CSE 311 Lecture 04: Boolean Algebra - University of Washington › courses › cse311 › 18au ›...

CSE 311 Lecture 04: Boolean AlgebraEmina Torlak and Kevin Zatloukal

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http://127.0.0.1:4000/courses/cse311/18au/lectures/lecture04.html?print-pdf#/

TopicsEquivalence and proofs

A brief review of .Boolean algebra

A notation for combinational circuits.Simplification and proofs

Optimizing circuits and proving theorems.

Lecture 03

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http://127.0.0.1:4000/courses/cse311/18au/lectures/lecture03.html


Equivalence and proofsA brief review of .Lecture 03

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http://127.0.0.1:4000/courses/cse311/18au/lectures/lecture03.html


Equivalence via truth tables and proofs is an assertion that two propositions and have the same

truth values in all possible cases.

and have the same meaning.

Checking equivalence has many real-world applications.Verification, optimization, and more!

There are two ways to check equivalence of propositional formulas.Brute-force: compare their truth tables.Proof-based: apply equivalences to transform one into the other.

A ≡ B A B

A ≡ B (A ↔ B) ≡ # $%&$'(')*

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Boolean algebraA notation for combinational circuits.

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Recall the relationship between logic and circuits …Digital circuits implement propositional logic:

corresponds to 1 or high voltage. corresponds to 0 or low voltage.

Digital circuits are functions that

take values 0/1 as inputs and produce 0/1 as output;, where is built out of wires and gates; and

every bit of output is computed from some bits of the input.

#+

out = F(input) F

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Recall the relationship between logic and circuits …Digital circuits implement propositional logic:

corresponds to 1 or high voltage. corresponds to 0 or low voltage.

Digital circuits are functions that

take values 0/1 as inputs and produce 0/1 as output;, where is built out of wires and gates; and

every bit of output is computed from some bits of the input.

#+

out = F(input) F

We call these types of digital circuits combinational logic circuits.There are other kinds of digital circuits (called sequential circuits) butwe’ll focus on combinational circuits in this course.

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Boolean algebra is a notation for combinational logicThink of it as a notation for propositional logic used in circuit design.

Boolean algebra consists of the following elements and operations:

a set of elements ,binary operations ,a unary operation .

B = {0, 1}{+, ⋅}{ }′

These correspond to the truthvalues , and the logicalconnectives .

{+, #}∨, ∧, ¬

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Boolean algebra is a notation for combinational logicThink of it as a notation for propositional logic used in circuit design.

Boolean algebra consists of the following elements and operations:


Boolean operations satisfy the following axioms for any :

Closure

Commutativity

Associativity

Distributivity

Identity

Complementarity

Null

Idempotency

Involution

B = {0, 1}{+, ⋅}{ }′


{+, #}∨, ∧, ¬

a, b, c ∈ B

a + b ∈ Ba ⋅ b ∈ B

a + b = b + aa ⋅ b = b ⋅ a

a + (b + c) = (a + b) + ca ⋅ (b ⋅ c) = (a ⋅ b) ⋅ c

a + (b ⋅ c) = (a + b) ⋅ (a + c)a ⋅ (b + c) = (a ⋅ b) + (a ⋅ c)

a + 0 = aa ⋅ 1 = a

a + = 1a′

a ⋅ = 0a′

a + 1 = 1a ⋅ 0 = 0

a + a = aa ⋅ a = a

( = aa′)′

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Example: from spec to code to logic to circuitsSuppose that we want to compute the number of lectures or sectionsremaining at the start of a given day of the week.

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The function for this computation has the following signature:

Inputs: day of the week (integers from 0 to 6), lecture flag (boolean).Output: number of sessions le! (integer from 0 to 3).

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The function for this computation has the following signature:


Here are some examples of the function’s input/output behavior:

Input: (Wednesday, Lecture), Output: 2Input: (Monday, Section), Output: 1

How would you implement this function in Java?

