Boolean Algebra

Boolean Assertions

• Statements that will result in true or false outcomes

a > 5 0 = = b a <= b

Negating Boolean Assertions

Rewriting code for easier readability is usually why we negate Boolean assertions

if(!(x < 5))becomes

if( x > = 5)

Boolean Algebra

• Operands(values): true, false• Operators: and (&&) or( | |) not(!)

DeMorgan’s Laws

• not(A or B) = not A and not B!(A || B) = !A && !B

• not(A and B) = not A or not B!(A && B) = !A || ! B

Application of DeMorgan’s Law“Craps”

• If you roll a 7 or 11 on the first roll, you win• If you roll a 2, 3, or 12 on the first roll, you lose• Otherwise on subsequent rolls you want to

roll your original number before you roll a 7 to win

Here is the truth table that proves the first DeMorgan’s Law. not(A or B) = not A and not B

!(A || B) = !A && !B

A B !(A||B) !A !B !A&&!B

true true false false false false

true false false false true false

false true false true false false

false false true true true True

Following is the truth table that proves the second DeMorgan's Law.not(A and B) = not A or not B

!(A && B) = !A || ! B

A B !(A&&B) !A !B !A||!B

true true false false false false

true false true false true true

false true true true false true

false false true true true true

Notice that columns with the titles ! (A && B) and ! A || ! B result in the same answers.

• initial roll – compare• subsequent rolls until you won or lost the game:

do –while with the sentinel: while( !((sum == point) || (sum == 7)) );

while( (sum != point) && (sum != 7) );

Proving Demorgan’s LawI

A B not(A or B) not A not B not A and not B

1 1 0 0 0 0

1 0 0 0 1 0

0 1 0 1 0 0

0 0 1 1 1 1

Proving Demorgan’s LawII

A B not(A and B) not A not B not A or not B

1 1

1 0

0 1

0 0