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mathbooleanboolean-logictruthtableboolean-algebra

Simplify boolean equation from truth table

I need help simplifying the following to the simplest terms. Boolean algebra just doesn't quite click with me yet, any help is appreciated. (!A!B!C)+(!AB!C)+(!ABC)+(A!B!C)+(A!BC)+(AB!C) I got it to the following, but I don't know where to go from here: !A(!B!C + B!C + BC) + A(!B!C + B(XOR)C)

If you are curious and want to check my previous work, I got the original equation from the truth table: enter image description here

Solution

  • Initially we have A(~B~C + ~BC + ~CB) + ~A(~B~C + B~C + BC)

    First Term: A(~B~C + ~BC + ~CB)

    = A(~B(~C + C) + ~CB)
= A(~B(True) + ~CB)
= A(~B + ~CB)
= A((~B + ~C)(~B + B))
= A((~B + ~C)(True))
= A(~B + ~C)

    Second Term: ~A(~B~C + B~C + BC)

    = ~A(~C(~B + B) + BC)
= ~A(~C(True) + BC)
= ~A(~C + BC)
= ~A((~C + C) (~C + B))
= ~A((True) (~C + B))
= ~A(~C + B)

    So First Term + Second Term becomes: ~A(~C + B) + A(~B + ~C)

    = ~A~C + ~AB + A~B + A~C
= AxorB + ~A~C + A~C
= AxorB + ~C(~A + A)
= AxorB + ~C(True)
= AxorB + ~C

    Hence we end up with AxorB + ~C