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From spec to …codepublic class Sessions {

public static int classesLeft(int day, boolean lecture) { switch (day) { case 0: // SUNDAY case 1: // MONDAY return lecture ? 3 : 1; case 2: // TUESDAY case 3: // WEDNESDAY return lecture ? 2 : 1; case 4: // THURSDAY return lecture ? 1 : 1; case 5: // FRIDAY return lecture ? 1 : 0; default: //case 6: // SATURDAY return lecture ? 0 : 0; } }

public static void main(String []args){ System.out.println("(W, L) -> " + classesLeft(3,true)); System.out.println("(M, S) -> " + classesLeft(1,false)); }}

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http://tpcg.io/hXxbka


From spec to …

Suppose that we need this function run really fast …

codepublic class Sessions {



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From spec to …

Suppose that we need this function run really fast … To do that, we’llimplement a custom circuit (hardware accelerator!).

codepublic class Sessions {



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From code to combinational logic …Recall the signature of our function:


How many bits for each input/output?

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From code to combinational logic …Recall the signature of our function:


How many bits for each input/output?

Inputs: 3 bits for day of the week, 1 bit for the lecture flag.Output: 2 bits for the number of sessions le!.

day lecture

count

d 2 d 1 Ld 0c1 c 0

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From code to combinational logic via a truth table

day lecture

count

d 2 d 1 Ld 0c1 c 0

DaySUN 000 0SUN 000 1MON 001 0MON 001 1TUE 010 0TUE 010 1WED 011 0WED 011 1THU 100 0THU 100 1FRI 101 0FRI 101 1SAT 110 0SAT 110 1

- 111 0- 111 1

switch (day) { case 0: // SUNDAY case 1: // MONDAY return lecture ? 3 : 1; case 2: // TUESDAY case 3: // WEDNESDAY return lecture ? 2 : 1; case 4: // THURSDAY return lecture ? 1 : 1; case 5: // FRIDAY return lecture ? 1 : 0; default: // SATURDAY etc. return lecture ? 0 : 0;}

d2d1d0 L c1c0

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day lecture

count

d 2 d 1 Ld 0c1 c 0

DaySUN 000 0 01SUN 000 1 11MON 001 0MON 001 1TUE 010 0TUE 010 1WED 011 0WED 011 1THU 100 0THU 100 1FRI 101 0FRI 101 1SAT 110 0SAT 110 1

- 111 0- 111 1


d2d1d0 L c1c0

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day lecture

count

d 2 d 1 Ld 0c1 c 0

DaySUN 000 0 01SUN 000 1 11MON 001 0 01MON 001 1 11TUE 010 0TUE 010 1WED 011 0WED 011 1THU 100 0THU 100 1FRI 101 0FRI 101 1SAT 110 0SAT 110 1

- 111 0- 111 1


d2d1d0 L c1c0

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day lecture

count

d 2 d 1 Ld 0c1 c 0

DaySUN 000 0 01SUN 000 1 11MON 001 0 01MON 001 1 11TUE 010 0 01TUE 010 1 10WED 011 0WED 011 1THU 100 0THU 100 1FRI 101 0FRI 101 1SAT 110 0SAT 110 1

- 111 0- 111 1


d2d1d0 L c1c0

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day lecture

count

d 2 d 1 Ld 0c1 c 0

DaySUN 000 0 01SUN 000 1 11MON 001 0 01MON 001 1 11TUE 010 0 01TUE 010 1 10WED 011 0 01WED 011 1 10THU 100 0THU 100 1FRI 101 0FRI 101 1SAT 110 0SAT 110 1

- 111 0- 111 1


d2d1d0 L c1c0

11



day lecture

count

d 2 d 1 Ld 0c1 c 0

DaySUN 000 0 01SUN 000 1 11MON 001 0 01MON 001 1 11TUE 010 0 01TUE 010 1 10WED 011 0 01WED 011 1 10THU 100 0 01THU 100 1 01FRI 101 0FRI 101 1SAT 110 0SAT 110 1

- 111 0- 111 1


d2d1d0 L c1c0

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day lecture

count

d 2 d 1 Ld 0c1 c 0

DaySUN 000 0 01SUN 000 1 11MON 001 0 01MON 001 1 11TUE 010 0 01TUE 010 1 10WED 011 0 01WED 011 1 10THU 100 0 01THU 100 1 01FRI 101 0 00FRI 101 1 01SAT 110 0SAT 110 1

- 111 0- 111 1


d2d1d0 L c1c0

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day lecture

count

d 2 d 1 Ld 0c1 c 0

DaySUN 000 0 01SUN 000 1 11MON 001 0 01MON 001 1 11TUE 010 0 01TUE 010 1 10WED 011 0 01WED 011 1 10THU 100 0 01THU 100 1 01FRI 101 0 00FRI 101 1 01SAT 110 0 00SAT 110 1 00

- 111 0- 111 1


d2d1d0 L c1c0

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day lecture

count

d 2 d 1 Ld 0c1 c 0


- 111 0 00- 111 1 00


d2d1d0 L c1c0

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From code to combinational logic via a truth table:


- 111 0 00- 111 1 00

To find an expression for , look at the rows where .

