Wrong number of $4\times 4$ domino tilings, but why?

Question

From the internet, I know that a domino tiling of a $4\times4 $ checker board can be arranged in $36$ different ways.

With the following reasoning, I conclude that it must be $37$, which is one more than the correct result. Where is my mistake?

Tiling a $4\times 2$ or $2\times 4$ board can be done in $5$ ways, where

-- is a horizontal domino and

| is a vertical one:
|

||    ||    --    --    --
||    ||    --    --    ||
||    --    ||    --    ||
||    --    ||    --    --

Now, when you look at the $4\times 4$ board, you can find

$25$ tilings by arranging two $4\times 2$ tilings $x$ and $y$ side by side

xxyy
xxyy
xxyy
xxyy

$5$ tilings by combining a structure which is not present in the previous $25$ tilings with the $5$ different $2\times 4$ tilings (horizontal versions of the above $4\times 2$ tilings):

|--|
|--|
xxxx
xxxx

$5$ tiling by putting the previous $5$ upside down

xxxx
xxxx
|--|
|--|

$2$ tiling of this kind:

|--|             ----
|||| and rotated |--|
||||             |--|
|--|             ----

$25 + 5 + 5 + 2 = 37$.

Answer 1

The doubled one is the following, which is present in both sets of 5 tilings:

|--|
|--|
|--|
|--|