I have this boolean expression: $$x'y'w' + yz + xzw'.$$

It should lend itself to being simplified to $$x'y'w' + yz + zw'$$ (I also checked the result with Mathematica) algebraically (i.e. axioms of Boolean Algebras).

I've tried every possible way I could think of but I cannot simplify it, so I ask if any of you can do this.

  • $\begingroup$ THis formula can't be satisfied (assuming $x^{'} \equiv \lnot x$ and $ab \equiv a \land b$ also assuming $+\equiv \lor$. To see this, not that you have $\lnot x \land \lnot w$ in the first term, and $\lnot x \land w$ in the third term. Atleast one of these has to be false. $\endgroup$ – mm8511 Apr 21 '18 at 19:16
  • $\begingroup$ @mm8511: What about $x = F$ and $y = z = T$? That seems to satisfy the formula. Perhaps you were misreading "or" as "and"? $\endgroup$ – John Hughes Apr 21 '18 at 19:17
  • $\begingroup$ Those terms, dear @mm8511 don't both have to be true. The statement is true if one or more of the disjuncts is true. $\endgroup$ – Namaste Apr 21 '18 at 19:35

$$x'y'w'+yz+xzw'\overset{Adjacency \times 3}=$$


$$x'y'zw'+x'y'z'w'+yzw+yzw'+xy'zw'\overset{Idempotence \times 2}=$$


$$x'y'zw'+x'y'z'w'+yzw+yzw'+yzw'+y'zw'\overset{Adjacency \times 3}=$$


  • $\begingroup$ Nice job, @Bram28. Just checking back to see if there's been any progress/successful answer to this question. $\endgroup$ – Namaste Apr 21 '18 at 21:13

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.