Simplify Product of Sums
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!
$\endgroup$ 51 Answer
$\begingroup$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 $$F(x,y,z)=(x+y)(x+z)(y+z).$$
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