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Boolean Algebra Simplifier

Minimise any logic expression to its simplest SOP and POS forms

Updated · Free, no signup

Simplified (sum of products)

AB + C

Simplified (product of sums)

(A + C)(B + C)

Minterms

Σm(1, 3, 5, 6, 7)

Variables

A, B, C

Literals in your expression

15

Literals in simplified SOP

3

Gates for the SOP circuit

1 AND, 1 OR, 0 NOT

Prime implicants found

2

Your expression (parsed)

A'·B'·C + A'·B·C + A·B'·C + A·B·C + A·B·C'
  • Simplification cuts the expression from 15 to 3 literals.

Prime implicants (Quine–McCluskey)

PatternTermCovers mintermsStatus
11-AB6, 7Essential
--1C1, 3, 5, 7Essential

About the Boolean Algebra Simplifier

This Boolean algebra simplifier takes a logic expression and reduces it to the simplest equivalent form. Type something like A'B'C + A'BC + AB'C + ABC + ABC' or !(A & B) | (A & !C) and it returns the minimal sum of products (SOP), the minimal product of sums (POS), the minterm list, the prime implicants it found and how many literals and gates the simplification saves.

It is meant for digital electronics students checking homework, engineers reducing the gate count of combinational logic before building it in TTL, CMOS or an FPGA, and programmers tidying complicated if-conditions. The result can be written in textbook notation (AB' + C), with dots (A·B' + C), in programming syntax (A & !B | C) or in words.

Rather than applying algebra rules one at a time (which can stall in a local minimum), the tool builds the full truth table and runs the Quine–McCluskey algorithm, then picks the smallest set of prime implicants that covers every minterm. That guarantees the result is equivalent to your input and minimal in number of terms; when several minimal answers exist, one of them is shown. Up to eight single-letter variables are supported.

How to use the boolean algebra simplifier

  1. 1Type your Boolean expression using single-letter variables (A', !A or NOT A for complements).
  2. 2Choose how you want the answer written.
  3. 3Read the minimal sum of products and product of sums.
  4. 4Check the prime implicant table to see which terms were essential.
  5. 5Compare the literal and gate counts to see what the simplification saves.

Formula and method

Combine: XY + XY' = X · Absorption: X + XY = X · De Morgan: (XY)' = X' + Y', (X + Y)' = X'Y'

Every simplification in Boolean algebra rests on a few identities: complementary terms merge (XY + XY' = X), absorption removes redundant terms (X + XY = X), and De Morgan's laws move a NOT through AND and OR. The Quine–McCluskey method applies the merging rule systematically: it lists all minterms, repeatedly combines pairs that differ in exactly one variable, and keeps the terms that can no longer be combined — the prime implicants.

It then selects essential prime implicants (the only ones covering some minterm) and searches for the smallest set of the rest that covers every remaining minterm, preferring fewer literals on ties. The minimal product of sums is found the same way from the zeros of the function and converted with De Morgan's law. Gate counts assume two-level logic with one multi-input OR gate and one inverter per complemented variable.

X'
NOT X (complement)
XY
X AND Y (product)
X + Y
X OR Y (sum)
Prime implicant
A product term that cannot be enlarged by dropping a variable
Pattern
1 = variable, 0 = complemented, – = eliminated

Worked examples

Five minterms that reduce to two terms

The four terms containing C cover every combination of A and B, so they merge into C. ABC' combines with ABC to give AB. The result AB + C uses 3 literals instead of 15; its POS form is (A + C)(B + C).

Absorption law: A(A + B)

Expanding gives AA + AB = A + AB, and A + AB = A by absorption. The output simply follows A, so B is redundant.

XOR hidden in a product of sums

Expanding gives AA' + AB' + A'B + BB'. AA' and BB' are always 0, leaving AB' + A'B — the exclusive-OR of A and B, true when exactly one input is 1.

Code-style input: !(A & B) | (A & !C)

!(A & B) is false only when A and B are both 1; then A & !C adds the case C = 0. The output is 0 only for A = B = C = 1, so the function is a three-input NAND: !A | !B | !C.

Four corners of a 4-variable K-map

All four terms share B' and D' while A and C take every combination, so they merge into the single term B'D' — the classic 'corners' group on a Karnaugh map.

Frequently asked questions

How do you simplify a Boolean expression?+

Apply identities such as XY + XY' = X, X + XY = X and De Morgan's laws, or use a Karnaugh map for up to four or five variables. For a guaranteed minimal result, the Quine–McCluskey algorithm used here combines minterms systematically and picks the fewest prime implicants.

What is the difference between SOP and POS?+

Sum of products (SOP) is an OR of AND terms, like AB + C, and maps to AND–OR or NAND–NAND logic. Product of sums (POS) is an AND of OR terms, like (A + C)(B + C), and maps to OR–AND or NOR–NOR logic. Both can describe the same function.

What is a prime implicant?+

A prime implicant is a product term that implies the function and cannot be made smaller by removing a variable. An essential prime implicant is the only one covering some minterm, so it must appear in every minimal solution.

Can a Boolean expression have more than one simplest form?+

Yes. Some functions have several different minimal SOPs with the same number of terms and literals, for example when the remaining minterms can be covered by different prime implicants. Any of them is equally correct; this tool shows one.

What notation does the simplifier accept?+

Single-letter variables with AND (AB, A·B, A*B, A & B, AND), OR (A + B, A | B, OR), NOT (A', !A, ~A, NOT A), XOR (^, ⊕, XOR), NAND, NOR, XNOR, implication (->) and parentheses. Constants 0 and 1 are allowed.

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