5

In order to evaluate the truth value of $∃x \ ∀y \ ∀z \ P(x,y,z)$, it is useful to "read" it : "there is a positive real $x$ such that, for every (positive reals) $y$ and $z$ it is true that $xyz=1$. The reasoning is : assume that $x > 0$ exists such that .... From $xyz=1$ we get $yz= \dfrac 1 x$ (we can do it because we have $x > 0$) for every $y,...


4

Ok, several ways to do this: Probably the most straightforward way is to say that there are three distinct orange apples and no more: $\exists x \exists y \exists z (A(x) \land O(x) \land A(y) \land O(y) \land A(z) \land O(z) \land x \neq y \land x \neq z \land y \neq z\land \neg \exists w (A(w) \land O(w) \land w \neq x \land w \neq y \land w \neq z))$ ...


3

would the first proposition indicate that there is an x that is both Blue(x) and a Circle"(x) or could the x of Blue(x) be distinct of the x that satisfies Circle(x)?" The first proposition says that there is an $x$ such that both $Blue(x)$ and $Circle (x)$ holds simultaneously. The second proposition says that there is an $x$ such that $Bed(x)$ holds AND ...


3

I think of interpretation as a two-stage process. First, translate the symbols into "mathematical language" without referencing the quantified terms, and then coax it into natural language. For instance, $$\forall x\ E(T,x)$$ is "for everything, Tom eats it", which I revise to "Tom eats everything." But if it's something where $x$ is referenced in both ...


3

The first statement says: for each $j$ in $\{1,2,3\}$, you can find a real number $a$ (which number may depend on which $j$) with the property that the $j$th function maps $a$ to $1$. Presumably, you have three functions called $f_1$, $f_2$, and $f_3$, and the first statement tells you that each and every one of these three functions "hits" the value $1$ ...


2

The first one says that for every $j$ in the set $\{1,2,3\}$, there exists an $a$ in the set of real numbers dependent on $j$ such that $f_j(a)=1$. The second, that there exists an $a$ in the set of real numbers such that for every $j$ in the $\{1,2,3\}$, no matter which one, we have $f_j(a)=1$. Can you take it from here? Hint:


2

I assume you meant $\forall x(P(x) \land S(x))$ (with $S(x)$ inside the scope of the quantifier $\forall x$), not $\forall x(P(x)) \land S(x)$. If you say $\forall x(P(x) \land S(x))$, this means "For all entities in the domain, $P$ holds and $S$ holds". This statement is false in the domain $\{1,2,3\}$, since neither $P$ nor $S$ hold of $3$. If you want ...


2

Statement 1 is true and statement 2 is false, because the order of quantifiers matter. As the existence quantifier in statement 1 comes after the "for all" quantifers, it claims existence after assignments have been made to the variables named in those. Contrary to that the existence quantifer in statement 2 comes first, and therefore claims existence of ...


1

1) There are at least two people who everyone knows. Domain = {People} My take in this.... for the 1) part is it valid to do something like this ∃𝒙∃𝒚∀z𝑷(𝒙, 𝒚,z). Where in my words i could be totally wrong.. there exist a pair (x,y) who everyone (z) knows. You must say: "There are some $x$ and some $y$ who are not the same people and every $z$ ...


1

One could use this tree proof generator to determine if these formulas are true or false. If a tree is generated then the formula is true. If a countermodel is generated then it is false. These are simple enough that they will reach a conclusion quickly. Here is the first one with a countermodel showing that it is false: https://www.umsu.de/trees/#%E2%88%...


1

Your best bet is to come up with interpretations of $P$ that will let you think about whether the implications are true. For instance, in the first problem, you could let $P(x,y)$ be the statement $|x|+|y|=0$ where $x$ and $y$ are real numbers. Then it is true that $\exists x\exists y P(x,y)$. Is it also true that $\forall x \exists y P(x,y)$? Then try ...


1

So you have three functions, named by the indices $\{1,2,3\}$, which accept real values as arguments (aka. their domain is points on the real number line). $(\forall j \in \{1,2,3\}) ( \exists a \in \Bbb R)~~f_j(a)=1$ Each index has at least one argument which makes its function equal one. These arguments need not be the same. Each function will ...


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