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From: Ben Bacarisse <ben.usenet@bsb.me.uk>
Newsgroups: comp.theory
Subject: Re: Simply defining =?iso-8859-1?Q?G=F6del?= Incompleteness and Tarski Undefinability away V33 (Mendelson Satisfiability)
Date: Tue, 28 Jul 2020 00:22:37 +0100
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olcott <NoOne@NoWhere.com> writes:

> On 7/24/2020 8:28 PM, Ben Bacarisse wrote:
>> olcott <NoOne@NoWhere.com> writes:
>> Do the exercises first.  That way I can be sure you really have a sound
>> intuitive understanding of what the formal stuff is intended to pin
>> down.  It's much easier to understand the next two pages if you are 100%
>> sure about the informal notion of satisfiable.
>>
>
>  ii.(b) A^2_1(x1, x2) ⇒ A^2_1(x2, x1)
>         A(y = z) ⇒ A(z = y) is always true
>         because of the symmetric Property of equality there
>         can never be a time when the LHS is different than the RHS.

So the answer is...?

>  ii.(c) A^2_1(x1, x2) ⇒ A^2_1(x2, x1)
>         A(y ⊆ z) ⇒ A(z ⊆ y)
> only false when (y ⊆ z) and  ¬(z ⊆ y)
> (z ⊃ y) I think that this is the only case where A(y ⊆ z) is true.
> Which only seems to leave y = z as the solution set.

This one in directly analogous to ii.(a) and I think you've made the
same slip.

> In this case it doesn't seem to matter if the demain is integers or
> hamsters.

It certainly would matter.  If the domain were integers or hamsters,
intersection would (I hope) be meaningless.  (I know what you meant to
say -- I'm just pointing out you need to be more careful with your
words.)

> iii.(b) (∀x1)(∀x2)(∀x3) (A^2_1(x1,x2) ∧ A^2_1(x2,x3) ⇒ A^2_1(x1,x3))
> (A(x1 = x2) ∧ A(x2 = x3) ⇒ A(x1 = x3)) is always true.

Yes.  It's true (in this interpretation).

> iii.(c) (∀x1)(∀x2)(∀x3) (A^2_1(x1,x2) ∧ A^2_1(x2,x3) ⇒ A^2_1(x1,x3))
> (A(x1 ⊆ x2) ∧ A(x2 ⊆ x3) ⇒ A(x1 ⊆ x3))
> Is only false when A(x1 ⊃ x3) and A(x1 ⊆ x2) ∧ A(x2 ⊆ x3)
>
> To make is simpler we can think of it as integers instead of sets
> Is only false when A(x1 > x3) and A(x1 <= x2) ∧ A(x2 <= x3)
> This can't ever happen so it is always true.

You can't change the interpretation and be sure to get the right
answer.  It works here, but you'd have to explain why the two
interpretations work in similar ways to use it as your main argument.

-- 
Ben.