Alright, have another one which the solutions manual is vague on.

I have to reduce \$f =(x1+x3+x4)(x1+x2'+x3)(x1+x2'+x3'+x4)\$. Right in the first step, though, the solution does something I don't understand.

They go to \$(x1+x3+x4)(x1+x2'+x3)(x1+x2'+x3+x4)(x1+x2'+x3'+x4)\$. Where did that 4th term come from? I thought we could add a sum if it was the same as one of the current terms? How are they allowed to change the x3?


1 Answer 1


I assume your question is about boolean algrbra and you are using \$'\$ to represent a not operation \$+\$ to indicate a "inclusive or" operation and no symbol at all to denote a "and" operation.

I think they are replacing




This will not change the result because whenever \$(x1+x2'+x3)\$ is 1

\$(x1+x2'+x3+x4)\$ will also be 1.

  • \$\begingroup\$ Mathematically, I can see how that works out, but how would I have known to do that? I've done a few problems and understand the basic rules but the solutions seem to include these extra terms that I wouldn't have guessed. Is synthesis something that comes from experience or is there a more formal path for deriving these solutions? \$\endgroup\$
    – PoGaMi
    Oct 11, 2015 at 23:58
  • \$\begingroup\$ There are various formal techniques. For example Karnaugh maps, en.wikipedia.org/wiki/Karnaugh_map and en.wikipedia.org/wiki/Quine%E2%80%93McCluskey_algorithm but I expect that they were just expecting you to spot that (x1+x2′+x3+x4) would be useful (since in the next step it can be used to simplify the last term) and would not change the result because of it's interactions with the third term. \$\endgroup\$ Oct 12, 2015 at 0:16
  • \$\begingroup\$ Oops, the last sentance of that comment should have said "interactions with the middle term" \$\endgroup\$ Oct 12, 2015 at 3:06

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