Negation Of A Compound Statement

Question

How can I write the negation of the following compound statements with "QUANTIFIERS"

Actually it is easy to do but I feel like I'm missing some logic and cant understand fully.

1) Every student that solves all given questions will get a passing grade in this course and will love mathematics forever.

2) There are some gnats that when seeing a gnu turn red and chase all bross that at least five meters away from their gros.

3) In every month there is on day during which it is rains Ron plays the flute and if it does not rain then Ron plays the harp.

I just wrote the 3 questions but the only 1 of them is enough to be solve. These statements needs to be negate with quantifiers but I dont even know how to state such a statement with quantifiers.

I'll be waiting for your answers & helps, thank you.

Answer 1

• Identify the thing(s) that's being quantified.
• Break a compound statement into its constituent statements.
• Use the general rules for negating quantified statements: \begin{align} \lnot \forall x\, Px &\quad\Leftrightarrow\quad \exists x\,\lnot Px\\ \lnot \exists x\, Px &\quad\Leftrightarrow\quad \forall x\,\lnot Px \end{align}

Use your first statement as an example. Let

• $Sxy=$ $x$ solves a given question $y$.
• $Px=$ $x$ gets a passing grade in this course.
• $Lxt=$ $x$ loves mathematics at time $t$.

Then the statement is symbolized as \begin{equation} \forall x\,(\forall y\, Sxy \Rightarrow (Px\wedge \forall t\,Lxt)) \end{equation}

Negating this, we get \begin{align} \lnot(\forall x\,(\forall y\, Sxy \Rightarrow (Px \wedge \forall t\,Lxt))) &=\exists x\,\lnot(\forall y\, Sxy \Rightarrow (Px \wedge \forall t\,Lxt)))\\ &=\exists x\,(\forall y\, Sxy\wedge (\lnot Px \vee \exists t\,\lnot Lxt)) \end{align}

Translated into English, the negated statement is

There is at least one student who solves all given questions but either he does not get a passing grade in this course or there is a time at which he does not love mathematics.

Answer 2

Take the first statement. Let $S$ be all students, $Q(s)$ be the predicate "student $s$ solves all questions", $P(s)$ be the predicate "student $s$ passes the course", and $L(s)$ the predicate "student $s$ will love mathematics forever". Then we can translate the original sentence this way: $$(\forall \, s\in S)(Q(s)\to(P(s)\land L(s))).$$ Negating it looks like this: \begin{align*} \neg(\forall \, s\in S)(Q(s)\to(P(s)\land L(s)))&= (\exists \, s\in S)\neg(Q(s)\to(P(s)\land L(s)))\\ &=(\exists \, s\in S)(Q(s)\land\neg(P(s)\land L(s))) \\ &=(\exists \, s\in S)(Q(s)\land(\neg P(s)\lor \neg L(s))). \end{align*} This scheme isn't the only way to phrase the statement in first-order logic, but it's one way.

Answer 3

What do you think about this one ? Let P(x) denote all the gnats(Universal set) and "x" denotes a single gnat. Then, S(x) holds for "Seeing a gnu, turn red" Finnaly C(x) "and chase all bros that at least five meters away from their gros " Well probably C(x) is wrong and I should divide it into "2" parts, such as : chasing statement is the 1st part and the "at least five meters" is second part because of "AT LEAST" ?

$$ \exists x \in P(x):Q(x) \implies C(x)$$

$$ \neg(\exists x \in P(x):Q(x) \implies C(x))$$

hence, $$\forall x \in P(x) : Q(x) \land \neg(C(x)) $$