Given the Boolean expression: \[ F = (x + y' + z')(x + y' + z)(x + y + z') \] First, combine the first two parenthetical terms. We use the distributive law \( (A+B)(A+C) = A + BC \), where \( A = (x+y') \), \( B = z' \), and \( C = z \): \[ (x + y' + z')(x + y' + z) = (x + y') + (z' \cdot z) \] Since the logical AND of a variable and its complement is 0 (i.e., \( z' \cdot z = 0 \)), this simplifies to: \[ x + y' \] Now, substitute this result back into the full expression and multiply it by the third term: \[ F = (x + y')(x + y + z') \] Apply the distributive law again: \( (x + A)(x + B) = x + AB \), where \( A = y' \) and \( B = (y + z') \): \[ F = x + (y' \cdot (y + z')) \] Expand the inner parenthesis: \[ y' \cdot (y + z') = (y' \cdot y) + (y' \cdot z') \] Since \( y' \cdot y = 0 \), this leaves us with: \[ 0 + y'z' = y'z' \] Substitute this final result back into the expression for F: \[ F = x + y'z' \]