> >> Exactly - the groups are essentially the same, but with the roles of 0 and 1 reversed.
> >>
> >> The isomorphism takes (b0, b1, b2, ...bn) to (b0', b1', b2', ...bn'), where 0' = 1, 1' = 0.
> >> Mike.
> >>>
> >>> https://oeis.org/wiki/Number_of_groups_of_order_n
> >>>
> >>>
> >>>>>
> >>>>> https://accu.org/journals/overload/20/109/lewin_1915/
> >>>>>
> >>>>> "We have already seen that XOR is associative, that the vector (F, … F) is the identity element and that every element has itself as an inverse. It’s easy to see that it is also closed over the set. Hence (S, XOR) is a group."
> >>>>>
> >>>>> https://math.stackexchange.com/questions/2599027/is-there-a-logic-gate-nand-or-etc-which-forms-a-group-under-the-set-0-1
> >>>>>
> >
> > I see.. so it would be completely analogous to the abelian group of addition modulo 2 on the set {0,1} being isomorphic to the group of multiplication on the set {-1, 1}? And in fact all four groups being isomorphic.
> >
> > 0 + 0 = 0
> > 0 + 1 = 1
> > 1 + 0 = 1
> > 1 + 1 = 0
> >
> > -1 * -1 = 1
> > -1 * 1 = -1
> > 1 * -1 = -1
> > 1 * 1 = 1
> >
> Yes. Sometimes our knowledge of particular examples (operations like addition/multiplication/other
> combined with the meanings we already know for particular elements) can get in the way of seeing the
> underlying structure. In the case of XOR, XNOR perhaps the way to be convinced they are
> "structurally" the same is to (1) write out the operation in a table, and (2) the further rewrite
> those tables using a neutral a,b for the elements, with a representing the identity (and b the other
> element of course):
>
> XOR:
> 0 1
> --+------
> 0 | 0 1
> 1 | 1 0
>
> using a=identity=0 , b=1:
>
> a b
> --+------
> a | a b
> b | b a
>
>
> XNOR:
> 0 1
> --+------
> 0 | 1 0
> 1 | 0 1
>
> using a=identity=1 , b=0:
>
> b a
> --+------
> b | a b
> a | b a
>
> (same as for XOR, but just in a different order)
>
> Mike.

In general group theory is conceptually very confusing, for instance in the way
operations that act on the set of elements are themselves elements of that set.
So there is no clear distinction like you have in traditional arithmetic or algebra where
expressions like 4 + 6 or (8 - a) * 2b make sense because there is a clear distinction
between the numbers/constants/variables that operations act on and the operations
that act on those numbers/constants/variables.
But it wouldn't make sense to have expressions like (* + /) - -, where
you add multiplication and division and subsequently subtract subtraction.

xor 00 01 10 11
00 00 01 10 11
01 01 00 11 10
10 10 11 00 01
11 11 10 01 00

xnor 00 01 10 11
00 11 10 10 00
01 10 11 00 10
10 01 00 11 10
11 00 01 10 11

a b c d
a a b c d
b b a d c
c c d a b
d d c b a

I dunno.. somehow I can't really see how they are both isomorphic to that abstract
structure. It seems that if you map (a,b,c,d) to (00,01,10,11), you get xor, but
if you map (a,b,c,d) to (11,10,01,00) you don't seem to get xnor.

???? 11 10 01 00
11 11 10 10 00
10 10 11 00 10
01 01 00 11 10
00 00 01 10 11