The first “fact” cannot be proven because it is false! Given the definition:

```
containsS :: Shape -> Coordinate -> Bool
(Rectangle s1 s2) `containsS` (x,y)
= let t1 = s1 / 2
t2 = s2 / 2
in -t1 <= x && x <= t1 && -t2 <= y && y <= t2
(Ellipse r1 r2) `containsS` (x,y)
= (x/r1)^2 + (y/r2)^2 <= 1
```

It is not true that:

Rectangle s1 s2 `containsS` p = Rectangle (-s1) s2 `containsS` p

This can be easily seen e.g. when `s1 = 4`, `s2 = 2`, `p = (1,1)`. However, it's true that:

rectangle s1 s2 `containsS` p = rectangle (-s1) s2 `containsS` p

Given our updated definition of ``containsS`` for polygons from Exercise 8.4. To prove the second fact is trivial:

Ellipse (-r1) r2 `containsS` (x,y) = (x/(-r1))^{2}+ (y/r2)^{2}<= 1 = (x^{2}/(-r1)^{2}) + (y/r2)^{2}<= 1 = (x^{2}/r1^{2}) + (y/r2)^{2}<= 1 = (x/r1)^{2}+ (y/r2)^{2}<= 1 = Ellipse r1 r2 `containsS` (x,y)

## One response to “Exercise 8.10”

IMHO

Rectangle s1 s2 `containsS` p = Rectangle (-s1) s2 `containsS` p

is not true only because you didn’t updated definition of Rectangle `containsS`p for negative sidelengths. I didn’t updated it as well because we were not asked to do it but I think we should. If a side length x is e.g. 10 then p should meet

-5<=px && px<=5

but if x = – 10then p should meet

5<=px && px<=-5

which has no sense.