# Simplifying logic circuit [closed]

I've been trying to simplify a combinational circuit and wanted to know whether there was any simpler way of solving such questions. Do we have to find all the outputs and such or we can use a simpler method ?

• Welcome to EE.SE. Please include the image in your question rather than as an external link. Commented Feb 16, 2019 at 12:37
• Here are some search terms which may help you: "de morgan's law" , "truth table", "karnaugh map" Commented Feb 16, 2019 at 14:24
• Why not start by making a table with columns for A to G? Just list out a row for every permutation of A to D, and then by hand work out for F and G. Perhaps you might see some obvious simplifications by just looking over the table. It is a way to start. Also, you could then show that work here. It might encourage others to add some thoughts for you.
– jonk
Commented Feb 16, 2019 at 19:33
• No simpler method. But there are tricks you pick up with experience. I'd use deMorgan's symbols for $T_4$, $T_5$ and $T_6$. But at this point that is meaningless to you. Commented Feb 17, 2019 at 1:47

Using deMorgan's symbols for $$\T_4\$$, $$\T_5\$$ & $$\T_6\$$. A NAND is a negative OR.

$$F = \bar A + \overline {\overline {ABC}} + \overline {\overline {BC}}$$

$$F = \bar A + ABC + BC$$

$$F = \bar A + BC$$

$$G = \overline {(\bar A + BC)(\overline {\overline {ABC}} + \bar B + \overline {\overline {BCD}})(\overline {\overline {BC}} + \overline {\overline {BCD}} + \bar D)}$$ $$G = \overline {(\bar A + BC)(ABC + \bar B + BCD)(BC + BCD + \bar D)}$$ $$G = \overline {(\bar A + BC)(ABC + \bar B + BCD)(BC + \bar D)}$$ $$G = \overline {(\bar A D + BC)(ABC + \bar B + BCD)}$$

G can still be simplified. Otherwise check the math.

But deMorgan's can simplify the process if applied correctly.