==000 && L==1

==001 && L==1

==010 && L==1

==011 && L==1

c1

d2d1d0 L c1c0 c1 = 1c1

d2d1d0d2d1d0d2d1d0d2d1d0

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- 111 0 00- 111 1 00


==000 && L==1

==001 && L==1

==010 && L==1

==011 && L==1

Split up the bits of the day to get a formula for each row.

==0 && ==0 && ==0 && L==1

==0 && ==0 && ==1 && L==1

==0 && ==1 && ==0 && L==1

==0 && ==1 && ==1 && L==1

c1

d2d1d0 L c1c0 c1 = 1c1


d2 d1 d0d2 d1 d0d2 d1 d0d2 d1 d0

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- 111 0 00- 111 1 00


==000 && L==1

==001 && L==1

==010 && L==1

==011 && L==1


==0 && ==0 && ==0 && L==1

==0 && ==0 && ==1 && L==1

==0 && ==1 && ==0 && L==1

==0 && ==1 && ==1 && L==1

Translate to Boolean algebra to get an expression for .

c1

d2d1d0 L c1c0 c1 = 1c1


d2 d1 d0d2 d1 d0d2 d1 d0d2 d1 d0

c1

⋅ ⋅ ⋅ Ld ′2 d ′

1 d ′0

⋅ ⋅ ⋅ Ld ′2 d ′

1 d0⋅ ⋅ ⋅ Ld ′

2 d1 d ′0

⋅ ⋅ ⋅ Ld ′2 d1 d0

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- 111 0 00- 111 1 00


==000 && L==1

==001 && L==1

==010 && L==1

==011 && L==1


==0 && ==0 && ==0 && L==1

==0 && ==0 && ==1 && L==1

==0 && ==1 && ==0 && L==1

==0 && ==1 && ==1 && L==1

Translate to Boolean algebra to get an expression for .

c1

d2d1d0 L c1c0 c1 = 1c1


d2 d1 d0d2 d1 d0d2 d1 d0d2 d1 d0

c1

⋅ ⋅ ⋅ Ld ′2 d ′

1 d ′0

⋅ ⋅ ⋅ Ld ′2 d ′

1 d0⋅ ⋅ ⋅ Ld ′

2 d1 d ′0

⋅ ⋅ ⋅ Ld ′2 d1 d0

=c1 ⋅ ⋅ ⋅ L +d ′2 d ′

1 d ′0

⋅ ⋅ ⋅ L +d ′2 d ′

1 d0⋅ ⋅ ⋅ L +d ′

2 d1 d ′0

⋅ ⋅ ⋅ Ld ′2 d1 d0

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- 111 0 00- 111 1 00

Now we repeat this process to get .

