playground
`Q(i+1)=bar{bar S cdot bar Q}= S+Q`
allora se
`bar S = 1 -> S = 0`
quindi
`Q(i+1)=bar{bar S cdot bar Q}= S+Q=0+Q=Q(i)`
playground.txt · Ultima modifica: 2020/07/03 15:58 da 127.0.0.1
`Q(i+1)=bar{bar S cdot bar Q}= S+Q`
allora se
`bar S = 1 -> S = 0`
quindi
`Q(i+1)=bar{bar S cdot bar Q}= S+Q=0+Q=Q(i)`