Why 1 + 0' = 1'

[bvssvni]

The reason 1 + 0' = 1':

1 + 2' = 1 + 0 = 1 = 3'

1 + n' equals the nth minus 1 prime in base 0, because 2' is the first prime.

So, 1 + 1' = 2'.

Notice that 1 is 3', so 3' + 1' = 2'.

This introduces the somewhat counter-intuitive idea that addition in Algexenotation can result in a smaller original number than any of the original arguments.

One can also write 1 + 1' = 0. Notice the similarity to 1 + (-1) = 0.

Other identities:

0 + 2' = 0    trivial, since `0 + 0 = 0`
0 + 1' = 1'   non-trivial, but fits with `1 + 1' = 2'`
0 + 0' = 0'   non-trivial, but fits with `0 + x = x` for all `x`

Now, think about these as corners of a square:

0 + 0' = 0'         1 + 0' = ?
0 + 1' = 1'         1 + 1' = 2'

It seems natural that ? is larger than 0' but smaller than 2'.
There is only one number with this property, which is 1'.