Boolean Simplification of AB + A'+B'

Question

Is there any way to simplify this function? Or is this the simplest equation? : AB + A'+B'

Answer 1

A general technique in boolean algebra is to add or multiply some term by "$P+P$'" to make it similar to other terms so that they can be simplified. In this case:

$AB+A'(B+B')+B'$

$=(A+A')B+A'B'+B'$

$=B+A'B'+(A+A')B'$

$=B+A'B'+AB'+A'B'$

$=B+A'B'+AB'$

$=B+(A+A')B'$

$=B+B'$.

But in complicated cases it may not simplify immediately.

Note that this solution uses only identity, inverse, idempotency and distributivity and so avoids using De Morgan's laws unlike the other solutions. De Morgan's laws, which are extremely useful of course, can be proven in a similar way if desired. Alternatively one can use truth tables to verify all basic rules.

Answer 2

F = AB + A'+B'
F' = (A'+B').AB = A'AB + ABB' = 0

Again if I take F = F'' = 1 , 1 will the the result.

Answer 3

$$AB+A'+B' = AB+ \overline{AB} = 1$$

Answer 4

$$0\cdot0+1+1=1\\1\cdot0+0+1=1\\0\cdot1+1+0=1\\1\cdot1+0+0=1$$

Answer 5

AB + A' + B'

Expand missing terms.

AB + A' (B' + B) + B' (A' + A) - Complement Law

AB + A'B' + A'B + A'B' + AB'

Eliminate duplicates and rearrange.

A'B' + A'B + AB' + AB

A' (B' + B) + A (B' + B) - Complement Law

A' + A

1

http://www.mi.mun.ca/users/cchaulk/misc/boolean.htm