# Using DeMorgan's Theorem to Simplify Logic

concept

So far we have a good handful of techniques for simplifying logic expressions but right now we're going to have a hard time trying to simplify something like
$$\not{\not{A + BC} + \not{A\not{B}}}$$
But in this topic we're going to learn DeMorgan's Theorem which will let us simplify that monster into the very simple \(A\not{B}\).
Which is way simpler.

fact

DeMorgan's Theorem has two equations to it:
$$\not{A+B} = \not{A}\not{B}$$
$$\not{AB} = \not{A}+\not{B}$$

The way we think about this is when there's a bar over either a sum or a times we break the bar, change the operation (sum becomes times, times becomes sum) and then we put a bar over each term

example

Simplify \(\not{\not{AB} + \not{C}}\)

We have a bar over a sum of two terms so we break it, change the sum to a product, and write each term with an additional line over them. \(\not{\not{AB}}\not{\not{C}}\) Now we know that \(\not{\not{X}} = X\) so: \(\not{\not{AB} + \not{C}} = ABC\) If an expression has several bars always remember to only work with one at a time, students trying to skip some working and jump through several steps is a huge source of mistakes.

example

Simplify \(\not{\not{A+BC}+\not{A\not{B}}}\)

Break the top bar first. \(\not{\not{(A + BC)}}\not{\not{A\not{B}}}\) Simplify those double bars. \((A+BC)(A\not{B})\) Expand out the multiplication using the distributive property. \(AA\not{B}+BCA\not{B}\) Now \(AA = A\) in the first term and we know that a term that contains \(B\) and \(\not{B}\) must be 0 so we see our expression simplifies to: \(A\not{B}\)fact

When applying DeMorgan's Theorem to three or more terms at a time simply do to each term what you would normally do when there are two terms

example

Simplify \(\not{ABC}\)

So since \(\not{AB} = \not{A}+\not{B}\) we'll extend that to three terms to get: \(\not{ABC} = \not{A} + \not{B} + \not{C}\)
practice problems