# Have I expressed this proposition correctly?

Up until I started writing this question, I have been attempting to teach myself these discrete math concepts, but now I want clarification on a question.

From the book:

Suppose that the domain of $$Q$$(x, y, z) consists of triples x, y, z, where x = 0, 1, or 2, y = 0 or 1, and z = 0 or 1. Write out these propositions using disjunctions and conjunctions.

c) $$\exists z\neg Q(0, 0, z)$$

d) $$\exists x\neg Q(x, 0, 1)$$

c) $$\neg [Q(0, 0, 0) \lor Q(0, 0, 1)]$$

d) $$\neg [Q(0, 0, 1) \lor Q(1, 0, 1) \lor Q(2, 0, 1)]$$

I probably did not need to show both answers, as they perhaps make the same mistake.

What I did was rearrange the statement $$\exists z\neg Q(0, 0, z)$$ into $$\neg\forall xQ(0,0,z)$$ and since the "for all" statement I used the conjunction between each predicate, and then since everything in the parentheses is negated I switched them all to disjunctions (De Morgan's laws).

I switched to the answers in the back:

c) $$\neg Q(0, 0, 0) \lor \neg Q(0, 0, 1)$$

Is this not equivalent to:

$$\neg [Q(0,0,0) \land Q(0,0,1)]$$

Their answer to (d) is very much the same.

Take the first. Your translation says that $Q(0,0,z)$ fails at both $z=0$ and $z=1$. But the existential statement you are translating says that $Q(0,0,z)$ fails somewhere.
You ask whether $\lnot Q(0, 0, 0) \lor \lnot Q(0, 0, 1)$ is equivalent to $\lnot [Q(0,0,0) \land Q(0,0,1)]$. Indeed they are equivalent. However, you wrote earlier that your answer to (c) was $\lnot [Q(0,0,0) \lor Q(0,0,1)]$.