Boolean Algebra Rule 10

Question

Hi I have a question about the following algebra rule

A + AB = A

My textbook explains this as follows A + AB = A This rule can be proved as such:

• Step 1:

  Dustributive Law:

  A + AB = A*1 = A(1+B) Huh...? Where do they get the one(1) from?

• Step 2:

  1 + B = 1

• Step 3:

: A + 1 = A

Thus A + AB = A

If anyone can clarify this for me it would be greatly appreciated

Answer 1

The 1 would stand for ⊤ or true. To prove the rule they're assuming the right side is true, distributing this information into the left side, and reducing.

Starting with A ∨ (A ∧ B) ↔ A,

A + AB = A

Call A ⊤,

⊤ ∨ (⊤ ∧ B) ↔ ⊤

1 + 1 * B = 1 now it reads "true or true and B equals true" which could just as easily be "with gravy or gravy and something else, you will have gravy"

1 + B = 1 AND having higher precedence...

1 = 1 OR is true if at least one operand is true

A no further reduction possible.

It could just as easily be done using ⊥ (false) instead

⊥ ∨ (⊥ ∧ B) ↔ ⊥

0 + 0 * B = 0 which would read "false or false and B equals false" which could just as easily be "without bananas or bananas and something else, you will not have bananas"

0 + 0 = 0 AND having higher precedence...

0 = 0 OR is false

A no further reduction possible

It might help to construct a truth table and then review the rules for distribution. The 1 appearing in your formula is distributed into the terms to facilitate simplifying the statement.

Since rule 0101 (10) maps to (P ∧ Q) ∨ Q ↔ Q

P Q x
0 0 0
0 1 1
1 0 0
1 1 1

or a Karnaugh map

   ~Q Q
~P 0  1
 P 0  1

see also http://en.wikipedia.org/wiki/Truth_function#Table_of_binary_truth_functions : Proposition Q

Answer 2

• A = A*1 (A AND TRUE has the same truth value as A, by the law of identity for AND)
• 1 = 1+B (TRUE OR B is TRUE, by the identity for OR)

So you can always replace X with X*1, and you can always replace 1 with X+1.

See http://en.wikipedia.org/wiki/Boolean_algebra