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Just to reference the question:
A' = not A
AB = A AND B
A+B = A OR B
Simplify the following expression using Boolean logic:
AB'C + A'B'C' + A'C + BC' + ABC
To start, we start seeing what can be grouped using Boolean identities.
We see the first and last terms, AB'C and ABC, both have AC in the term, and we can use distribution to remove them:
AC(B' + B) + A'B'C' + A'C + BC'
B' OR B is simply TRUE by boolean identities, so we can get rid of that term to give the following:
AC + A'B'C' + A'C + BC'
Now we do the same two steps by recognizing the first and third terms, AC and A'C, both share a C paired with A and A'. We can then get rid of the (A + A') by Boolean identity:
C(A + A') + A'B'C' + BC'
C + A'B'C' + BC'
Now, we can redistribute the C':
C + C'(A'B' + B)
Now we need to take a detour. When you have a situation where you're saying J + J'K, that's the same thing as saying J + K. To make sense of this, think of how the first part says "If J is true or if J is false and K is true, this statement is true." The portion that says "if J is false" is redundant because we wouldn't even look at K if J were true. Another way to look at this is to make a truth table to verify that "J + J'K" and "J + K" are equivalent expressions.
Back to the problem:
C + C'(A'B' + B)
We have the above situation here, where we're looking at C OR C' AND (A'B' + B). We established above that the C' is redundant, and we can now drop it out of the expression:
C + (A'B' + B)
We now have the same situation we're we're looking at B OR B' AND A'. Again, we can now drop the redundant B':
C + (A' + B)
Now we can rearrange and make things look nice because of the commutative and associative properties of Boolean algebra:
A' + B + C
You can easily make a truth table because you only have 3 inputs. When you go through checking each of the 8 cases, you end up finding that there is only one possibility that outputs 0:
You can now use DeMorgan's Law to expand this to the answer we got before:
(AB'C')' = A' + (B')' + (C')' = A' + B + C
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