c0

d2d1d0 L c1c0=c1 ⋅ ⋅ ⋅ L +d ′

2 d ′1 d ′

0⋅ ⋅ ⋅ L +d ′

2 d ′1 d0

⋅ ⋅ ⋅ L +d ′2 d1 d ′

0⋅ ⋅ ⋅ Ld ′

2 d1 d0

c0

=c0

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- 111 0 00- 111 1 00


c0

d2d1d0 L c1c0=c1 ⋅ ⋅ ⋅ L +d ′

2 d ′1 d ′

0⋅ ⋅ ⋅ L +d ′

2 d ′1 d0

⋅ ⋅ ⋅ L +d ′2 d1 d ′

0⋅ ⋅ ⋅ Ld ′

2 d1 d0

c0

=c0 ⋅ ⋅ ⋅ +d ′2 d ′

1 d ′0 L ′

⋅ ⋅ ⋅ L +d ′2 d ′

1 d ′0

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- 111 0 00- 111 1 00


c0

d2d1d0 L c1c0=c1 ⋅ ⋅ ⋅ L +d ′

2 d ′1 d ′

0⋅ ⋅ ⋅ L +d ′

2 d ′1 d0

⋅ ⋅ ⋅ L +d ′2 d1 d ′

0⋅ ⋅ ⋅ Ld ′

2 d1 d0

c0

=c0 ⋅ ⋅ ⋅ +d ′2 d ′

1 d ′0 L ′

⋅ ⋅ ⋅ L +d ′2 d ′

1 d ′0

⋅ ⋅ ⋅ +d ′2 d ′

1 d0 L ′

⋅ ⋅ ⋅ L +d ′2 d ′

1 d0

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- 111 0 00- 111 1 00


c0

d2d1d0 L c1c0=c1 ⋅ ⋅ ⋅ L +d ′

2 d ′1 d ′

0⋅ ⋅ ⋅ L +d ′

2 d ′1 d0

⋅ ⋅ ⋅ L +d ′2 d1 d ′

0⋅ ⋅ ⋅ Ld ′

2 d1 d0

c0

=c0 ⋅ ⋅ ⋅ +d ′2 d ′

1 d ′0 L ′

⋅ ⋅ ⋅ L +d ′2 d ′

1 d ′0

⋅ ⋅ ⋅ +d ′2 d ′

1 d0 L ′

⋅ ⋅ ⋅ L +d ′2 d ′

1 d0⋅ ⋅ ⋅ +d ′

2 d1 d ′0 L ′

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- 111 0 00- 111 1 00


c0

d2d1d0 L c1c0=c1 ⋅ ⋅ ⋅ L +d ′

2 d ′1 d ′

0⋅ ⋅ ⋅ L +d ′

2 d ′1 d0

⋅ ⋅ ⋅ L +d ′2 d1 d ′

0⋅ ⋅ ⋅ Ld ′

2 d1 d0

c0

=c0 ⋅ ⋅ ⋅ +d ′2 d ′

1 d ′0 L ′

⋅ ⋅ ⋅ L +d ′2 d ′

1 d ′0

⋅ ⋅ ⋅ +d ′2 d ′

1 d0 L ′

⋅ ⋅ ⋅ L +d ′2 d ′

1 d0⋅ ⋅ ⋅ +d ′

2 d1 d ′0 L ′

⋅ ⋅ ⋅ +d ′2 d1 d0 L ′

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- 111 0 00- 111 1 00


c0

d2d1d0 L c1c0=c1 ⋅ ⋅ ⋅ L +d ′

2 d ′1 d ′

0⋅ ⋅ ⋅ L +d ′

2 d ′1 d0

⋅ ⋅ ⋅ L +d ′2 d1 d ′

0⋅ ⋅ ⋅ Ld ′

2 d1 d0

c0

=c0 ⋅ ⋅ ⋅ +d ′2 d ′

1 d ′0 L ′

⋅ ⋅ ⋅ L +d ′2 d ′

1 d ′0

⋅ ⋅ ⋅ +d ′2 d ′

1 d0 L ′

⋅ ⋅ ⋅ L +d ′2 d ′

1 d0⋅ ⋅ ⋅ +d ′

2 d1 d ′0 L ′

⋅ ⋅ ⋅ +d ′2 d1 d0 L ′

⋅ ⋅ ⋅ +d2 d ′1 d ′

0 L ′

⋅ ⋅ ⋅ L +d2 d ′1 d ′

0

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- 111 0 00- 111 1 00


c0

d2d1d0 L c1c0=c1 ⋅ ⋅ ⋅ L +d ′

2 d ′1 d ′

0⋅ ⋅ ⋅ L +d ′

2 d ′1 d0

⋅ ⋅ ⋅ L +d ′2 d1 d ′

0⋅ ⋅ ⋅ Ld ′

2 d1 d0

c0

=c0 ⋅ ⋅ ⋅ +d ′2 d ′

1 d ′0 L ′

⋅ ⋅ ⋅ L +d ′2 d ′

1 d ′0

⋅ ⋅ ⋅ +d ′2 d ′

1 d0 L ′

⋅ ⋅ ⋅ L +d ′2 d ′

1 d0⋅ ⋅ ⋅ +d ′

2 d1 d ′0 L ′

⋅ ⋅ ⋅ +d ′2 d1 d0 L ′

⋅ ⋅ ⋅ +d2 d ′1 d ′

0 L ′

⋅ ⋅ ⋅ L +d2 d ′1 d ′

0⋅ ⋅ ⋅ Ld2 d ′

1 d0

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From combinational logic to circuits

