Put $f(A,B,C) = A+B'C$ in $Σ$ $\pi$ notation

Question

As far as I know I need first to convert $A+B'C$ to canonical disjunctive form and canonical conjunctive form, but I think I'm doing it wrong.

my attempt

conjunctive form.

$A+B'C$

$A(B'C+BC')+B'C(A+A')$

$AB'C+ABC'+AB'C+A'B'C$

$AB'C+ABC'+A'BC$

disjunctive form

$(A+B'+C)(A+B+C')(A'+B+C)$

But as I have 3 variables I will have $2^3$ lines in the truth table, but the 2 forms have only 3 minterms and maxterms, There are 2 lines left.

Answer 1

You have gotten CNF and DNF mixed up. Using $A+B'C$ (aka $A\lor (\lnot B \land C)$)

$A+B'C = (A+B')(A+C)$ is in conjunctive normal form.

• Use distributive property of "or" over "and" to get $A+B'C = (A+B')(A+C)$
• Conjunctive normal form is often referred to as "Products of Sums".

$A+B'C$ is already in disjunctive normal form..

• Disjunctive normal form is often referred to as "Sums of Products". In the above case note that $A= 1A$, and hence can be seen as a product, and clearly, then $A + B'C = 1A+B'C$ is clearly a sum of two products.

In the Wikipedia links I provide, note that $+$ is $\lor$, and concantenation or $*$ is $\land$. Also, instead of $B'$, for example, it uses $\lnot B$.

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To try to address your comment, we can read right off the truth table for $A + B'C$ to find the disjunctive normal form you seek.

Disjunctive normal form:

In which cases is $A+(B'C)$ true? Here we just read off the truth values under A, B, C, for which $A+B'C$ is true. Note, it is true in any case $A$ is true, and it is also true when $A$ is false, but $B'C$ is true.

This yields the sum of products: $ABC + ABC' + AB'C+ AB'C' + A'B'C$.