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#### References to other Issues or PRs

fixes #21269 

#### Brief description of what is fixed or changed


#### Other comments

* to return or to raise

Symbolic Range is tricky because you can't assume that it isn't EmptSet when making queries like `.inf`, e.g. neither `Range(x,y,z).inf` nor Range(x,y,z)[0] can return `x` unless we know that (y - x)*z > 0

And, technically, even `Range(x, x + 2, 1)[0]` should not return `x` unless we know that it is an integer.

- [x] this now raises an error

* the relational form

cf #21285 For better protection, a two more Mods could be added to give: `And(Eq(Mod(x,1),a%s), Eq(Mod(a,1),0), Eq(Mod(s,1),0))` where `a` is a representative point in the Range that is not infinite, `s` is the step and `x` is the relational variable. We should also make sure that there is not a clash between the symbols of the Range and the relational variable as in `Range(x, x + 5, 2).as_relational(x)`.

- [x] this is done but the user can clash symbols if they want to
```python
>>> var('x')
x
>>> Range(x, x + 3, 2).as_relational(x)
(x >= x) & Eq(Mod(x, 1), 0) & (x <= x + 2)
>>> var('x',real=True)
x
>>> Range(x, x + 3, 2).as_relational(x)
Eq(Mod(x, 1), 0)
>>> var('x',integer=1)
x
>>> Range(x, x + 3, 2).as_relational(x)
True

>>> var('x')
x
>>> Range(1, 1+ 2*x, x).as_relational(x)
Eq(Mod(-1, x), 0) & Eq(Mod(x, 1), 0) & (((x >= 1) & (x <= x + 1)) | ((x <= -1) &
 (x <= 1) & (x >= x + 1)))
>>> var('x',real=1)
x
>>> Range(1, 1+ 2*x, x).as_relational(x)
(x >= 1) & Eq(Mod(-1, x), 0) & Eq(Mod(x, 1), 0)
>>> var('x',integer=1)
x
>>> Range(1, 1+ 2*x, x).as_relational(x)
(x >= 1) & Eq(Mod(-1, x), 0)
```
#### Release Notes

<!-- BEGIN RELEASE NOTES -->
* sets
  * Range involving symbols will be made more canonical (correcting a bug in calculation of `sup`) and errors will be raised when trying to compute attributes that are not known due to limitations of assumptions on the arguments.
<!-- END RELEASE NOTES -->