Here is as a circuit …

d 2

d 1

L

d 0 c1

=c0 ⋅ ⋅ ⋅ +d ′2 d ′

1 d ′0 L ′

⋅ ⋅ ⋅ L +d ′2 d ′

1 d ′0

⋅ ⋅ ⋅ +d ′2 d ′

1 d0 L ′

⋅ ⋅ ⋅ L +d ′2 d ′

1 d0⋅ ⋅ ⋅ +d ′

2 d1 d ′0 L ′

⋅ ⋅ ⋅ +d ′2 d1 d0 L ′

⋅ ⋅ ⋅ +d2 d ′1 d ′

0 L ′

⋅ ⋅ ⋅ L +d2 d ′1 d ′

0⋅ ⋅ ⋅ Ld2 d ′

1 d0

=c1 ⋅ ⋅ ⋅ L +d ′2 d ′

1 d ′0

⋅ ⋅ ⋅ L +d ′2 d ′

1 d0⋅ ⋅ ⋅ L +d ′

2 d1 d ′0

⋅ ⋅ ⋅ Ld ′2 d1 d0

c1

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What can we do with the logic encoding?

Create hardware implementations!

And perform program verification …

Example: verify that classesLeftreturns 3 only if lecture is true.


- 111 0 00- 111 1 00

=c0 ⋅ ⋅ ⋅ +d ′2 d ′

1 d ′0 L ′

⋅ ⋅ ⋅ L +d ′2 d ′

1 d ′0

⋅ ⋅ ⋅ +d ′2 d ′

1 d0 L ′

⋅ ⋅ ⋅ L +d ′2 d ′

1 d0⋅ ⋅ ⋅ +d ′

2 d1 d ′0 L ′

⋅ ⋅ ⋅ +d ′2 d1 d0 L ′

⋅ ⋅ ⋅ +d2 d ′1 d ′

0 L ′

⋅ ⋅ ⋅ L +d2 d ′1 d ′

0⋅ ⋅ ⋅ Ld2 d ′

1 d0

=c1 ⋅ ⋅ ⋅ L +d ′2 d ′

1 d ′0

⋅ ⋅ ⋅ L +d ′2 d ′

1 d0⋅ ⋅ ⋅ L +d ′

2 d1 d ′0

⋅ ⋅ ⋅ Ld ′2 d1 d0

d2d1d0 L c1c0

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What can we do with the logic encoding?

Create hardware implementations!

And perform program verification …

Example: verify that classesLeftreturns 3 only if lecture is true.

Check that .


- 111 0 00- 111 1 00

∧ → L ≡ #c1 c0

=c0 ⋅ ⋅ ⋅ +d ′2 d ′

1 d ′0 L ′

⋅ ⋅ ⋅ L +d ′2 d ′

1 d ′0

⋅ ⋅ ⋅ +d ′2 d ′

1 d0 L ′

⋅ ⋅ ⋅ L +d ′2 d ′

1 d0⋅ ⋅ ⋅ +d ′

2 d1 d ′0 L ′

⋅ ⋅ ⋅ +d ′2 d1 d0 L ′

⋅ ⋅ ⋅ +d2 d ′1 d ′

0 L ′

⋅ ⋅ ⋅ L +d2 d ′1 d ′

0⋅ ⋅ ⋅ Ld2 d ′

1 d0

=c1 ⋅ ⋅ ⋅ L +d ′2 d ′

1 d ′0

⋅ ⋅ ⋅ L +d ′2 d ′

1 d0⋅ ⋅ ⋅ L +d ′

2 d1 d ′0

⋅ ⋅ ⋅ Ld ′2 d1 d0

d2d1d0 L c1c0

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Simplification and proofsOptimizing circuits and proving theorems.

