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

My answers to these:

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.

So is my answer wrong?

Also, I am not sure which tags to use for this question.. this subject is very new to me.

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The answers you suggest are not, each time for basically the same reason.

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 $\lnot [Q(0,0,0) \lor Q(0,0,1)]$.

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  • $\begingroup$ Okay, I think my miscalulation occurred when I said "not for all" and then I magically threw in another negation and said "not not for all," thanks for the feedback. $\endgroup$
    – Leonardo
    Dec 13, 2012 at 4:52

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