Implicant

Also called: implicants, implicant of a function

A product term that is 1 only on rows where the function is 1. Whenever the implicant is 1, the function is guaranteed to be 1.

A product term P is an implicant of a function F if P = 1 forces F = 1. In other words, every row that P covers is a 1 of F. The name comes from logic: P implies F.

A helpful picture: each product term is a patch laid over some rows of the truth table. An implicant is a safe patch, one that lands only on 1s. It does not have to cover every 1; it just must not cover a 0.

Examples for F(A, B, C) = Σm(0, 4, 5, 7):

Every minterm of F is an implicant: covers only row 5.
covers rows 4 and 5, both 1s: an implicant.
covers rows 4–7, and row 6 is a 0: not an implicant.

Every term of a correct SOP for F is an implicant, since a term that covered a 0 would make F wrong there. Combining adjacent minterms always produces implicants.

The interesting implicants are the biggest ones, which cannot grow without hitting a 0: the prime implicants.

A\BC	00	01	11	10
0	1m0	0m1	0m3	0m2
1	1m4	1m5	1m7	0m6

Worked example

Example

Testing candidate terms

F(A, B, C) = Σm(0, 4, 5, 7). Is each term an implicant: , , ?

1.
covers 1-1: rows 5 and 7. Both are 1s, so yes.
2.
covers -00: rows 0 and 4. Both are 1s, so yes.
3.
covers 11-: rows 6 and 7. Row 6 is a 0, so no.

Common mistakes

Thinking an implicant must cover all of F's 1s. It only must avoid F's 0s.
Believing that removing a literal keeps a term an implicant. Removing a literal doubles the rows covered and may take in a 0.
Forgetting the reverse is safe: adding a literal to an implicant always gives another implicant, because it covers fewer rows.

Practice Implicant

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Learn it step by step

Implicant is taught in Boolean Simplification and Karnaugh Maps.

Boolean Simplification →Karnaugh Maps →

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