Simply the following using de morgans law: a) F = xy + x'y' + y b) F = (xy + x'y') (x+y'z')

Here's how to simplify the given Boolean expressions using De Morgan's Law: a) Simplify F = xy + x'y' + y Step 1: Take the complement of the function F. F' = (xy + x'y' + y)' Step 2: Apply De Morgan's Law (A+B+C)' = A'B'C'. F' = (xy)' · (x'y')' · y' Step 3: Apply De Morgan's Law again to (xy)' = x'+y' and (x'y')' = x''+y'' = x+y. F' = (x'+y') · (x+y) · y' Step 4: Expand the product (x'+y')(x+y) and then multiply by y'. (x'+y')(x+y) = x'x + x'y + y'x + y'y = 0 + x'y + xy' + 0 = x'y + xy' Now, multiply by y': F' = (x'y + xy') · y' = x'yy' + xy'y' Apply the Complement Law (yy'=0) and Idempotent Law (y'y'=y'): F' = x'(0) + x(y') = 0 + xy' = xy' Step 5: Take the complement of F' to find F using De Morgan's Law (AB)' = A'+B'. F = (xy')' = x' + (y')' = x' + y The simplified expression is: F = x' + y b) Simplify F = (xy + x'y)(x+yz') Step 1: Simplify the initial expression using Distributive and Complement Laws. (xy + x'y) = y(x+x') = y(1) = y The expression becomes: F = y(x+yz') Distribute y and apply the Idempotent Law (yy=y): F = yx + y(yz') = xy + yy z' = xy + yz' Step 2: Take the complement of F and apply De Morgan's Law (A+B)' = A'B'. F' = (xy + yz')' = (xy)' · (yz')' Step 3: Apply De Morgan's Law again to (xy)' = x'+y' and (yz')' = y'+(z')' = y'+z. F' = (x'+y')(y'+z) Step 4: Expand and simplify F'. F' = x'y' + x'z + y'y' + y'z Apply Idempotent Law (y'y'=y') and Absorption Law (A+AB=A): F' = x'y' + x'z + y' + y'z F' = y'(x'+1) + x'z + y'z = y'(1) + x'z + y'z = y' + x'z + y'z F' = y'(1+z) + x'z = y'(1) + x'z = y' + x'z Step 5: Take the complement of F' to find F using De Morgan's Law (A+B)' = A'B'. F = (y' + x'z)' = (y')' · (x'z)' Step 6: Apply De Morgan's Law again to (x'z)' = (x')' + z' = x+z'. F = y · (x+z') Distribute y: F = yx + yz' = xy + yz' The simplified expression is: F = xy + yz' That's 2 down. 3 left today — send the next one.