DeMorgans Laws

Warm-upA Little Boolean Review

DeMorgan’s Laws:!(P&&Q) = ___________________!(P || Q) = ____________________


DeMorgan’s Laws:!(P&&Q) = !P || !Q!(P || Q) = !P && !Q


! (x > = y ) = _______________

! ( x > y ) = _______________


Use DeMorgan’s Law to change these:

! (x > = y && a < b ) = ____________

! ( x >= y || a >= b ) = _____________


Use DeMorgan’s Law to change these:

! (x > = y && a < b ) = x<y || a>=b

! ( x >= y || a >= b ) = x<y || a<b

More Boolean!

(y>10000) || ( x>1000 && x<1500 )

a. (y>10000 || x>1000) && (y>10000 || x<1500)

b. (y>10000 || x>1000) || (y>10000 || x<1500)

c. (y>10000) && (x>1000 || x<1500 )

d. (y>10000 && x>1000) || (y>10000 && x<1500)

e. (y > 10000 && x>1000) && (y>10000 && x<1500)

Use the old distributive property:

(y>10000) || ( x>1000 && x<1500 )

A * (B + C) = A * B + A * C

a. (y>10000 || x>1000) && (y>10000 || x<1500)

b. (y>10000 || x>1000) || (y>10000 || x<1500)

c. (y>10000) && (x>1000 || x<1500 )

d. (y>10000 && x>1000) || (y>10000 && x<1500)

e. (y > 10000 && x>1000) && (y>10000 && x<1500)


(y>10000) || ( x>1000 && x<1500 )

A * (B + C) = A * B + A * C

(y>10000 || x>1000)

a. (y>10000 || x>1000) && (y>10000 || x<1500)

b. (y>10000 || x>1000) || (y>10000 || x<1500)

c. (y>10000) && (x>1000 || x<1500 )

d. (y>10000 && x>1000) || (y>10000 && x<1500)

e. (y > 10000 && x>1000) && (y>10000 && x<1500)


(y>10000) || ( x>1000 && x<1500 )

A * (B + C) = A * B + A * C

(y>10000 || x>1000) &&

a. (y>10000 || x>1000) && (y>10000 || x<1500)

b. (y>10000 || x>1000) || (y>10000 || x<1500)

c. (y>10000) && (x>1000 || x<1500 )

d. (y>10000 && x>1000) || (y>10000 && x<1500)

e. (y > 10000 && x>1000) && (y>10000 && x<1500)


(y>10000) || ( x>1000 && x<1500 )

A * (B + C) = A * B + A * C

(y>10000 || x>1000) && (y>10000 || x<1500)

a. (y>10000 || x>1000) && (y>10000 || x<1500)

b. (y>10000 || x>1000) || (y>10000 || x<1500)

c. (y>10000) && (x>1000 || x<1500 )

d. (y>10000 && x>1000) || (y>10000 && x<1500)

e. (y > 10000 && x>1000) && (y>10000 && x<1500)

(x==y && x!=z) || (x!=y && x==z)

(x==y || x==z) && (x!=y || x!= z)

(x==y) != (x==z)Let’s see what happens if the last statement is true.

(x==y && x!=z) || (x!=y && x==z)

(x==y || x==z) && (x!=y || x!= z)

(x==y) != (x==z)Let’s see what happens if the last statement is true.

The value of the left side and the right side can’t be the same. They must be “Not Equal”. One side must be true and the other side must be false.

(x==y && x!=z) || (x!=y && x==z)

(x==y || x==z) && (x!=y || x!= z)

(x==y) != (x==z)Since x = y or x = z, but x doesn’t equal y and z. If all three are equal, then both the left and right would be true, and would not meet the != condition

(x==y && x!=z) || (x!=y && x==z)

(x==y || x==z) && (x!=y || x!= z)

(x==y) != (x==z)In other words, they can’t all be the same.

Furthermore, if all three are different, then both the left and right side would be false, which means they don’t meet the != condition.

(x==y && x!=z) || (x!=y && x==z)

(x==y || x==z) && (x!=y || x!= z)

(x==y) != (x==z)Conclusion: x is equal to either y or z. Does this condition make the two statements above true?

(x==y && x!=z) || (x!=y && x==z)

(x==y || x==z) && (x!=y || x!= z)

(x==y) != (x==z)

Knowing this, let’s look at this statement.

Do the same conditions make it true?

(x==y && x!=z) || (x!=y && x==z)

(x==y || x==z) && (x!=y || x!= z)

(x==y) != (x==z)

If x equals y and does not equal z, the left side is true

Since the operator is OR, the statement is true no matter what is on the right side.

(x==y && x!=z) || (x!=y && x==z)

(x==y || x==z) && (x!=y || x!= z)

(x==y) != (x==z)

If x equals z and does not equal y, the right side is true

Since the operator is OR, the statement is true no matter what is on the left side.

(x==y && x!=z) || (x!=y && x==z)

(x==y || x==z) && (x!=y || x!= z)

(x==y) != (x==z)

So, the third and first meet the same conditions

when evaluating to “true”. How about the middle?

(x==y && x!=z) || (x!=y && x==z)

(x==y || x==z) && (x!=y || x!= z)

(x==y) != (x==z)

If x equals either ‘y’ or ‘z’, the left is true. If the ‘x’

is different than either ‘y’ or ‘z’, the right is true.

(x==y && x!=z) || (x!=y && x==z)

(x==y || x==z) && (x!=y || x!= z)

(x==y) != (x==z)

So given the exact conditions again, that x equals

either y or z, but not both, the middle statement is true.

(x==y && x!=z) || (x!=y && x==z)

(x==y || x==z) && (x!=y || x!= z)

(x==y) != (x==z)You may figure out the condition that make the last statement false and follow the same steps to compare those conditions to the other two statements

(x==y && x!=z) || (x!=y && x==z)

(x==y || x==z) && (x!=y || x!= z)

(x==y) != (x==z)Hint: This statement is false only if

* all three are the same, * all three are different, * or y = z but not x.

De Morgan’s Laws

• !(a&&b) = !a || !b

• !(a || b) = !a && !b

Using DeMorgans LawsThe expression !((x<=y) && (y>5)) is

equivilant to which of the following?

• A. (x<=y) && (y>5)

• B. (x<=y) || (y>5)

• C. (x>=y) || (y<5)

• D. (x>y) || (y<=5)

• E. (x>y) && (y<=5)

Using DeMorgans LawsThe expression !(x>y && y<=0) is equivilant

to which of the following?

• A. !(x<=y) || (y>0)

• B. x>y && y<=0

• C. x<=y || y>0

• D. x>y || y<0

• E. x<=y && y<=0

Given that x is true, y is true, and z is false, which of the following expressions will evaluate to false?

A. (x&&y) || zB. (x || y) && zC. y || (x&&z)D. x || (y && z)E. x && (y || z)

Assuming that c and d are boolean variables, the expression !c || d is equivalent to which of the following?

• A. !(c && d)

• B. !(c && !d)

• C. c && !d

• D. !(c || !d)

• E. !(!c && d)

Assuming that a and b are boolean when is the following expression true:

!(!a || b) || (!a && b)

• A. If and only if a and b have different values

• B. If and only if a and b have the same value

• C. If and only if both a and b are true

• D. If and only if both a and b are false

• E. None

DeMorgans Laws

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