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So far, we’ve used the basics of Boolean algebra …Boolean algebra consists of the following elements and operations:


Boolean operations satisfy the following axioms for any :

Closure

Commutativity

Associativity

Distributivity

Identity

Complementarity

Null

Idempotency

Involution

B = {0, 1}{+, ⋅}{ }′


{+, #}∨, ∧, ¬

a, b, c ∈ B

a + b ∈ Ba ⋅ b ∈ B

a + b = b + aa ⋅ b = b ⋅ a

a + (b + c) = (a + b) + ca ⋅ (b ⋅ c) = (a ⋅ b) ⋅ c

a + (b ⋅ c) = (a + b) ⋅ (a + c)a ⋅ (b + c) = (a ⋅ b) + (a ⋅ c)

a + 0 = aa ⋅ 1 = a

a + = 1a′

a ⋅ = 0a′

a + 1 = 1a ⋅ 0 = 0

a + a = aa ⋅ a = a

( = aa′)′

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We can use the basics to prove some useful theoremsUniting

Absorption

Factoring

Consensus

DeMorgan’s

a ⋅ b + a ⋅ = ab′

(a + b) ⋅ (a + ) = ab′

a + a ⋅ b = aa ⋅ (a + b) = a(a + ) ⋅ b = a ⋅ bb′

(a ⋅ ) + b = a + bb′

(a + b) ⋅ ( + c) = a ⋅ c + ⋅ ba′ a′

a ⋅ b + ⋅ c = (a + c) ⋅ ( + b)a′ a′

(a ⋅ b) + (b ⋅ c) + ( ⋅ c) = a ⋅ b + ⋅ ca′ a′

(a + b) ⋅ (b + c) ⋅ ( + c) = (a + b) ⋅ ( + c)a′ a′

(a + b + … = ⋅ ⋅ …)′ a′ b′

(a ⋅ b ⋅ … = + + …)′ a′ b′

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Example: proving theorems using the axioms

Uniting

Absorption

Closure

Commutativity

Associativity

Distributivity

Identity

Complementarity

Null

Idempotency

Involution

X ⋅ Y + X ⋅ Y ′ === X

X + X ⋅ Y ===== X

a + b ∈ Ba ⋅ b ∈ B

a + b = b + aa ⋅ b = b ⋅ a

a + (b + c) = (a + b) + ca ⋅ (b ⋅ c) = (a ⋅ b) ⋅ c

a + (b ⋅ c) = (a + b) ⋅ (a + c)a ⋅ (b + c) = (a ⋅ b) + (a ⋅ c)

a + 0 = aa ⋅ 1 = a

a + = 1a′

a ⋅ = 0a′

a + 1 = 1a ⋅ 0 = 0

a + a = aa ⋅ a = a

( = aa′)′

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Uniting Distributivity

Absorption

Closure

Commutativity

Associativity

Distributivity

Identity

Complementarity

Null

Idempotency

Involution

X ⋅ Y + X ⋅ Y ′ = X ⋅ (Y + )Y ′

== X

X + X ⋅ Y ===== X

a + b ∈ Ba ⋅ b ∈ B

a + b = b + aa ⋅ b = b ⋅ a

a + (b + c) = (a + b) + ca ⋅ (b ⋅ c) = (a ⋅ b) ⋅ c

a + (b ⋅ c) = (a + b) ⋅ (a + c)a ⋅ (b + c) = (a ⋅ b) + (a ⋅ c)

a + 0 = aa ⋅ 1 = a

a + = 1a′

a ⋅ = 0a′

a + 1 = 1a ⋅ 0 = 0

a + a = aa ⋅ a = a

( = aa′)′

19



Uniting DistributivityComplementarity

Absorption

Closure

Commutativity

Associativity

Distributivity

Identity

Complementarity

Null

Idempotency

Involution

X ⋅ Y + X ⋅ Y ′ = X ⋅ (Y + )Y ′

= X ⋅ 1= X

X + X ⋅ Y ===== X

a + b ∈ Ba ⋅ b ∈ B

a + b = b + aa ⋅ b = b ⋅ a

a + (b + c) = (a + b) + ca ⋅ (b ⋅ c) = (a ⋅ b) ⋅ c

a + (b ⋅ c) = (a + b) ⋅ (a + c)a ⋅ (b + c) = (a ⋅ b) + (a ⋅ c)

a + 0 = aa ⋅ 1 = a

a + = 1a′

a ⋅ = 0a′

a + 1 = 1a ⋅ 0 = 0

a + a = aa ⋅ a = a

( = aa′)′

19



Uniting DistributivityComplementarityIdentity

Absorption

Closure

Commutativity

Associativity

Distributivity

Identity

Complementarity

Null

Idempotency

Involution

X ⋅ Y + X ⋅ Y ′ = X ⋅ (Y + )Y ′

= X ⋅ 1= X

X + X ⋅ Y ===== X

a + b ∈ Ba ⋅ b ∈ B

a + b = b + aa ⋅ b = b ⋅ a

a + (b + c) = (a + b) + ca ⋅ (b ⋅ c) = (a ⋅ b) ⋅ c

a + (b ⋅ c) = (a + b) ⋅ (a + c)a ⋅ (b + c) = (a ⋅ b) + (a ⋅ c)

