Minterms and Maxterms

A minterm is an AND of every input variable, with each variable appearing exactly once (either complemented or not). A maxterm is the dual: an OR of every input variable, again with each appearing once.

For $n$ variables there are $2^n$ minterms, one for each row of the truth table. Same for maxterms. Each minterm $m_i$ is true for exactly one input combination (row $i$); each maxterm $M_i$ is false for exactly one input combination (row $i$).

The two are complements of each other: $\overline{m_i}=M_i$. They’re the atoms from which any Boolean Function can be built.

Minterms

A minterm is a product term whose value is $1$ for exactly one input row. To build the minterm for a given row, take each variable: if its row value is $1$, write the variable; if its row value is $0$, write its complement.

$X$	$Y$	Product term	Symbol
0	0	$\overline{X}\overline{Y}$	$m_0$
0	1	$\overline{X}Y$	$m_1$
1	0	$X\overline{Y}$	$m_2$
1	1	$XY$	$m_3$

Three-variable example: $m_0=\overline{x_1}\overline{x_2}\overline{x_3}$ and $m_6=x_1{x}_2\overline{x_3}$.

The index $i$ in $m_i$ is the decimal value of the input row in binary. Row $6$ in three variables is $110$: variables $x_1$ and $x_2$ are $1$, $x_3$ is $0$, giving $m_6=x_1{x}_2\overline{x_3}$.

If any one minterm is true, all the others are false. They partition the input space into $2^n$ disjoint cases.

Maxterms

A maxterm is a sum term whose value is $0$ for exactly one input row. The convention is opposite to minterms: if the row value is $1$, write the complement; if it’s $0$, write the variable as-is. (That’s the only way to get the sum to be zero on that row.)

$X$	$Y$	Sum term	Symbol
0	0	$X+Y$	$M_0$
0	1	$X+\overline{Y}$	$M_1$
1	0	$\overline{X}+Y$	$M_2$
1	1	$\overline{X}+\overline{Y}$	$M_3$

Three-variable example: $M_4=\overline{x_1}+x_2+x_3$.

If any one maxterm is false, all the others are true.

Why this matters

Minterms feed into the sum-of-products canonical form: write the OR of every minterm where $f=1$. Maxterms feed into the product-of-sums form: write the AND of every maxterm where $f=0$. Either form gives a circuit you could build directly — usually wasteful, but it’s the starting point for minimization with Karnaugh Map or Boolean Algebra identities.

Worked example. Suppose $f(x_1,x_2)=m_0+m_1+m_3$, that is, $f=1$ for rows $0,1,3$ and $f=0$ for row $2$. Then $\overline{f}=m_2=x_1\overline{x_2}$, and applying De Morgan’s Laws:

$f=\overline{x_1\overline{x_2}}=\overline{x_1}+x_2=M_2$

So the same function can be expressed as $\sum m(0,1,3)$ in minterm form or $\prod M(2)$ in maxterm form. Same behavior, just listing the rows where $f$ is true vs. listing the rows where $f$ is false.