The simplified form of the Boolean expression F = A(B+C(AB+AC)’) is

The given Boolean expression is F = A(B+C(AB+AC)’).

Let's simplify it step by step:

Step 1: Apply De Morgan's law to the expression inside the brackets. De Morgan's law states that the complement of a conjunction is the disjunction of the complements, and vice versa. So, (AB+AC)’ becomes A’B’ + A’C’.

Step 2: Now, the expression becomes F = A(B+C(A’B’ + A’C’)).

Step 3: Distribute C over the expression inside the brackets. This gives us F = A(B + CA’B’ + CA’C’).

Step 4: Notice that CA’B’ + CA’C’ simplifies to CA’B’ because in Boolean algebra, A.A = A. So, the expression becomes F = A(B + CA’B’).

Step 5: Distribute A over the expression inside the brackets. This gives us F = AB + ACA’B’.

Step 6: Notice that ACA’B’ simplifies to 0 because A.A’ = 0 in Boolean algebra. So, the expression becomes F = AB.

So, the simplified form of the given Boolean expression is F = AB.