Logic gates and Boolean algebra: the gates AND, OR, NOT, NAND, NOR, XOR and their truth tables, Boolean expressions, the laws of Boolean algebra, De Morgan's laws, and universal gates.

Simplification (up to 3 marks): factor out $A$ by the distributive law: $A \cdot B + A \cdot \overline{B} = A \cdot (B + \overline{B})$. By the complement law $B + \overline{B} = 1$, so this is $A \cdot 1$. By the identity law $A \cdot 1 = A$. The simplified expression is $A$.

Verification (up to 2 marks): a truth table for $A,B = 00,01,10,11$ gives $A \cdot B = 0,0,0,1$ and $A \cdot \overline{B} = 0,0,1,0$, so the OR is $0,0,1,1$, which is exactly the column for $A$.

Markers reward the named distributive, complement and identity laws and a truth table whose output equals $A$.

De Morgan's laws (up to 2 marks): $\overline{A + B} = \overline{A} \cdot \overline{B}$ and $\overline{A \cdot B} = \overline{A} + \overline{B}$. In words: break the bar over the bracket and swap the operator.

Universal NAND (up to 3 marks): NAND is universal because every other gate can be made from it. A NOT is a NAND with both inputs tied together. An AND is a NAND followed by a NAND-inverter. An OR is built by inverting both inputs (each with a NAND) and feeding them into a NAND, since $\overline{\overline{A} \cdot \overline{B}} = A + B$ by De Morgan. Combining these gives a NOR as an OR followed by a NAND-inverter.

Markers reward both De Morgan's laws stated correctly, and a valid construction showing NAND can make NOT, AND and OR (hence any gate, including NOR).