On 27 May 2012, at 08:20, Bill Richter wrote:
> John, I'm over 400 lines now formalizing my Hilbert axiom paper
> http://www.math.northwestern.edu/~richter/hilbert.pdf (code below) and
> proved that being on the same side of a line defines an equivalence
> relation. This is really a lot of fun, and my son likes it too: the
> formalized mathematical proofs become more "concrete".
>
> I know you're real busy, but I have three questions. My first 6 lemmas
> here are all set-theory, and I wonder if I'm going about it the right
> way. Also, I couldn't figure out how to define an equivalence
> relation. It seems that there ought to be a way to define function
> ER: point#point->bool ---> bool
> (or something) testing if a relation is an equivalence relation.
HOL effectively lets you define the class of all equivalence relations. Thus:
let refl_def = new_definition
`Refl R = !x:'a. R x x`;;
let symm_def = new_definition
`Symm R = !x y:'a. R x y ==> R y x`;;
let trans_def = new_definition
`Trans R = !x y z:'a. R x y /\ R y z ==> R x z`;;
let equiv_rel_def = new_definition
`EquivRel R = (Refl R /\ Symm R /\ Trans R)`;;
(Here the fact that HOL is polymorphic is giving you much more expressivity in
practice than raw ZF set theory would give.)
After the above definitions, type_of`EquivRel` will evaluate to something like
`:(?83981->?83981->bool)->bool` where ?83981 is a type variable that can be
instantiated to any other type. In your case you can write `equiv_rel (\A B.
(A, B) same_side l)` to assert that the relationship of being on the same side
of the line l is an equivalence relation. However, you are going to need to do
something a bit trickier, since your axioms seem to be loose about whether `(A,
B) same_side l` holds if A or B is on the line l. So you will either need to
strengthen your axioms so that `(\A B. (A, B) same_side l)`becomes an
equivalence relation on the set of all points or generalise the above
definitions to the notion of an equivalence relation on a subset:
let refl_on_def = new_definition
`ReflOn R P = !x:'a. P x ==> R x x`;;
let symm_on_def = new_definition
`SymmOn R P = !x y:'a. P x /\ P y /\ R x y ==> R y x`;;
let trans_on_def = new_definition
`TransOn R P = !x y z:'a. P x /\ P y /\ P z /\ R x y /\ R y z ==> R x z`;;
let equiv_rel_on_def = new_definition
`EquivRelOn R P = (ReflOn R P /\ SymmOn R P /\ TransOn R P)`;;
So you can write `EquivRelOn (\A B. (A, B) same_side l) l`. (Given a free
choice, I would strengthen the axioms as it will simplify the proofs.)
> Below you can see my partly successful effort. Also, my axiom B3 is
> bulkier than I'd like. I want to say that exactly one of three things
> is true, so I wrote
> a \/ b \/ c /\ ~(a /\ b) /\ ~(a /\ c) /\ ~(b /\ c).
> Is there a nice HOL Light way to say this?
>
Well you could define a function on lists of truth values and use that. Thus:
let all_false_def = new_recursive_definition list_RECURSION
` (AllFalse [] = T)
/\ (AllFalse (CONS b bs) = (~b /\ AllFalse bs)) `;;
let one_true_def = new_recursive_definition list_RECURSION
` (OneTrue [] = F)
/\ (OneTrue (CONS b bs) = ((b /\ AllFalse bs) \/ (~b /\ OneTrue bs))) `;;
And then you can say things like `OneTrue [A = B; B = C; A = C]`. But I
suspect this won't make life any simpler for you because to reason with this
notation, you will probably just end up rewriting with the definitions as in:
REWRITE_CONV[one_true_def; all_false_def] `OneTrue [A = B; B = C; A = C]`;;
which returns:
|- OneTrue [A = B; B = C; A = C] <=>
A = B /\ ~(B = C) /\ ~(A = C) \/
~(A = B) /\ (B = C /\ ~(A = C) \/ ~(B = C) /\ A = C)
and so doesn't help with reducing the number of cases. John's "Without Loss Of
Generality" paper may give you some ideas
(http://www.cl.cam.ac.uk/~jrh13/papers/wlog.html) if case explosion is a big
problem for you, but I don't know how well that fits in with miz3.
Regards,
Rob.
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