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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.

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

$$x'y'zw'+x'y'z'w'+yzw+yzw'+xyzw'+xy'zw'\overset{Absorption}=$$

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

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

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

$$x'y'w'+yz+zw'$$

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  • $\begingroup$ 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

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