1. Statement of the problems. (a) Simplify the expression $ (x + y')(x + z) $ using the laws of Boolean algebra. 3a Let $B$ be a Boolean algebra and $a,b\in B$. Prove that the following statements are equivalent: (i) $a\le b$, (ii) $a\wedge b' = 0$, (iii) $a'\vee b = 1$. b If $A=\begin{bmatrix}1 & 2\\1 & 1\end{bmatrix}$, $B=\begin{bmatrix}3 & 0\\1 & 2\end{bmatrix}$ and $C=\begin{bmatrix}3 & 1\\1 & 2\end{bmatrix}$ find (i) $CB + A^2B$ and (ii) $B^2 + AC$. 2. Solution to (a): simplify $ (x + y')(x + z) $. 1. Start by expanding using the distributive law: $ (x + y')(x + z) = xx + xz + y'x + y'z $. 2. Use idempotence $xx = x$ to get $ x + xz + y'x + y'z $. 3. Apply absorption $x + xz = x$ to remove the $xz$ term, giving $ x + y'x + y'z $. 4. Use absorption again $x + y'x = x$ to remove $y'x$, leaving $ x + y'z $. 5. Therefore the simplified form is $$x + y'z$$. 3. Solution to 3a: equivalence of (i), (ii) and (iii). 1. We prove (i)$\Rightarrow$(ii). Assume $a\le b$, which means $a = a\wedge b$ by definition of the order in a Boolean algebra. 2. Then $a\wedge b' = (a\wedge b)\wedge b' = a\wedge(b\wedge b')$ by associativity and commutativity. 3. Since $b\wedge b' = 0$, we have $a\wedge(b\wedge b') = a\wedge 0 = 0$. 4. Thus (i) implies (ii). 5. Next prove (ii)$\Rightarrow$(i). Assume $a\wedge b' = 0$. 1. Use the distributive law with $b\vee b' = 1$: $a = a\wedge 1 = a\wedge(b\vee b') = (a\wedge b)\vee(a\wedge b')$. 2. By assumption $a\wedge b' = 0$, so $a = (a\wedge b)\vee 0 = a\wedge b$. 3. Hence $a = a\wedge b$, which means $a\le b$, so (ii) implies (i). 6. Now show equivalence between (ii) and (iii) using De Morgan. 1. Note De Morgan: $(a\wedge b')' = a'\vee b$. 2. If (ii) holds then $a\wedge b' = 0$, so taking complements gives $(a\wedge b')' = 0' = 1$. 3. By De Morgan that means $a'\vee b = 1$, so (ii)$\Rightarrow$(iii). 4. Conversely, if $a'\vee b = 1$ then $(a\wedge b')' = 1$, so $a\wedge b' = 0$ on taking complements again, giving (iii)$\Rightarrow$(ii). 7. Combining the implications we have (i)$\Leftrightarrow$(ii)$\Leftrightarrow$(iii), as required. 4. Solution to b: matrix computations with $A,B,C$ as given. 1. Compute $A^2 = A\cdot A$. $A = \begin{bmatrix}1 & 2\\1 & 1\end{bmatrix}$, so $A^2 = \begin{bmatrix}1\cdot 1 + 2\cdot 1 & 1\cdot 2 + 2\cdot 1\\1\cdot 1 + 1\cdot 1 & 1\cdot 2 + 1\cdot 1\end{bmatrix} = \begin{bmatrix}3 & 4\\2 & 3\end{bmatrix}$. 2. Compute $CB = C\cdot B$. $C = \begin{bmatrix}3 & 1\\1 & 2\end{bmatrix}$, $B = \begin{bmatrix}3 & 0\\1 & 2\end{bmatrix}$, so $CB = \begin{bmatrix}3\cdot 3 + 1\cdot 1 & 3\cdot 0 + 1\cdot 2\\1\cdot 3 + 2\cdot 1 & 1\cdot 0 + 2\cdot 2\end{bmatrix} = \begin{bmatrix}10 & 2\\5 & 4\end{bmatrix}$. 3. Compute $A^2B = (A^2)\cdot B$. $A^2 = \begin{bmatrix}3 & 4\\2 & 3\end{bmatrix}$, so $A^2B = \begin{bmatrix}3\cdot 3 + 4\cdot 1 & 3\cdot 0 + 4\cdot 2\\2\cdot 3 + 3\cdot 1 & 2\cdot 0 + 3\cdot 2\end{bmatrix} = \begin{bmatrix}13 & 8\\9 & 6\end{bmatrix}$. 4. Now compute (i) $CB + A^2B$ by entrywise addition. $CB + A^2B = \begin{bmatrix}10 & 2\\5 & 4\end{bmatrix} + \begin{bmatrix}13 & 8\\9 & 6\end{bmatrix} = \begin{bmatrix}23 & 10\\14 & 10\end{bmatrix}$. 5. Next compute $B^2 = B\cdot B$. $B^2 = \begin{bmatrix}3 & 0\\1 & 2\end{bmatrix}\cdot\begin{bmatrix}3 & 0\\1 & 2\end{bmatrix} = \begin{bmatrix}3\cdot 3 + 0\cdot 1 & 3\cdot 0 + 0\cdot 2\\1\cdot 3 + 2\cdot 1 & 1\cdot 0 + 2\cdot 2\end{bmatrix} = \begin{bmatrix}9 & 0\\5 & 4\end{bmatrix}$. 6. Compute $AC = A\cdot C$. $AC = \begin{bmatrix}1 & 2\\1 & 1\end{bmatrix}\cdot\begin{bmatrix}3 & 1\\1 & 2\end{bmatrix} = \begin{bmatrix}1\cdot 3 + 2\cdot 1 & 1\cdot 1 + 2\cdot 2\\1\cdot 3 + 1\cdot 1 & 1\cdot 1 + 1\cdot 2\end{bmatrix} = \begin{bmatrix}5 & 5\\4 & 3\end{bmatrix}$. 7. Finally compute (ii) $B^2 + AC$ by entrywise addition. $B^2 + AC = \begin{bmatrix}9 & 0\\5 & 4\end{bmatrix} + \begin{bmatrix}5 & 5\\4 & 3\end{bmatrix} = \begin{bmatrix}14 & 5\\9 & 7\end{bmatrix}$. 5. Final answers. (a) Simplified Boolean expression: $$x + y'z$$. (3a) The three statements (i) $a\le b$, (ii) $a\wedge b' = 0$, and (iii) $a'\vee b = 1$ are equivalent, with the proof given above. (b)(i) $CB + A^2B = \begin{bmatrix}23 & 10\\14 & 10\end{bmatrix}$. (b)(ii) $B^2 + AC = \begin{bmatrix}14 & 5\\9 & 7\end{bmatrix}$. Venn diagram note: two overlapping circles inside a rectangle with the left circle labeled "2a" and the right circle labeled "2a" and the overlap region unlabeled, positioned top-right.