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`Q(i+1)=bar{bar S cdot bar Q}= S+Q`

allora se

`bar S = 1 → S = 0`


`Q(i+1)=bar{bar S cdot bar Q}= S+Q=0+Q=Q(i)`

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playground.txt · Ultima modifica: 2020/07/03 17:58 (modifica esterna)