Inequality flips the recipe
Last lesson, equality used cntlzw to spot a zero result. Inequality wants the opposite, a non-zero result, and it needs its own trick to get there. MWCC subtracts the two inputs both ways round, ORs the pair, and then keeps only the top bit of what comes out.
subf r5, r3, r4 # one direction
subf r0, r4, r3 # the other direction
or r0, r5, r0 # combine
srwi r3, r0, 31 # pull bit 31 down to 0/1
blr
Why both directions? Because that pair of subtractions guarantees bit 31 is set the moment the inputs differ. At least one of the two will overflow into that bit. When the inputs match, both differences are zero and so is the OR. The closing srwi r3, r0, 31 then drags bit 31 down to a tidy 0 or 1. There's a corner case that still behaves: if one difference wraps to INT_MIN, the other wraps to INT_MIN as well, the OR stays INT_MIN, bit 31 stays lit, and the answer holds.
Your task
Write not_equal, taking two ints, to reproduce the assembly above.