If the number is binary as byte1H and byte1L (that is 16-bit number), the subtraction could be performed as:
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MOV A, byte1L
CPL A
ADD A, #1
MOV byte2L, A
MOV A, byte1H
CPL A
ADDC A, #0
MOV byte2H, A
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Now byte2H and byte2L hold 65336-byte1H/L
But, if you like instead that byte2H/L = 65335-byte1H/L
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MOV A, byte1L
CPL A
MOV byte2L, A
MOV A, byte1H
CPL A
MOV byte2H, A
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