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There are two addition Terms and Three inputs
Inputs are a,b,c
for b to b’ conversion one NAND gate is required
For adding two numbers 3 NAND Gates are required
Therefore split the term as Y = a + ab’ + ab’c as
Z = b’ —- One NAND Gate
Z1 = ab’ is an and operation . this requires 3 NAND gates of 2 inputs
Z2 = ab’ + ab’c = Z1(1+c) = Z1 since 1 + c in digital is 1
theretofore Y becomes a + Z1 Totally 4 NAND gates of two input is the Minimal requirement
There are other different methods too for implementation of this function
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