Exercise DET-02.05 – Boolean algebra

Prof. Dr. Thorsten Jungmann (last updated 2026-10-13)

Relates to DET-02 Logic operations

Task

With the rules of Boolean algebra, you can transform logic expressions without drawing a circuit. Give all results in the notation of electrical engineering that the lesson uses. Subtasks b) and d) are based on exercises in the textbook by LaMeres [5, p. 146].

a) Write the following expressions in the notation of electrical engineering: $\neg A\wedge (B\vee C)$ from mathematics, !(A && B) || C from computer science and $\neg (A\vee \neg B)\wedge C$ from mathematics.

b) Name the theorem of Boolean algebra that each of the following statements describes, and write it as an equation:

• (1) A variable combined with itself by OR yields the variable itself.
• (2) A variable that is negated twice yields the variable itself.
• (3) A variable combined with 1 by OR yields 1.
• (4) A variable that appears in several product terms can be factored out.
• (5) A term that another term of the expression already covers can be removed.
• (6) An AND gate with inverted inputs is equivalent to an OR gate with an inverted output.

c) Insert all parentheses into $Y_1=\overline{A}+B\cdot \overline{C}$ that show the order in which the expression is evaluated. Then evaluate $Y_1$ for $A=0$, $B=0$ and $C=1$.

d) Prove with a proof by exhaustion that an AND gate with inverted inputs is equivalent to an OR gate with an inverted output: $\overline{A}\cdot \overline{B}=\overline{A+B}$.

e) Prove with a truth table that $A+B\cdot C=(A+B)\cdot (A+C)$. With OR, the distributive property has no counterpart in ordinary arithmetic.

f) Simplify $Y_2=A\cdot \left(\overline{A}+B\right)$ step by step, and name the theorem that you use in each step.

g) Simplify $Y_3=A\cdot B+A\cdot B\cdot C+A\cdot \overline{B}$ step by step, and name the theorem that you use in each step.

h) Simplify $Y_4=\overline{\overline{A}+B}+\overline{A+B}$ step by step, and name the theorem that you use in each step.

Solution

a) $\overline{A}\cdot (B+C)$; $\overline{A\cdot B}+C$; $\overline{A+\overline{B}}\cdot C$
b) (1) idempotence, $A+A=A$; (2) involution, $\overline{\overline{A}}=A$; (3) null element, $A+1=1$; (4) distributive property, $A\cdot B+A\cdot C=A\cdot (B+C)$; (5) absorption, $A+A\cdot B=A$; (6) De Morgan’s laws, $\overline{A}\cdot \overline{B}=\overline{A+B}$
c) $Y_1=\left(\overline{A}\right)+\left(B\cdot \left(\overline{C}\right)\right)$; for $A=0$, $B=0$ and $C=1$: $Y_1=1$
d) The columns $\overline{A}\cdot \overline{B}$ and $\overline{A+B}$ agree in all four rows.
e) The columns $A+B\cdot C$ and $(A+B)\cdot (A+C)$ agree in all eight rows.
f) $Y_2=A\cdot B$
g) $Y_3=A$
h) $Y_4=\overline{B}$

▶️ Full worked solution