so I have spent nearly 5 hours trying to simplify this particular expression but I keep going round and round in circles. I have my hard copy notes to show you where I scribbled for hours and hours to no end. so please can someone please show me how can this be done?


I tried expansion, common, DeMorgan but I can't find my way out.


1 Answer 1


We can do it by brute force, expanding all 108 product terms and combining like terms carefully. Many of the terms can be eliminated immediately.

In hindsight or by luck or by sharp observation that D and E appear most frequently, we can proceed as follows.


Let $D=1$ and $E = 1$. The formula becomes 11111 = 1
Let $D=1$ and $E = 0$. The formula becomes $(\overline{A}+\overline{B})1 1 1 A = \overline{B}A$
Let $D=0$ and $E =1$. The formula becomes $1(\overline{A}+\overline{C}) C \overline{B} 1 = \overline{A}C\overline{B}$
Let $D=0$ and $E = 0$. The formula becomes $(\overline{A}+\overline{B})(\overline{A}+\overline{C})1 \overline{B} A = \overline{C}\overline{B}A$

So the original formula is $$DE + (D\overline{E}A + \overline{D}(E\overline{A}C + \overline{E}\overline{C}A))\overline{B}$$

(Computationally, it is debatable which formula is simpler.)

  • $\begingroup$ thanks for your reply. can you please look into this image i uploaded and let me know what i'm doing wrong? I have spent nearly 24 hours now. and I need to simplify it using common rules. please forgive the bad hand writing and cutting and im sorry I don't know how to type in LATEX. photos.app.goo.gl/L2Epwh2U1S67qBL68 $\endgroup$
    – ali farhad
    Nov 20, 2018 at 19:32
  • $\begingroup$ That image is not easier to read and to make sense. If you have understood my answer, you should be able to modify it so that it just uses "common rules". Hint, multiply the original formula with (de+de'+d'e+d'e'), which is 1 and expand the formula as shown in the answer. $\endgroup$
    – John L.
    Nov 20, 2018 at 23:26
  • $\begingroup$ sorry couldn't make it up.. $\endgroup$
    – ali farhad
    Nov 21, 2018 at 7:13
  • $\begingroup$ I will use ' to mean negation. (a'+b'+e)(a'+c'+d)(c+d+e')(b'+d)(a+e)de=(a'+b'+e)(a'+c'+d)(c+d+e')(b'+d)(a+e)ddddee=(a'+b'+e)e(a'+c'+d)d(c+d+e')d(b'+d)d(a+e)ede=((a'+b'+e)e)((a'+c'+d)d)((c+d+e')d)((b'+d)d)((a+e)e)de=edddede=de. See if you can figure out the rest. $\endgroup$
    – John L.
    Nov 21, 2018 at 11:30

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