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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