Hi Jon,

One thought which may help you:

I would start from the end, resolving your last statement using a ‘mpjaodan’, 
‘pm2.61dane’, or, if you use induction, ‘pm2.61i’.

This will create new sub goals for your different cases, which you can then 
proceed to prove.

BR,
_
Thierry

> Le 2 juin 2019 à 16:46, Jon P <[email protected]> a écrit :
> 
> Thanks for all help, area quad is now included in set.mm so that is pretty 
> nice :)
> 
> If I have a chance I am thinking of these as next steps, if anyone is 
> interested and has thoughts that would be appreciated. 
> 
> First take any 3 points in R^2 and make a triangle. I think if you order the 
> points by x coord there are 5 cases (as RR is well ordered). Here is an image 
> of them.  Computing the area of Case 1 (all x coords the same) is trivial 
> (though needs to be covered), cases 2 and 3 (two x coords the same) are 
> covered by area quad directly, which is nice. Cases 4 and 5 are a bit 
> trickier. 
> 
> The approach I am thinking of is to split the domain at the point Q e. RR. 
> This will make the triangle in to two triangles which are case 2 and 3 so are 
> computable. 
> 
> I started down this road but I am not sure about the direction, this general 
> result is what I was thinking of trying to prove first. 
> 
> *hypothesis
> 54::  |- A e. dom area 
> 55::  |- Q e. RR
> 56::  |- B = { <. x , y >. | ( <. x , y >. e. A /\ x <_ Q ) }
> 57::  |- C = { <. x , y >. | ( <. x , y >. e. A /\ Q <_ x ) }
> *resutls
> 600:: |- B e. dom area
> 601:: |- C e. dom area
> qed:: ( area ` A ) = ( area ` B ) + ( area ` C )
> 
> Here is how far I have got, it is not so easy. I was then going to try to 
> write an "area of a triangle" theorem which covers all cases. I do not have 
> so much energy to work on it but hopefully I can do something :)
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