Boolean Algebra How gates get picked

Boolean Arithmetic Boolean ≠ Binary Boolean addition: Boolean – 1/0 only, no places Boolean addition:

Boolean Arithmetic Boolean addition: + means OR

Boolean Multiplication

Boolean Multiplication Multiplication is AND

Boolean Variables Capital letter used for variables Inverted A, B, C… A or A'

Identities Additive:

Identities Multiplicative:

Double Complement Two negations cancel: A = A

Break …

Commutative Property Order of variables not important:

Associative Property + and · associative Just like "normal" algebra

Distributive Can distribute/factor Just like "normal" algebra

Theorem 1 A + AB = A I have A Or I have A and B I really just need A

Theorem 2 A + A B = A + B I have A Or I have B but not A I need A or B (A satisfies first term - only care about second if first is not satisfied)

Theorem 3 (A + B)(A + C) = A + BC I have A or B AND I have A or C I need A or both B and C

Example: A + AC + BC A(1 + C) + BC A(1) + BC A + BC

Example: AB + AB' A(B + B') A(1) A Note : A' = A

Example: A + A'B' + B' A + B'(A' + 1) A + B'(1) A + B'

Example: A + B + A' (A + A') + B + B 1

Example: A(B +AB) + AC AB + AAB + AC AB + AB + AC AB + AC A(B + C)

Truth Table Truth table defines Boolean function When in doubt, check the truth table…

Proof By Exhaustion Proof by exhaustion : prove equivalence by comparing truth tables Ex: AB = A + B A B AB 1 A B A + B 1

Not The Same AB != A · B A + B != A + B A B AB 1 A B A · B 1 A B A + B 1 A B A · B 1 A B A + B 1 A B A+B 1

DeMorgan's Theorems A+B = A ∙ B AB = A + B If you don't have both of A and B …you do not have A or do not have B A+B = A ∙ B If you don't have either of A or B …you do not have A and you do not have B

DeMorgan's Break up solid bar by switching operation:

Samples AB + A B = A + B + A B DeMorgan's = A + A B + B Commutative = A (1 + B) + B Distributive = A (1) + B 1+ anything = 1 = A + B 1 · anything = self

Samples ( A +B) ·A = ( A · B )·A DeMorgan's = (A· B )·A Double negative cancel = A· B ·A Associative = A·A· B Commutative = A· B Anything · self = self