Sometimes it is nice to make code that is straight-line without a extra jumps in it. Here is another way to compare bits 3 & 5 of the accumulator to determine if the two bits match using straight-line code. Clearly not "more efficient" than other examples given this does show how bit level computations are done. This implements the concept given bit a an bit b that:
a xor b = (a and /b) or (/a and b)
....and of course most know that (a xor b) = 0 if (a == b)

org 030h
mov a, 50h ;as example using data at addr 50h
mov c, acc.5
anl c, /acc.3
mov psw.1, c
mov c, acc.3
anl c, /acc.5
orl c, psw.1
jc mismatch ;go to deal with mismatch condition
match:
.... code here for match
....


mismatch:
.... code here for no match
....

Michael Karas