In binary algebra, if $x,y,z\in\mathbb B$ such that $\mathbb B=\{0, 1\}$:
The **identity element** for **AND** $1$ is such that any $x\cdot 1 = x$.
The **identity element** for **OR** $0$ is such that any $x + 0 = x$.
In this space, it can be deduced that $x+x'=1$ and $x\cdot x'=0$.
**De Morgan's laws** are much easier to express in boolean algebra, and denote distributing a negation by flipping the operator:
$$
(x\cdot y)'=x'+y' \\
(x+y)=x'\cdot y'
$$
Please see [ECE 108: Discrete Math 1#Operator laws](/1b/ece108/#operator-laws) for more information.
The **synthesis** of an algebraic formula represents its implementation via logic gates. In this course, its total cost is the sum of all inputs to all gates and the number of gates, *excluding* initial inputs of "true" or an initial negation.
In order to deduce an algebraic expression from a truth table, **OR** all of the rows in which the function returns true and simplify.
The **minterm** $m$ is a **product** term where all variables in the function appear once. There are $2^n$ minterms for each function, where $n$ is the number of input variables.
To determine the relevant function, the subscript can be converted to binary and each function variable set such that:
- if the digit is $1$, the complement is used, and
- if the digit is $0$, the original is used.
$$m_j=x_1+x_2+\dots x_n$$
!!! example
For a function that accepts three variables:
- there are eight minterms, from $m_0$ to $m_7$.
- the sixth minterm $m_6=xyz'$ because $6=0b110$.
For a sample function defined by the following minterms:
$$
\begin{align*}
f(x_1,x_2,x_3)&=\sum m(1,2,5) \\
&=m_1+m_2+m_5 \\
&=x_1x_2x_3' + x_1x_2'x_3 + x_1'x_2x_3'
\end{align*}
$$
The **maxterm** $M$ is a **sum** term where all variables in the function appear once. It is more or less the same as a minterm, except the condition for each variable is **reversed** (i.e., $0$ indicates the complement).
$$M_j=x_1+x_2+\dots +x_n$$
!!! example
For a sample function defined by the following maxterms:
\begin{align*}
f(x_1,x_2,x_3,x_4)&=\prod M(1,2,8,12) \\
&=M_1M_2M_8M_{12} \\
\end{align*}
??? example
Prove that $\sum m(1,2,3,4,5,6,7)=x_1+x_2+x_3$: **(some shortcuts taken for visual clarity)**