Similar question to:
Boolean Algebra – Product of Sums
I was given a truth table and asked to give the sums-of-products and the product-of-sums expressions.
I reduced the sums-of-products expression to this, which I believe is correct:
$$F(x,y,z) = xy + yz + xz$$
I have so far reduced my product-of-sums expression to this:
$$F(x,y,z) = (x+y+z)(x+y+z’)(x+y’+z)(x’+y+z)$$
but I can't figure out how to further reduce my product-of-sums, whilst still retaining the "product-of-sums" form and not converting back to "sums-of-products" form. If someone could show me how to further reduce the product-of-sums, I would be appreciative if you could explain which Boolean identity is being used at each stage of of the simplification (i.e. Distributive Law, DeMorgan's Law etc). Thanks!
Best Answer
Note that $(x+y+z)(x+y+z')=x+y$ etc.
Enter twice the dummy term $(x+y+z)$ into $F(x,y,z)=(x+y+z)(x+y+z')(x+y'+z)(x'+y+z).$ Then $$F(x,y,z)=\underbrace{(x+y+z)(x+y+z')}\underbrace{\color{blue}{(x+y+z)}(x+y'+z)}\underbrace{\color{blue}{(x+y+z)}(x'+y+z)}$$ so $$F(x,y,z)=(x+y)(x+z)(y+z).$$