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There are a number of identities that we can use to simplify Boolean expressions:

• ̅A = A   (two NOTs cancel out)


• A · 0 = 0   (anything AND 0 is just 0)
• A + 1 = 1   (anything OR 1 is just 1)
• A · 1 = A   (anything AND 1 is itself)
• A + 0 = A   (anything OR 0 is itself)


• A · B = B · A   (order of AND doesn't matter)
• A + B = B + A   (order of OR doesn't matter)
• (A · B) · C = A · (B · C)   (AND is associative)
• (A + B) + C = A + (B + C)   (OR is associative)


• A · A = A   (anything AND itself is itself)
• A + A = A   (anything OR itself is itself)
• A · A = 0   (anything AND its complement is 0)
• A + A = 1   (anything OR its complement is 1)


• A · (B + C) = (A · B) + (A · C)   (AND can be distributed into the brackets, like multiplication)
• A + (B · C) = (A + B) · (A + C)   (OR can also be distributed into the brackets)


• A + (A · B) = A   (we can factorise the LHS to get A · (1 + B), which simplifies to A)
• A · (A + B) = A   (we expand to get (A · A) + (A · B), which is A + (A · B) which we know is just A)


• A · B = A + B   (De Morgan's Law: split the bar and flip the sign)
• A + B = A · B   (De Morgan's Law: split the bar and flip the sign)

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Simplify the Boolean expression A · B̅ + (A · B) + (B · C)

Apply De Morgan's Law to the first term:   A · ̅B = A + ̅B = A + B

Therefore the expression is   A + B + (A · B) + (B · C)


We can rearrange to get   B + (B · C) + A + (A · B)


We know that   B + (B · C) = B

Therefore we have   B + A + (A · B)


We can distribute into the brackets, getting   B + ((A + A) · (A + B))

But   A + A = 1   and   1 · (A + B) = A + B

So now we have   B + A + B


And finally   B + B = 1

Thus we are left with   A + 1 which is simply 1

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