Let's say we have a truth table with two inputs (A, B) and one output (X).

0 0 0
0 1 1
1 0 1
1 1 0

I know that I can write a boolean function by (a) choosing rows where X is 1, (b) writing the minterms for these lines and (c) write cannonical sum of products (CSOP) like this: $$ F(A,B) = (\overline{A}\cdot B) + (A \cdot \overline{B}) $$

But how do we use the same principle if there is only one input (A) and one output (X) i.e.:

0 1
1 0

Does minterm even exist in this example?

Or maybe even this (X is constantly 0):

0 0
1 0

Note: I want the correct mathematical derivation of these simple examples.


2 Answers 2


I don't know what you mean by "correct mathematical derivation" but by inspection for your second case: \$X = \overline{A}\$

and for your third case: \$X = 0\$

  • \$\begingroup\$ Yeah I know this... But how do you handle the minterms in these scenarios? I am questioning the minterms method and I am questioning it is not quite universal (does not work for 1 input). \$\endgroup\$
    – 71GA
    Commented May 28, 2021 at 19:40
  • 1
    \$\begingroup\$ Maybe elaborate more on your confusion? In my eyes, you are using the same method to get the correct answer. "Choose rows where X is 1". Okay, so select the first row and write the equation: Not A. There is no B column so B can't possibly be part of the equation. There are no more rows where X is 1, so we're done: it's already minimized. For constant zero: "Choose rows where X is 1". There aren't any, so X is always zero. All done. \$\endgroup\$ Commented May 28, 2021 at 21:12
  • \$\begingroup\$ Ok thanks. I was curious about the mind process that you clearly showed me. \$\endgroup\$
    – 71GA
    Commented May 28, 2021 at 22:35

In fact the true answer is:

We first write the short minterm form of a boolean function for "ones" (where X = 1) of the truth table:

$$ X(A) = \sum m(0) $$

In this case we only have one minterm i.e

$$ m_0 = \overline{A} $$

So we can simply write:

$$ X(A) = m_0\\ \boxed{X(A) = \overline{A}} $$

For the other example we can simmilarly write the short minterm form of the boolean function for "zeroes" (where X = 0):

$$ \overline{X}(A) = \sum m(0,1) $$

In this case we have two minterms i.e

$$ m_0 = \overline{A}\\ m_1 = A $$

So we can simply write:

$$ \overline{X}(A) = m_0 + m_1\\ \overline{X}(A) = \overline{A} + A\\ \overline{X}(A) = 1\\ \boxed{X(A) = 0} $$

This is the correct derivation.

Principles described here are from the book:

"Fundamentals of digital electronics and computer design with VHDL (Sandige)", chapter 1.3.


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