Consensus theorem: AB + A'C + BC = AB + A'C, with proof
The consensus theorem says that the consensus term of two terms is redundant: (A AND B) OR (NOT A AND C) OR (B AND C) = (A AND B) OR (NOT A AND C).
The third term, B AND C, is the consensus of the first two: it is what you get by taking the terms that contain a variable and its negation (A and NOT A), removing that variable, and combining what is left. Whenever B AND C is true, either A is true (so A AND B holds) or A is false (so NOT A AND C holds), so the term adds nothing and can be dropped.
The theorem matters in two directions. Dropping a consensus term is the step Karnaugh maps perform when a group is covered by two others, and it is how a cover is reduced to its essential prime implicants. Adding a consensus term, the other way round, is a way to find a shared factor that then allows a different simplification, which is why some hand simplifications first add B AND C and then remove something else.
The engine has no named consensus rule, and its simplifier, which applies one local law at a time, leaves the three-term form as it is: removing the term is not a local rewrite. The proof below is an exact equivalence check on the eight rows over A, B and C, and the worked example compares the two forms row by row.
Statement
(A AND B) OR (NOT A AND C) OR (B AND C) = (A AND B) OR (NOT A AND C) Proof by truth table: (A AND B) OR (NOT A AND C) OR (B AND C) = (A AND B) OR (NOT A AND C)
| # | A | B | C | lhs | rhs |
|---|---|---|---|---|---|
| 0 | FALSE | FALSE | FALSE | FALSE | FALSE |
| 1 | FALSE | FALSE | TRUE | TRUE | TRUE |
| 2 | FALSE | TRUE | FALSE | FALSE | FALSE |
| 3 | FALSE | TRUE | TRUE | TRUE | TRUE |
| 4 | TRUE | FALSE | FALSE | FALSE | FALSE |
| 5 | TRUE | FALSE | TRUE | FALSE | FALSE |
| 6 | TRUE | TRUE | FALSE | TRUE | TRUE |
| 7 | TRUE | TRUE | TRUE | TRUE | TRUE |
The same law in four notations
(A AND B) OR (NOT A AND C) OR (B AND C)(A && B) || (!A && C) || (B && C)(A ∧ B) ∨ (¬A ∧ C) ∨ (B ∧ C)(A·B) + (A'·C) + (B·C)\left(A \cdot B\right) + \left(\overline{A} \cdot C\right) + \left(B \cdot C\right)(A AND B) OR (NOT A AND C)(A && B) || (!A && C)(A ∧ B) ∨ (¬A ∧ C)(A·B) + (A'·C)\left(A \cdot B\right) + \left(\overline{A} \cdot C\right)The common slip
Dropping a term that is not a consensus. (A AND B) OR (NOT A AND C) OR (B AND D) keeps all three terms; B AND D is not the consensus of the first two because D is not C.
Questions
Does the theorem have a product-of-sums form?
Yes, its dual: (A OR B) AND (NOT A OR C) AND (B OR C) = (A OR B) AND (NOT A OR C). The equivalence checker proves it in the same way.
Related
Lesson that teaches this
Pitfalls
Tools
- Boolean algebra calculator and simplifier with steps (every law named and verified)
- Logical equivalence checker: two expressions, exact verdict, counterexample
Practice
- Boolean simplification exercises with answers: basic
- Boolean simplification exercises with answers: intermediate
- Boolean simplification exercises with answers: advanced
See also
- Consensus in practice: AB + A'C + BC simplified
- Absorption law: A OR (A AND B) = A, with proof
- Distributive law: A AND (B OR C) = (A AND B) OR (A AND C), both forms
- Prime implicants and essential prime implicants on a K-map
Laws
Generated from the Tangibool engine, semantics tangibool-semantics/1, engine 1.0.0, build . Every table, map, simplification and verdict on these pages was computed by the Tangibool engine when the site was built, and every claim was re-checked by the same exact checker the app uses. Limits are stated where they apply. © 2026 Foreranger LTD.