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SOP/POS forms and canonical form of A+BC+ABC

Short answer

SOP is OR of AND terms; POS is AND of OR terms. Canonical SOP: F = Σm(3,4,5,6,7). Canonical POS: F = ΠM(0,1,2).

Define SOP and POS form and convert F = A + BC + ABC into its canonical form.

SOP and POS forms

  • Sum of Products (SOP): a logical sum (OR) of product (AND) terms, e.g. F=AB+A‾CF = AB + \overline{A}C. It is built as an AND-OR (or NAND-NAND) circuit. If every term has all the variables, it is the canonical SOP (sum of minterms).
  • Product of Sums (POS): a logical product (AND) of sum (OR) terms, e.g. F=(A+B)(A‾+C)F = (A+B)(\overline{A}+C). It is built as an OR-AND (or NOR-NOR) circuit. If every term has all the variables, it is the canonical POS (product of maxterms).

Canonical form of F=A+BC+ABCF = A + BC + ABC

Expand each term to contain A, B and C:

A=A(B+B‾)(C+C‾)=ABC+ABC‾+AB‾C+AB‾ C‾BC=BC(A+A‾)=ABC+A‾BCABC=ABC\begin{aligned} A &= A(B+\overline{B})(C+\overline{C}) = ABC + AB\overline{C} + A\overline{B}C + A\overline{B}\,\overline{C} \\ BC &= BC(A+\overline{A}) = ABC + \overline{A}BC \\ ABC &= ABC \end{aligned}

Add them and keep each term once (X+X=XX+X=X):

F=A‾BC+AB‾ C‾+AB‾C+ABC‾+ABCF = \overline{A}BC + A\overline{B}\,\overline{C} + A\overline{B}C + AB\overline{C} + ABC
TermABCMinterm
A‾BC\overline{A}BC011m3m_3
AB‾ C‾A\overline{B}\,\overline{C}100m4m_4
AB‾CA\overline{B}C101m5m_5
ABC‾AB\overline{C}110m6m_6
ABCABC111m7m_7

Canonical SOP:

F=Σm(3,4,5,6,7)F = \Sigma m(3,4,5,6,7)

Canonical POS: the missing numbers 0, 1, 2 are the maxterms:

F=ΠM(0,1,2)=(A+B+C)(A+B+C‾)(A+B‾+C)F = \Pi M(0,1,2) = (A+B+C)(A+B+\overline{C})(A+\overline{B}+C)

(The given function itself simplifies to F=A+BCF = A + BC, since ABCABC is absorbed by AA.)

✓ Answer checked by Claude Opus 5.5 (high effort)

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