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From: Mitchell Hockley on 2 Jun 2010 06:02
Hi,

Could anyone help me with this question (from Bostock's Intermediate
Logic)

"The theory of groups can be presented as having in its vocabulary
just identity and a single two-place function f(x,y) which we write as
'x.y'. The usual laws for identity apply, and in addition these three
axioms:"

(A1) Axyz(x.(y.z) = (x.y).z)

(A2) AxyEz(x = z.y)

(A3) AxyEz(x = y.z)

Prove:

a = a.c |= c = c.c

(Hint: Use Ez(c = z.a)

Thanks for any help,
Mitch.
From: Frederick Williams on 2 Jun 2010 08:58
Mitchell Hockley wrote:
>
> Hi,
>
> Could anyone help me with this question (from Bostock's Intermediate
> Logic)
>
> "The theory of groups can be presented as having in its vocabulary
> just identity and a single two-place function f(x,y) which we write as
> 'x.y'. The usual laws for identity apply, and in addition these three
> axioms:"
>
> (A1) Axyz(x.(y.z) = (x.y).z)
>
> (A2) AxyEz(x = z.y)
>
> (A3) AxyEz(x = y.z)
>
> Prove:
>
> a = a.c |= c = c.c

Hint: that means that every model of a = a.c is a model of c = c.c.

> (Hint: Use Ez(c = z.a))

Eh? What if a = 0 and c =/= 0?


--
I can't go on, I'll go on.
From: Arturo Magidin on 2 Jun 2010 15:31
On Jun 2, 7:58 am, Frederick Williams <frederick.willia...(a)tesco.net>
wrote:
> Mitchell Hockley wrote:
>
> > Hi,
>
> > Could anyone help me with this question (from Bostock's Intermediate
> > Logic)
>
> > "The theory of groups can be presented as having in its vocabulary
> > just identity and a single two-place function f(x,y) which we write as
> > 'x.y'. The usual laws for identity apply, and in addition these three
> > axioms:"
>
> > (A1) Axyz(x.(y.z) = (x.y).z)
>
> > (A2) AxyEz(x = z.y)
>
> > (A3) AxyEz(x = y.z)
>
> > Prove:
>
> > a = a.c   |=   c = c.c
>
> Hint: that means that every model of a = a.c is a model of c = c.c.
>
> > (Hint: Use Ez(c = z.a))
>
> Eh?  What if a = 0 and c =/= 0?

I think you are confused; when we use "a=0" in a group, the operation
is usually taken to be addition, not multiplication. In a group, if
you have an element a such that Ay (ay=ya=a) (i.e., a "zero" element
in the sense of semigroups), then you have that the group consists of
only one element.

The Hint refers to (A2) in which the x has been particularized to
equal a.

To the OP: suppose you have a structure satisfying (A1)-(A3), and in
addition that Eac(a.c=a). You want to show that in that case, c.c=c.
By A2, you have that Ez(c=z.a). So in c.c, you can replace the first c
by z.a. Do that and see what happens.

--
Arturo Magidin
From: George Greene on 2 Jun 2010 17:53
On Jun 2, 8:58 am, Frederick Williams <frederick.willia...(a)tesco.net>
wrote:
>
> Eh?  What if a = 0 and c =/= 0?

It doesn't matter what a is, or what c is.
You are being asked to prove a universal generalization.
This is going to hold for ALL a and ALL c.
From: George Greene on 2 Jun 2010 17:58
On Jun 2, 3:31 pm, Arturo Magidin <magi...(a)member.ams.org> wrote:
> To the OP: suppose you have a structure satisfying (A1)-(A3), and in
> addition that Eac(a.c=a)

This is sort of a misuse of the existential quantifier.
You are actually trying to prove that the implication holds
for ALL a and ALL c.

The original problem is arguably mis-phrased (in the book).
What you ACTUALLY want to prove is NOT
> a = a.c |= c = c.c
BUT RATHER
Aac[ a=a.c --> c=c.c ]

The fact that the original is stated withOUT quantifiers really
is problematic unless the book has a convention regarding
how they ought to be put back in.
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