# Simplify $(¬x∨y∨z)∧(x∨¬(y∧z))$

I've been asked to construct a proposition based on its truth table. The truth table is:

x y z F(x,y,z)
1 1 1 1
1 1 0 1
1 0 1 1
1 0 0 0
0 1 1 0
0 1 0 1
0 0 1 1
0 0 0 1


I've gone this far with constructing the proposition:
$(x∧¬y∧¬z)∨(¬x∧y∧z) ≡ F$
$¬((x∧¬y∧¬z)∨(¬x∧y∧z)) ≡ T$
$≡(¬(x∧¬y∧¬z))∧(¬(¬x∧y∧z))$
$≡(¬x∨y∨z)∧(x∨¬(y∧z))$

But now I don't know how to proceed. The previous question was about the XOR connective (⊕) so I think the final proposition will also have the XOR connective. I've noticed from the table that $F(x,y,z)$ is false precisely when only one of either $x$ or $(y,z)$ is true, which tells me that the proposition is probably "$x$ not XOR $(y,z)$" but I've tested the proposition $¬(x⊕(y∨z))$ and it does not produce that truth table. Please help!

## 2 Answers

That's as simplified as you can get it, though personally I prefer statements in CNF or DNF and find those 'simpler', and so I would stick with:

$$(\neg x\lor y \lor z)\land (x\lor \neg y \lor \neg z)$$

• Yeah, thanks! I guess I was too caught up on trying to include the XOR connective in the expression. It took me a good night's sleep to realize I already had my final answer. Commented Nov 15, 2017 at 1:07

Hint. You can use your truth table and observe that $F(x,y,z) = 0$ if and only if $y = z = \neg x$.

• I'm not sure why I'm so baffled by this question, reading your answer I thought for sure it was going to be $¬x⊕y⊕z$, but it isn't. Thank you for your answer, though! Commented Nov 14, 2017 at 14:18