DeMorgan's Laws Explained with Simple Animations

Welcome to DeMorgan's Laws! Today we'll explore the first law, which shows us how to distribute NOT over AND.Let's start with a Venn diagram showing two sets: A and B.First, let's look at A AND B. This is the intersection of the two sets, shown here in blue.When we apply NOT to A AND B, we get everything EXCEPT the intersection. This is shown in red.Now, let's look at NOT A. This is everything outside circle A.And NOT B is everything outside circle B.When we combine NOT A OR NOT B, we get the same region as NOT of A AND B!Let's break down why this works. Everything except the intersection is the same as being outside at least one of the sets.This is DeMorgan's first law in action: NOT of A AND B equals NOT A OR NOT B.Now that we understand the first law, we're ready to explore the second law in our next section.Let's examine how NOT(A OR B) works in our Venn diagram.First, let's identify A OR B by highlighting the region covered by either circle.When we apply NOT to this region, we get everything outside both circles.This is exactly the same as taking NOT A AND NOT B.Let's see this step by step. First, NOT A is everything outside circle A.And NOT B is everything outside circle B.When we take the AND of these regions - meaning where they overlap - we get exactly the same region as NOT(A OR B).This demonstrates DeMorgan's Law: NOT of A OR B equals NOT A AND NOT B.Now that we understand NOT(A OR B), let's look at some practical applications.Let's see how DeMorgan's Laws apply to real-world situations, starting with a light bulb.Consider the statement: The light is NOT both ON AND working.This is equivalent to saying: The light is either OFF OR broken.Let's see how DeMorgan's Laws apply in digital circuits.NOT of AND can be replaced with OR of NOTs, simplifying our circuit design.Let's try some practice problems. Here's the first one:Using DeMorgan's Laws, we distribute the NOT across the AND, changing it to OR.Let's try a more challenging problem:For this nested expression, we apply DeMorgan's Laws twice to get our solution.