simplifying expressions - Simplify 2D polyhedral Regions like Interval

I don't think there's a built in "RegionSimplify" for polyhedral Regions and derived akin to how Interval automatically simplifies its argument expression:

Interval[{0, 1}, {1, 2}]

and

Interval[{0, 1}]~IntervalUnion~Interval[{1, 2}]

yield

Interval[{0, 2}]

In contrast, RegionUnion remains unevaluated, even in 2D:

Line[{{0, 0}, {1, 0}}]~ RegionUnion ~ Line[{{1, 0}, {2, 0}}]

(* RegionUnion[Line[{{0, 0}, {1, 0}}], Line[{{1, 0}, {2, 0}}]] *)

From

Rectangle[{0, 0}, {1, 1}] ~ RegionUnion ~ 
  Rectangle[{1, 0}, {2, 1}] // RegionBoundary // DiscretizeRegion

enter image description here

As expected, the Rectangles' common segment is not part of the output.

How to expose or compute the ideal normal form for 2D (at least) polyhedral Regions like these basic examples:

Overlapping lines:

Line[{{0, 0}, {2, 0}}]~ RegionUnion ~ Line[{{1, 0}, {3, 0}}] (* // RegionSimplify *)

(*  Line[{{0,0},{3,0}}] *)

Union of rectangles:

 Rectangle[{0, 0}, {1, 1}] ~ RegionUnion ~ 

      Rectangle[{1, 0}, {2, 1}] (* // RegionSimplify *)

(* Rectangle[{0, 0}, {2, 1}] *)

Boundary:

 Rectangle[{0, 0}, {1, 1}] ~ RegionUnion ~ 
      Rectangle[{1, 0}, {2, 1}] // RegionBoundary  (* // RegionSimplify *)

(* Line[{{0,0},{2,0},{2,1},{0,1},{0,0}}]  *)

The simplification should work for all 2D polyhedral regions including cones (which btw the current language does not support correctly):

ConicHullRegion[{{0, 0}}, {{1, 0}, {1, 1}}] ~RegionUnion ~ 
 ConicHullRegion[{{0, 0}}, {{0, 1}, {1, 1}}] (* // RegionSimplify *)

(* ConicHullRegion[{{0, 0}}, {{1, 0}, {0, 1}}] *)

For convex regions and convex-preserving operations (eg intersection) should suffice to consider the extreme set (vertices and rays), but in general regions aren't convex (and union doens't preserve it).

(Note: RegionIntersection of the same Rectangles hangs the kernel w/ msg: Last::nolast: "{} has a length of zero and no last element")

Comments

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I have several sigmoidal fits to 3 different datasets, with mean fit predictions plus the 95% confidence limits (not symmetrical around the mean) and the actual data. I would now like to show these different 2D plots projected in 3D as in but then using proper perspective. In the link here they give some solutions to combine the plots using isometric perspective, but I would like to use proper 3 point perspective. Any thoughts? Also any way to show the mean points per time point for each series plus or minus the standard error on the mean would be cool too, either using points+vertical bars, or using spheres plus tubes. Below are some test data and the fit function I am using. Note that I am working on a logit(proportion) scale and that the final vertical scale is Log10(percentage). (* some test data *) data = Table[Null, {i, 4}]; data[[1]] = {{1, -5.8}, {2, -5.4}, {3, -0.8}, {4, -0.2}, {5, 4.6}, {1, -6.4}, {2, -5.6}, {3, -0.7}, {4, 0.04}, {5, 1.0}, {1, -6.8}, {2, -4.7}, {3, -1.

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