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Workshop 4 – Digital Logic(Check worksheet as the questions on here don’t show the correct not values)
(i) 𝐴 + 𝐴𝐵
Use absorption law so it becomes A
(ii)𝐴𝐵 + 𝐴𝐵̅
Use inverse law get A(B+B_INVERSE) =A(1) = A TIMES 1 = A
(iii) 𝐴𝐵 + 𝐴(𝐶𝐷 + 𝐶𝐷̅)
We know from last question AB + AB = A ,so CD +CD must be C, so AB+A(C)
So we get A(B+C) as we factor out A
(iv) (𝐵𝐶̅+ 𝐴̅𝐷)(𝐴𝐵̅ + 𝐶𝐷̅)
(BC_ + A_D)(AB_ + CD_) = ABB_C_ + A_CDD_ + AA_B_D + BCC_D_ cancels to get 0
(v)𝐴(𝐴 + 𝐶)̅̅ (whole equation is not)
Absorption law so its just A_
(vi) ̅𝐴̅+̅𝐵𝐶̅̅
The whole equation is not but BC has another not, so not not BC is just goes to normal so
new equation for VI is A_ + BC.
(vii) 𝐴̅𝐵𝐶 + 𝐴𝐶
Factor by C to get C(A_B+A) then not A and A cancel each other due to inverse law so we
get just C(B) so answer is CB
(viii) 𝐴(𝐵 + 𝐴)(𝐶 + 𝐵)𝐵̅
Use absorption law to get A(C+B)B_ use inverse law to get AB_C
(ix) 𝐴𝐵 + 𝐵𝐶̅+ 𝐴𝐶
(x) 𝐴(𝐵 + 𝐴(𝐵 + 𝐶𝐵̅))
idempotent law = AB+AA(B+C!B)
Absorption law =AB+A(B+C!B)
Factorize = A(B+C!B)
Inverse of b = A+AC0
Idempotent = A+AC
AC
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