# Re: constraints in algebra instead of calculus

From: paul c <toledobythesea_at_oohay.ac>
Date: Sun, 17 Jun 2007 17:58:59 GMT
Message-ID: <DDedi.35098\$xq1.18030_at_pd7urf1no>

paul c wrote:
> Jon Heggland wrote:
>

```>> Jan Hidders wrote:
>>
>>> Assume R = { (A:1, B:2), (A:1, B:3) }
>>>
>>> R{B} = { (B:2}, (B:3) }
>>> R1 = R{B} GROUP {B} AS gB = { (gB:{ (B:2} }), (gB:{ (B:3) }) }
>>
>>
>>
>> Isn't that WRAP, not GROUP? According to (my reading of:) TTM, { (B:2),
>> (B:3) } GROUP {B} AS gB = { (gB:{ (B:2), (B:3) }) }.
>>
>>
>>> (R GROUP {B} AS gB) = { (A:1, gB:{ (B:2}, (B:3) }) }
>>> R2 = (R GROUP {B} AS gB){gB} = { (gB:{ (B:2}, (B:3) }) }
>>
>>
>>
>> But this is GROUP, not WRAP ... Is this an error, or are you using some
>> other definition of GROUP than TTM's?
```

>
>
> It's looking to me now that you are right about WRAP and that I too was
> confusing it with GROUP, so maybe the constraint should look like:
>
> R{B} WRAP {B} as gB = R{B} GROUP {B} as gB,
>
> also that as far as TTM/Tutorial D is concerned, Bob B was right about
>
> R{B} GROUP {B} as gB = (R GROUP {B} as gB} {gB} being a tautology.
>
> p (waiting to be corrected for the umpteenth time!)

Oops for the umpteenth time, should have said

R{B} WRAP {B} as gB = (R GROUP {B} as gB) {gB}

heh, umpteen + 1 and counting.

p Received on Sun Jun 17 2007 - 19:58:59 CEST

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