How does AB(1+C'D) simplify into AB in boolean algebra? I cannot compare their truth tables since literal number of these two expressions is not same. So can anyone explain this to me?
P.S: I found this in solution of a problem of Digital Logic book by Morris Mano and Micheal Ciletti. But since it is a brief solution of a problem, there are no explanations given.
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So you have AB(1+C'D)=AB
1+C'D is equal to 1.
We can make a truth table for 1+C'D
C|D|C'D|1+C'D
1|1|0*1=0|1+0=1
1|0|0*0=0|1+0=1
0|1|1*1=1|1+1=0
0|0|1*0=0|1+0=1
So it seems as we have an issue. If C=F and D=T this statement is not true. So, we can say that this isn't a tautology by counterexample.
In Boolean algebra, 1 represents True and + represents OR.
True OR anything = True.
1 + anything = 1.
1 + C'D = 1.
AB (1 + C'D) = AB.