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I have the following:

Considering three propositions $S,R,Q$, write the truth table of $(R \land S) \lor (Q \Rightarrow R)$.

And also check if the following proposition:
$$\forall x \in \mathbb{R}, \exists y \in \mathbb{Z} \mbox{ such that } \forall b \in \mathbb{N} \quad 2x - y - 2b = 0$$ is true or false and then write its negation.

What I have done is the following:
the truth table is $$\begin{array}{c|c} S & Q & R & R \land S & Q \Rightarrow R \\ \hline T & T & T & T & T \\ \hline F & F & F & F & T \\ \hline T & F & F & F & T \\ \hline F & T & F & F & F \\ \hline F & F & T & F & T \\ \hline F & T & T & F & T \\ \hline T & F & T & T & T \\ \hline T & T & F & F & F \\ \end{array}$$

I don't think that the table above it is related to what follows, but it has been requested in the same exercise.
Then I considered the proposition, and I have done the following reasoning:
$2x - y - 2b = 0 \iff 2x = y + 2b$

now for $x = 1, y = -3, b = 5$ it is true that
$2 = -3 + 5$

I could conclude that it is true, but, how can I prove for every element?

Then I wrote its negation considering:
$\forall x \in \mathbb{R}, \quad \nexists y \in \mathbb{Z} \mbox{ such that } \forall b \in \mathbb{N} \\ 2x - y - 2b \ne 0 \iff \\ 2x \ne y + 2b$

Please, can you give me any suggestions? Many thanks!

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    $2(x-b)$ is not an integer unless $x$ is. So proposition is false – 2017-02-19
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    It is missing the last column in the truth table, the one corresponding to the dijunction. – 2017-02-19

1 Answers 1

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The negation should be $\exists x \in \mathbb{R} \ \forall y \in \mathbb{Z} \ \exists b \in \mathbb{N} \quad 2x-y-2b \not= 0$
The way you wrote it has some problem. First, you have to negate the whole proposition, so you can't start by negating the existential quantifier because a universal quantifier is preceeding it, so the negation should be at the very beggining of the proposition.
Remember these equivalences: $$\neg \forall x \ \alpha \equiv \exists x \ \neg \alpha$$ $$\neg \exists x \ \alpha \equiv \forall x \ \neg \alpha$$

Second, be careful with that iff symbol, what you wrote means that $$\forall x \in \mathbb{R}, \quad \nexists y \in \mathbb{Z} \mbox{ such that } \forall b \in \mathbb{N} \ 2x - y - 2b \ne 0 $$ is equivalent to $$2x \ne y + 2b$$

As a general rule, when you have a proposition whis consists of a bunch of alternating quantifiers and then some subproposition $\alpha$, the negation is obtained by changing all the quantifiers (replace universal by existential and viceversa) and then $\neg \alpha$