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Boolean Algebra Calculator

Simplify boolean expressions step by step

The calculator will try to simplify/minify the given boolean expression, with steps when possible. Applies commutative law, distributive law, dominant (null, annulment) law, identity law, negation law, double negation (involution) law, idempotent law, complement law, absorption law, redundancy law, de Morgan's theorem. Supports all basic logic operators: negation (complement), and (conjunction), or (disjunction), nand (Sheffer stroke), nor (Peirce's arrow), xor (exclusive disjunction), implication, converse of implication, nonimplication (abjunction), converse nonimplication, xnor (exclusive nor, equivalence, biconditional), tautology (T), and contradiction (F).

It will also find the disjunctive normal form (DNF), conjunctive normal form (CNF), and negation normal form (NNF).

Related calculator: Truth Table Calculator

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Your Input

Simplify the boolean expression (A‾+B)⋅(B‾+C)‾\overline{\left(\overline{A} + B\right) \cdot \left(\overline{B} + C\right)}.

Solution

Apply de Morgan's theorem x⋅y‾=x‾+y‾\overline{x \cdot y} = \overline{x} + \overline{y} with x=A‾+Bx = \overline{A} + B and y=B‾+Cy = \overline{B} + C:

((A‾+B)⋅(B‾+C)‾)=(A‾+B‾+B‾+C‾){\color{red}\left(\overline{\left(\overline{A} + B\right) \cdot \left(\overline{B} + C\right)}\right)} = {\color{red}\left(\overline{\overline{A} + B} + \overline{\overline{B} + C}\right)}

Apply de Morgan's theorem x+y‾=x‾⋅y‾\overline{x + y} = \overline{x} \cdot \overline{y} with x=A‾x = \overline{A} and y=By = B:

(A‾+B‾)+B‾+C‾=(A‾‾⋅B‾)+B‾+C‾{\color{red}\left(\overline{\overline{A} + B}\right)} + \overline{\overline{B} + C} = {\color{red}\left(\overline{\overline{A}} \cdot \overline{B}\right)} + \overline{\overline{B} + C}

Apply the double negation (involution) law x‾‾=x\overline{\overline{x}} = x with x=Ax = A:

((A‾‾)⋅B‾)+B‾+C‾=((A)⋅B‾)+B‾+C‾\left({\color{red}\left(\overline{\overline{A}}\right)} \cdot \overline{B}\right) + \overline{\overline{B} + C} = \left({\color{red}\left(A\right)} \cdot \overline{B}\right) + \overline{\overline{B} + C}

Apply de Morgan's theorem x+y‾=x‾⋅y‾\overline{x + y} = \overline{x} \cdot \overline{y} with x=B‾x = \overline{B} and y=Cy = C:

(A⋅B‾)+(B‾+C‾)=(A⋅B‾)+(B‾‾⋅C‾)\left(A \cdot \overline{B}\right) + {\color{red}\left(\overline{\overline{B} + C}\right)} = \left(A \cdot \overline{B}\right) + {\color{red}\left(\overline{\overline{B}} \cdot \overline{C}\right)}

Apply the double negation (involution) law x‾‾=x\overline{\overline{x}} = x with x=Bx = B:

(A⋅B‾)+((B‾‾)⋅C‾)=(A⋅B‾)+((B)⋅C‾)\left(A \cdot \overline{B}\right) + \left({\color{red}\left(\overline{\overline{B}}\right)} \cdot \overline{C}\right) = \left(A \cdot \overline{B}\right) + \left({\color{red}\left(B\right)} \cdot \overline{C}\right)

Answer

(A‾+B)⋅(B‾+C)‾=(A⋅B‾)+(B⋅C‾)\overline{\left(\overline{A} + B\right) \cdot \left(\overline{B} + C\right)} = \left(A \cdot \overline{B}\right) + \left(B \cdot \overline{C}\right)