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The boolean function I want to convert is: F = xy' + yz'

I will first convert it to canonical form:

F = xy' + yz' ------> (1)

= xy'(z + z') + (x + x')yz'

= xy'z + xy'z' + xyz' + x'yz'

= Σ(2, 4, 5, 6) = π(0, 1, 3, 7)

Now, F' = π(2, 4, 5, 6) (Is this statement correct, why not?)

= (x + y' + z)(x' + y + z)(x' + y + z')(x' + y' + z) ------> (2)

I thought I had converted the function to POS form, but when I checked the truth tables for (1) and (2), they are the complete opposite. Truth table for 1 Truth table for 2

Why is this incorrect? According to this answer, I'll have to complement the input function twice to convert to POS form. But this seems very counter-intuitive because if I complement a function twice, will I not get back the original function?

Very Professional Username
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    This explains it far more thoroughly. https://electronics.stackexchange.com/questions/9817/how-to-convert-an-expression-from-sop-to-pos-and-back-in-boolean-algebra – nickalh Mar 10 '23 at 01:40

1 Answers1

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No, you don't simply complement twice.

You complement once, then use DeMorgan's theorem and the distributive property to get a SOP form for \$\overline{F}\$. Complement again, and use DeMorgan's theorem again to finally get the POS form.

In a sense, you do have the original equation back again because the result is logically equivalent to the original equation.

Elliot Alderson
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    Clarification- you do ultimately complement twice, but not just complement twice, also use each complement to apply DeMorgan's law first then again. – nickalh Mar 09 '23 at 19:05