Refer to Exercise 2.4 . Use the identities A=A ∩S and S=B ∪B and a distributive law to prove tha

step 1 Today we want to go over probability laws. We are given that A equals the intersection between A

Refer to Exercise 2.4 . Use the identities A=A ∩S and S=B ∪B...

Refer to Exercise $2.4 .$ Use the identities $A=A \cap S$ and $S=B \cup B$ and a distributive law to prove that a. $A=(A \cap B) \cup(A \cap \bar{B})$ b. If $B \subset A$ then $A=B \cup(A \cap \bar{B})$ c. Further, show that $(A \cap B)$ and $(A \cap \bar{B})$ are mutually exclusive and therefore that $A$ is the union of two mutually exclusive sets, $(A \cap B)$ and $(A \cap \bar{B})$ d. Also show that $B$ and $(A \cap B)$ are mutually exclusive and if $B \subset A, A$ is the union of two mutually exclusive sets, $B$ and $(A \cap \bar{B})$.

Refer to Exercise $2.4 .$ Use the identities $A=A \cap S$ and $S=B \cup B$ and a distributive law to prove that
a. $A=(A \cap B) \cup(A \cap \bar{B})$
b. If $B \subset A$ then $A=B \cup(A \cap \bar{B})$
c. Further, show that $(A \cap B)$ and $(A \cap \bar{B})$ are mutually exclusive and therefore that $A$ is the union of two mutually exclusive sets, $(A \cap B)$ and $(A \cap \bar{B})$
d. Also show that $B$ and $(A \cap B)$ are mutually exclusive and if $B \subset A, A$ is the union of two mutually exclusive sets, $B$ and $(A \cap \bar{B})$.

Key Concepts

Set Identities

This concept covers the basic principles that allow us to rewrite a set in different but equivalent forms using identity properties such as A = A ? S (where S is the universal set) and properties like B ? B = B. These identities are foundational in set theory and allow for the manipulation and simplification of expressions involving sets.

Distributive Laws in Set Theory

The distributive laws show how intersections distribute over unions and vice versa. They are key tools in transforming complex set expressions into simpler forms. For example, using these laws, one can expand or factor expressions to isolate common components, which is essential in proving identities such as A = (A ? B) ? (A ? B').

Complementation and the Role of a Set's Complement

Complementation deals with the elements not in a given set, typically represented by B' or ??. Understanding how a set interacts with its complement, particularly through intersections and unions, is critical for decomposing a set into mutually exclusive parts based on the presence or absence of certain elements.

Subset Relations

The concept of a subset, denoted as B ? A, is used to establish relationships between sets. In the context of these proofs, assuming that B ? A allows one to draw conclusions about how elements in A can be partitioned between those in B and those not in B, leading to the expression A = B ? (A ? B').

Mutually Exclusive (Disjoint) Sets

Mutually exclusive, or disjoint, sets are those that do not share any elements, meaning their intersection is empty. Demonstrating that two components, such as (A ? B) and (A ? B'), are mutually exclusive helps in understanding how a set can be partitioned into distinct, non-overlapping parts, which is a common theme in set theory proofs.

Transcript

00:01 Today we want to go over probability laws.

00:03 We are given that a equals the intersection between a and s, and s equals the union between b and b.

00:12 And the first thing that we want to do is demonstrate the distributive law.

00:18 Today we will do this by proving that a equals the intersect between a and b, unioned with the intersect between a complement b.

00:27 Today, we first want to prove that a equals a intersect b, union, a intersecting with the b complement.

00:36 So let's first with what we start with what we know a to equal.

00:40 So a equals the intersection between a and s.

00:45 We also know what s equals, which means that we can plug in the equation for s into this new equation.

00:52 So what happens is that we would get a intersecting with the quantity of the union between b and b complement.

01:04 What this suggests is that the distributive law is necessary to where this a intersect will be distributed to each term within our parentheses.

01:13 So we would end up with the intersection between a and b union with the intersection between a and b complement.

01:20 Now we want to prove how a will equal b unionized with the intersection of a and b complement, given that every element of b is also in a.

01:36 Today we want to, if we're being told that every element of b is also in a, that means that our a intersected with b is actually just going to be equal to b...

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