24 min read

The Laws of Boolean Algebra

Identity, null, idempotent, complement, commutative, associative, distributive and absorption laws — with an everyday meaning for each.

Why We Need Laws

Two different circuits can do exactly the same job. The laws of Boolean algebra let you rewrite an expression into a simpler one that gives the same truth table . A simpler expression needs fewer gates, which means a circuit that is cheaper, smaller, faster and uses less power. Each law has an AND form and an OR form . Try reading each one in words — most are common sense.

Copy this table into your notes. You will quote these laws by name when you simplify.

You can prove any law by comparing truth tables. If the two sides give the same output in every row, they are equal.

Worked example

Proving the absorption law A + A·B = A

A B A\cdotB A + A\cdotB A 00000010001001111111 The columns A + A\cdotB and A are identical in every row, so the law is true. In words: "the shop is open if the manager is in, or the manager is in and it is a weekday" — the second part adds nothing.

Worked example

Proving the second distributive law A + B·C = (A + B)·(A + C)

A B C B\cdotC A + B\cdotC A + B A + C (A + B)\cdot(A + C) 0000000000100010010001000111111110001111101011111100111111111111 The A + B\cdotC and (A + B)\cdot(A + C) columns match in all eight rows.

Practice

What is A + 1?

Why We Need Laws

Two different circuits can do exactly the same job. The laws of Boolean algebra let you rewrite an expression into a simpler one that gives the same truth table . A simpler expression needs fewer gates, which means a circuit that is cheaper, smaller, faster and uses less power. Each law has an AND form and an OR form . Try reading each one in words — most are common sense.

Copy this table into your notes. You will quote these laws by name when you simplify.

Checking a Law with a Truth Table

You can prove any law by comparing truth tables. If the two sides give the same output in every row, they are equal.

Worked example

Proving the absorption law A + A·B = A

A B A\cdotB A + A\cdotB A 00000010001001111111 The columns A + A\cdotB and A are identical in every row, so the law is true. In words: "the shop is open if the manager is in, or the manager is in and it is a weekday" — the second part adds nothing.

Worked example

Proving the second distributive law A + B·C = (A + B)·(A + C)

A B C B\cdotC A + B\cdotC A + B A + C (A + B)\cdot(A + C) 0000000000100010010001000111111110001111101011111100111111111111 The A + B\cdotC and (A + B)\cdot(A + C) columns match in all eight rows.

Practice

What is A + 1?

A	B	C	B·C	A + B·C	A + B	A + C	(A + B)·(A + C)
0	0	0	0	0	0	0	0
0	0	1	0	0	0	1	0
0	1	0	0	0	1	0	0
0	1	1	1	1	1	1	1
1	0	0	0	1	1	1	1
1	0	1	0	1	1	1	1
1	1	0	0	1	1	1	1
1	1	1	1	1	1	1	1

A	B	C	B·C	A + B·C	A + B	A + C	(A + B)·(A + C)
0	0	0	0	0	0	0	0
0	0	1	0	0	0	1	0
0	1	0	0	0	1	0	0
0	1	1	1	1	1	1	1
1	0	0	0	1	1	1	1
1	0	1	0	1	1	1	1
1	1	0	0	1	1	1	1
1	1	1	1	1	1	1	1

A	B	C	B·C	A + B·C	A + B	A + C	(A + B)·(A + C)
0	0	0	0	0	0	0	0
0	0	1	0	0	0	1	0
0	1	0	0	0	1	0	0
0	1	1	1	1	1	1	1
1	0	0	0	1	1	1	1
1	0	1	0	1	1	1	1
1	1	0	0	1	1	1	1
1	1	1	1	1	1	1	1

A	B	C	B·C	A + B·C	A + B	A + C	(A + B)·(A + C)
0	0	0	0	0	0	0	0
0	0	1	0	0	0	1	0
0	1	0	0	0	1	0	0
0	1	1	1	1	1	1	1
1	0	0	0	1	1	1	1
1	0	1	0	1	1	1	1
1	1	0	0	1	1	1	1
1	1	1	1	1	1	1	1

A	B	C	B·C	A + B·C	A + B	A + C	(A + B)·(A + C)
0	0	0	0	0	0	0	0
0	0	1	0	0	0	1	0
0	1	0	0	0	1	0	0
0	1	1	1	1	1	1	1
1	0	0	0	1	1	1	1
1	0	1	0	1	1	1	1
1	1	0	0	1	1	1	1
1	1	1	1	1	1	1	1

A	B	C	B·C	A + B·C	A + B	A + C	(A + B)·(A + C)
0	0	0	0	0	0	0	0
0	0	1	0	0	0	1	0
0	1	0	0	0	1	0	0
0	1	1	1	1	1	1	1
1	0	0	0	1	1	1	1
1	0	1	0	1	1	1	1
1	1	0	0	1	1	1	1
1	1	1	1	1	1	1	1