If M ever empties its stack, then 
when simulating M will empty its stack except the symbol X at the bottom. At this point, will enter its final state 
by using transition rule 2, thereby (correctly) accepting the input. We will prove that M and are equivalent.
when simulating M will empty its stack except the symbol X at the bottom. At this point, will enter its final state by using transition rule 2, thereby (correctly) accepting the input. We will prove that M and are equivalent.
Let M accepts 
. Then
for some 
. But then
( by transition rule 1)
( Since includes all the moves of M )
( by transition rule 2 )
Hence, also accepts .