a + 0 = aa ⋅ 1 = a

a + = 1a′

a ⋅ = 0a′

a + 1 = 1a ⋅ 0 = 0

a + a = aa ⋅ a = a

( = aa′)′

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Absorption Identity

Closure

Commutativity

Associativity

Distributivity

Identity

Complementarity

Null

Idempotency

Involution

X ⋅ Y + X ⋅ Y ′ = X ⋅ (Y + )Y ′

= X ⋅ 1= X

X + X ⋅ Y = X ⋅ 1 + X ⋅ Y==== X

a + b ∈ Ba ⋅ b ∈ B

a + b = b + aa ⋅ b = b ⋅ a

a + (b + c) = (a + b) + ca ⋅ (b ⋅ c) = (a ⋅ b) ⋅ c

a + (b ⋅ c) = (a + b) ⋅ (a + c)a ⋅ (b + c) = (a ⋅ b) + (a ⋅ c)

a + 0 = aa ⋅ 1 = a

a + = 1a′

a ⋅ = 0a′

a + 1 = 1a ⋅ 0 = 0

a + a = aa ⋅ a = a

( = aa′)′

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Absorption IdentityDistributivity

Closure

Commutativity

Associativity

Distributivity

Identity

Complementarity

Null

Idempotency

Involution

X ⋅ Y + X ⋅ Y ′ = X ⋅ (Y + )Y ′

= X ⋅ 1= X

X + X ⋅ Y = X ⋅ 1 + X ⋅ Y= X ⋅ (1 + Y)=== X

a + b ∈ Ba ⋅ b ∈ B

a + b = b + aa ⋅ b = b ⋅ a

a + (b + c) = (a + b) + ca ⋅ (b ⋅ c) = (a ⋅ b) ⋅ c

a + (b ⋅ c) = (a + b) ⋅ (a + c)a ⋅ (b + c) = (a ⋅ b) + (a ⋅ c)

a + 0 = aa ⋅ 1 = a

a + = 1a′

a ⋅ = 0a′

a + 1 = 1a ⋅ 0 = 0

a + a = aa ⋅ a = a

( = aa′)′

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Absorption IdentityDistributivityCommutativity

Closure

Commutativity

Associativity

Distributivity

Identity

Complementarity

Null

Idempotency

Involution

X ⋅ Y + X ⋅ Y ′ = X ⋅ (Y + )Y ′

= X ⋅ 1= X

X + X ⋅ Y = X ⋅ 1 + X ⋅ Y= X ⋅ (1 + Y)= X ⋅ (Y + 1)== X

a + b ∈ Ba ⋅ b ∈ B

a + b = b + aa ⋅ b = b ⋅ a

a + (b + c) = (a + b) + ca ⋅ (b ⋅ c) = (a ⋅ b) ⋅ c

a + (b ⋅ c) = (a + b) ⋅ (a + c)a ⋅ (b + c) = (a ⋅ b) + (a ⋅ c)

a + 0 = aa ⋅ 1 = a

a + = 1a′

a ⋅ = 0a′

a + 1 = 1a ⋅ 0 = 0

a + a = aa ⋅ a = a

( = aa′)′

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Absorption IdentityDistributivityCommutativityNull

Closure

Commutativity

Associativity

Distributivity

Identity

Complementarity

Null

Idempotency

Involution

X ⋅ Y + X ⋅ Y ′ = X ⋅ (Y + )Y ′

= X ⋅ 1= X

X + X ⋅ Y = X ⋅ 1 + X ⋅ Y= X ⋅ (1 + Y)= X ⋅ (Y + 1)= X ⋅ 1= X

a + b ∈ Ba ⋅ b ∈ B

a + b = b + aa ⋅ b = b ⋅ a

a + (b + c) = (a + b) + ca ⋅ (b ⋅ c) = (a ⋅ b) ⋅ c

a + (b ⋅ c) = (a + b) ⋅ (a + c)a ⋅ (b + c) = (a ⋅ b) + (a ⋅ c)

a + 0 = aa ⋅ 1 = a

a + = 1a′

a ⋅ = 0a′

a + 1 = 1a ⋅ 0 = 0

a + a = aa ⋅ a = a

( = aa′)′

19




Absorption IdentityDistributivityCommutativityNullIdentity

Closure

Commutativity

Associativity

Distributivity

Identity

Complementarity

Null

Idempotency

Involution

X ⋅ Y + X ⋅ Y ′ = X ⋅ (Y + )Y ′

= X ⋅ 1= X

X + X ⋅ Y = X ⋅ 1 + X ⋅ Y= X ⋅ (1 + Y)= X ⋅ (Y + 1)= X ⋅ 1= X

a + b ∈ Ba ⋅ b ∈ B

a + b = b + aa ⋅ b = b ⋅ a

a + (b + c) = (a + b) + ca ⋅ (b ⋅ c) = (a ⋅ b) ⋅ c

a + (b ⋅ c) = (a + b) ⋅ (a + c)a ⋅ (b + c) = (a ⋅ b) + (a ⋅ c)

a + 0 = aa ⋅ 1 = a

a + = 1a′

a ⋅ = 0a′

a + 1 = 1a ⋅ 0 = 0

a + a = aa ⋅ a = a

( = aa′)′

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Example: proving theorems using truth tables

DeMorgan’s law

NOR is equivalent to AND with inputscomplemented

0 0 1 1 1 10 1 1 0 0 01 0 0 1 0 01 1 0 0 0 0

DeMorgan’s law

NAND is equivalent to OR with inputscomplemented

0 0 1 1 1 10 1 1 0 1 11 0 0 1 1 11 1 0 0 0 0

(X + Y = ⋅)′ X ′ Y ′X Y X ′ Y ′ (X + Y)′ ⋅X ′ Y ′

(X ⋅ Y = +)′ X ′ Y ′X Y X ′ Y ′ (X ⋅ Y)′ +X ′ Y ′

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Example: simplifying (circuits) using Boolean algebra(from classesLe!)(HW2)(HW2)

Closure

Commutativity

Associativity

Distributivity

Identity

Complementarity

Null

Idempotency

Involution

c1 = ⋅ ⋅ ⋅ L + ⋅ ⋅ ⋅ L + ⋅ ⋅ ⋅ L + ⋅ ⋅ ⋅ Ld ′2 d ′

1 d ′0 d ′

2 d ′1 d0 d ′

2 d1 d ′0 d ′

2 d1 d0= …= ⋅ Ld ′

2

a + b ∈ Ba ⋅ b ∈ B

a + b = b + aa ⋅ b = b ⋅ a

a + (b + c) = (a + b) + ca ⋅ (b ⋅ c) = (a ⋅ b) ⋅ c

a + (b ⋅ c) = (a + b) ⋅ (a + c)a ⋅ (b + c) = (a ⋅ b) + (a ⋅ c)

a + 0 = aa ⋅ 1 = a

a + = 1a′

a ⋅ = 0a′

a + 1 = 1a ⋅ 0 = 0

a + a = aa ⋅ a = a

( = aa′)′

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Example: simplifying (circuits) using Boolean algebra(from classesLe!)(HW2)(HW2)

Here is the simplified circuit …

d 2L

c1

c1 = ⋅ ⋅ ⋅ L + ⋅ ⋅ ⋅ L + ⋅ ⋅ ⋅ L + ⋅ ⋅ ⋅ Ld ′2 d ′

1 d ′0 d ′

2 d ′1 d0 d ′

2 d1 d ′0 d ′

2 d1 d0= …= ⋅ Ld ′

2

c1

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SummaryBoolean algebra is a notation for combinational circuits.

It consists of elements and operations .The operations satisfy the axioms of Boolean algebra.

We can translate specs to code to logic and to circuits forfaster implementation in hardware, andprogram verification.

We can use axioms of Boolean algebra and truth tables toprove useful theorems, andsimplify and optimize combinational circuits.

{0, 1} {+, ⋅, }′

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CSE 311 Lecture 04: Boolean Algebra - University of Washington › courses › cse311 › 18au ›...

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Transcript of CSE 311 Lecture 04: Boolean Algebra - University of Washington › courses › cse311 › 18au ›...