Tatamibari: an introduction

Question

^{This is an entry for Fortnightly Topic Challenge #44: Introduce a new grid deduction genre to the community}

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This is a Tatamibari puzzle, where you divide the grid into rectangles and squares.

Rules taken from Wikipedia:

• Every partition must contain exactly one symbol in it.
• A + symbol must be contained in a square.
• A | symbol must be contained in a rectangle with a greater height than width.
• A - symbol must be contained in a rectangle with a greater width than height.
• Four pieces may never share the same corner.

An example puzzle and its solution:

Now, solve this puzzle:

Here is the puzzle in a playable form. The link leads to a puzz.link editor (which has a timer, if you care about that).

First answer with a full logical solution path gets the checkmark.

CSV version:

,,,,,,,,,|
,,,,,,,,,
,,,,-,,-,,,
,,-,-,,,|,,,
,,,,,,,,,|
+,,,,,,,,,
,,,+,,,+,+,,
,,,|,,+,,,,
,,,,,,,,,
-,,,,,,,,,

Answer 1

The basic deductions:

Now, we ask:

Which room touches the upper left corner? It turns out only one of the clues can reach there -- the clue in row 3, column 5.

And a similar question is helpful again:

Which room goes in row 4, column 1? Only one clue works there. And what about row 1, column 7? The ─ clue just underneath it must be the one to use that cell -- which requires us to cram the rooms in the top section together.

A similar process lets us fill up the left column:

R9C1 can only be reached by the clue below it, which fixes that as a 2×3 rectangle. Then R8C1 must be part of the square above it, and the left column is complete.

At the same time, we can use reachability on a bunch of cells in the bottom row, and fill up the bottom as well.

And similar arguments work to finish off the puzzle. The solution is: