Reevaluate TXA implementation

[JonnyWalker]

`The size of the destination register (i.e. the register transferred to) determines whether these instructions are 8-bit operations or 16-bit operations. When the destination register is 8 bits wide, 8 bits are transferred, and when the destination register is 16 bits wide, 16 bits are transferred.

The width of the accumulator is based on the m flag, and the width of the X and Y registers is based on the x flag, but the S register is always considered 16 bits wide. However, when the e flag is 1, SH is forced to $01, so in effect, TXS is an 8-bit transfer in this case since XL is transferred to SL and SH remains $01. Note that when the e flag is 0 and the x flag is 1 (i.e. 8-bit native mode), that XH is forced to zero, so after a TXS, SH will be $00, rather than $01. This is an important difference that must be accounted for if you want to run emulation mode code in (8-bit) native mode.`

Maybe add an assmebly file to check if the implementation is correct